{"loaderData":{"0-0-56":{"bookData":{"_id":"688c1a310cdd58e3928ca586","category_id":"66ceced8c3d3f3f05eac01a2","author_id":["69384582e8e393d9963956ea"],"book_name":"Mathematical Methods in Physics","book_image":"/books/CoverImagepng-1754012209380.png","book_description":"

This collection of notes outlines a fast-paced, practical, and problem-oriented course on the mathematical methods essential for a physicist. True to Feynman's style, the focus is on intuitive understanding and application rather than rigorous proof. The course serves as a \"physicist's toolkit\" for solving a wide range of real-world problems.

The key topics covered include:

Solving Equations: The course begins with practical, numerical methods for solving complex equations, including trial-and-error (interpolation), fixed-point iteration, and Newton's method. This immediately sets a tone of hands-on problem-solving.
Series and Summation: Feynman covers the manipulation of power series, including methods for finding their sum by differentiation and integration. He also introduces clever approximation techniques, like replacing the \"tail\" of a slowly converging series with an integral (Integral Approximation).
Advanced Integration Techniques: A significant portion of the course is dedicated to advanced methods for solving definite integrals, a common task in physics. Key techniques include:
- Differentiation under the integral sign (the \"Feynman trick\").
- Using complex numbers and contour integration to solve real integrals, including the calculation of residues at poles and handling branch points.
- Numerical Integration methods like the Trapezoidal Rule and Simpson's Rule, which were essential in the pre-computer era.
Essential Functions and Concepts for Physics:
- The Dirac Delta Function: Introduced as a practical tool for representing point sources or impulses, with the classic Feynman justification: \"The only justification we give for this is that it gives the correct answer.\"
- Special Functions: The course is a tour of the \"greatest hits\" of special functions, explaining where they come from and how to use them:
  - Gamma and Beta Functions: Including the Stirling approximation for factorials.
  - Bessel Functions: Presented as solutions to problems with cylindrical symmetry (e.g., vibrating drum heads, wave guides).
  - Legendre Polynomials: Presented as solutions to problems with spherical symmetry (e.g., potential theory, cavity resonators).
  - Elliptic, Error, and Sine/Cosine Integrals: Introduced as tabulated functions that appear in the solutions to specific physical problems (like diffraction).
Transform Methods: The final part of the course focuses on the powerful techniques of Fourier and Laplace transforms for solving linear differential equations and analyzing the response of linear systems (like electronic circuits or amplifiers). This section connects the abstract mathematics directly to concepts like impulse response, frequency response, and convolution.

In essence, these notes are a masterclass in how to think about and use mathematics to solve physical problems, emphasizing intuition, a wide variety of techniques, and the practical needs of a working scientist.

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The simplest way to solve any equation - except linear or quadratic equations - is by trial and error.

A. Interpolation

To solve \\[ \\dfrac{1}{1+x^2} = 2x \\]

Let:

x	LHS	RHS	DIFF
0	1	0	1	} 2.5
1	.5	2	-1.5
.4	.862	.8	.062
.5	….	….	….

Explanation

This document explains how to solve the equation \$1 / (1 + x²) = 2x\$ using a numerical method of trial and error, often called interpolation. To find the solution, we create a table to compare the Left-Hand Side (LHS) and Right-Hand Side (RHS) of the equation, aiming to find a value of \$x\$ where their difference (DIFF) is zero. We first test x=0, which gives a difference of +1, and then x=1, which gives a difference of -1.5. Since the sign of the difference changes between these two points, we know the solution must lie between 0 and 1. Because the absolute difference for x=0 (which is 1) is less than the difference for x=1 (which is 1.5), we can guess that the solution is closer to x=0, prompting our third guess of x=0.4. This trial yields a very small positive difference of +0.062. With a positive difference at x=0.4 and a negative one at x=1, we conclude that the solution is between 0.4 and 1. However, since the difference for x=0.4 is very small, we guess the solution is much closer to 0.4, so our fourth guess is x=0.5 to further narrow the interval and pinpoint the exact answer.

B. Iteration

Solve the equation for \$x\$: \\[ x = \\dfrac{1}{2}\\left(\\dfrac{1}{1+x^2}\\right) \\]

trial \$x\$	result \$x\$
0	\$\\to .5\$
.5	\$\\to .4\$
.4	\$\\to .431\$
…	…

Explanation

This example demonstrates another powerful numerical technique, known as Fixed-Point Iteration, to solve an equation. Notice that, this is the same equation as in the previous example (1 / (1 + x²) = 2x), just rearranged to isolate x on one side: \\[ x = \\frac{1}{2}\\frac{1}{1+x^2} \\] The core idea of this method is to find a value for x (a “fixed point”) that remains unchanged when we plug it into the expression on the right. We do this by starting with an initial guess and then iteratively feeding the result back into the equation until the numbers stop changing significantly.

Here is a step-by-step breakdown of the process shown:

Initial Guess (Trial 1): We start with an initial guess of trial x = 0. We plug this into the right side of the equation:
- Result = 1/2 * (1 / (1 + 0²)) = 1/2 * (1) = 0.5 This result, 0.5, becomes our next trial value.
First Iteration (Trial 2): Now, we use this result as our new trial x = 0.5:
- Result = 1/2 * (1 / (1 + 0.5²)) = 1/2 * (1 / 1.25) = 0.5 * 0.8 = 0.4
Second Iteration (Trial 3): We repeat the process with trial x = 0.4:
- Result = 1/2 * (1 / (1 + 0.4²)) = 1/2 * (1 / 1.16) ≈ 0.5 * 0.862 = 0.431

The process continues this way. We would next use 0.431 as our trial x and calculate a new result. With each step, the “trial x” and “result x” get closer to each other. The solution is found when the value we put in is the same as the value that comes out (i.e., when the sequence converges).

If the equation is solved in the form \$x = \\sqrt{\\dfrac{1}{2x}-1}\$, however, the results diverge. If the results do diverge solving the equation in the inverse manner will make them converge.

C. Newton’s Method

Solve the equation in the form \$f(x)=0\$. Let \$x_1\$ be the first trial, then the second trial is given by the relation \\[ x_2 = x_1 - \\dfrac{f(x_1)}{f'(x_1)} \\] This procedure is equivalent to extrapolating back along the tangent. If the curve is full of wiggles things may diverge.

[Let’s go back to the previous example]

\\[ f(x) = \\dfrac{1}{1+x^2} - 2x = 0 \\] \\[ f'(x) = \\dfrac{-2x}{(1+x^2)^2} - 2 \\]

x	f(x)	f’(x)
0	1	-2
.5	.2	….
…	….	….

Explanation

The table shows the step-by-step process of applying the method.

First Guess: You start with a simple initial guess: x₁ = 0.
Plug in the Guess: At x=0, you calculate the values of the function and its derivative:
- f(0) = 1/(1+0²) - 2(0) = 1
- f’(0) = -2(0)/(1+0²)² - 2 = -2
Calculate the Next Guess (x₂): Now, you use the Newton’s Method formula: \\[x_2=x_1-\\frac{f(x_1)}{f'(x_1)}=0-\\frac{1}{-2}=0.5\\]
- The Next Step (Row 2): Your new, improved guess is x₂ = 0.5. The second row of the table would use this value to calculate f(0.5) and f’(0.5) in order to find an even better guess, x₃. This process continues until f(x) is extremely close to zero.

Problem: Find first positive root of \$e^{-x} = \\cos x\$ by all three methods to 2 decimals. Estimate higher roots.

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A. Power Series

Problem: Prove \\[ 1+x+x^2+x^3+\\cdots = \\dfrac{1}{1-x} \\]

Hint

Recall the sum of a geometric sequence (progression): \\[ a+aq+aq^2+\\cdots+aq^{n-1}=\\frac{a(1-q^n)}{1-q} \\]

Problem: Sum \\[ 1+a\\cos\\theta + a^2\\cos 2\\theta + a^3\\cos 3\\theta + \\cdots \\]

Hint

Notice that \$(e^{i\\theta})^n=\\cos n\\theta + i \\sin n\\theta\$.

B. Summation by Integration and Differentiation

Consider the following series \\[ 1(x) - \\dfrac{1}{2}(x^2) + \\dfrac{1}{3}(x^3) - \\dfrac{1}{4}(x^4) + \\cdots = S(x) \\]

Explanation

This line defines an infinite series as a function, which we’ll call S(x). This type of series, which involves powers of x, is known as a power series. The goal is to find a more common function (like sin(x) or ln(x)) that is equal to this infinite sum.

\\[ 1 - x + x^2 - x^3 + x^4 + \\dots = S'(x) \\]

Explanation

Here, you’ve taken the derivative of the series S(x) term by term.

The derivative of x is 1.
The derivative of -(1/2)x² is -x.
The derivative of (1/3)x³ is x².
Taking the derivative often simplifies a series into a more recognizable pattern. The result is labeled S’(x), the derivative of S(x).

\\[ S'(x) = \\dfrac{1}{1+x} \\]

Explanation

This step recognizes that the new series (1 - x + x² - …) is a well-known pattern called a geometric series. The formula for a geometric series is a / (1 - r), where a is the first term and r is the common ratio. In this case, the first term a=1 and the ratio r = -x. Plugging this into the formula gives 1 / (1 - (-x)), which simplifies to 1 / (1+x).

\\[ S(x) = \\ln(1+x)+C \\]

Explanation

To get back to the original function S(x) from its derivative S’(x), you need to do the opposite of differentiating: integrating. The integral of 1 / (1+x) is ln(1+x). Whenever you find an indefinite integral, you must add a constant of integration, written as + C, because the derivative of any constant is zero.

evaluate constant for \$x=0\$ \\[ S(0) = \\ln 1 + C \\] \\[ 0=0+C \\]

Explanation

On the right side, we know that ln(1) = 0.
On the left side, we plug x=0 into the very first line of our notes: S(0) = 1(0) - (1/2)(0)² + …. Every term becomes zero, so S(0) = 0.
This gives us the simple equation 0 = 0 + C, which means C must be 0.

\\[ S(1) = \\ln 2 = .69315 \\]

[From calculus, recall the following theorem: ]

Convergence: Ratio Test If the \$\\lim_{n\\to\\infty} x_{n+1}/x_n < 1\$ the series converges.

C. Abel Summability.

Consider the series \\[ 1-2+3-\\cdots \\] Convert it into a power series in x as follows: \\[ S(x) = 1-2x+3x^2-\\cdots \\] Sum this power series: \\[ \\int S(x)dx = x-x^2+x^3-\\dots = \\dfrac{x}{1+x} \\] \\[ S(x) = \\dfrac{d}{dx}\\left(\\dfrac{x}{1+x}\\right) = \\dfrac{1}{(1+x)^2} \\] The Abel sum of the original series is defined as the \$\\lim_{x\\to 1} S(x)\$ \\[ \\text{Abel sum} = \\lim_{x\\to 1} \\left[\\dfrac{1}{(1+x)^2}\\right] = \\dfrac{1}{4} \\] In general, oscillating series can be summed, but the order of the terms must not be changed.

Explanation

While the series \$1 - 2 + 3 - \\dots\$ is divergent in the usual sense, it is Abel summable to \$\\frac{1}{4}\$. This is a different notion of convergence based on taking a limit of a related power series as \$x \\to 1^-\$. For more details, see Abel summability.

D. Summation of Series by Integration and Differentiation.

\\[ S = 1+\\dfrac{1}{4}+\\dfrac{1}{9}+\\dfrac{1}{16}+\\cdots \\]

Explanation

This is the infinite series we want to understand or potentially find a sum for. It’s the sum of the reciprocals of perfect squares, which can be written as \$\\sum_{n=1}^{\\infty} \\frac{1}{n^2}\$. This specific series is known as the Basel Problem, and its sum is famously \$\\frac{\\pi^2}{6}\$.

\\[ S(x) = x+x^2/4+x^3/9+\\cdots \\]

Explanation

To use calculus, we convert the sum into a power series function, \$S(x)\$. Notice that if you substitute \$x=1\$ into this function, you get the original series \$S\$. So, the goal is to find a closed-form expression for \$S(x)\$ and then evaluate it at \$x=1\$. This particular power series can be written as \$\\sum_{n=1}^{\\infty} \\frac{x^n}{n^2}\$.

\\[ S'(x) = 1+x/2+x^2/3+\\cdots \\]

Explanation

Here, you’ve taken the derivative of \$S(x)\$ term by term.

\\[ S'(x)= -\\dfrac{\\ln(1-x)}{x} \\]

Explanation

This is the crucial step where you recognize the simplified series as a known function. The series \$1 + \\frac{x}{2} + \\frac{x^2}{3} + \\dots\$ is exactly the expansion of \$-\\frac{\\ln(1-x)}{x}\$.

(This comes from the Taylor series expansion of \$\\ln(1-x) = -x - \\frac{x^2}{2} - \\frac{x^3}{3} - \\dots\$, and then dividing by \$x\$. For the expansion of \$\\ln(1-x)\$ see below.)

\\[ S(x) = -\\int\\dfrac{\\ln(1-x)}{x}dx \\]

Explanation

To get back to the original function \$S(x)\$ from its derivative \$S'(x)\$, you perform the antidifferentiation (integration). So, \$S(x)\$ is the indefinite integral of the expression you found for \$S'(x)\$.

One should know and recognize the series for the following expressions: (1) \\[\\dfrac{1}{1-x} = 1+x+x^2+x^3+\\cdots\\] (2) \\[-\\ln(1-x) = x+\\dfrac{x^2}{2}+\\dfrac{x^3}{3}+\\cdots\\] (3) \\[\\tan^{-1}x = x-\\dfrac{x^3}{3}+\\dfrac{x^5}{5}-\\dfrac{x^7}{7}+\\cdots\\] (4) \\[e^x = 1+\\dfrac{x}{1!}+\\dfrac{x^2}{2!}+\\dfrac{x^3}{3!}+\\cdots\\] (5) \\[\\sin x = x-\\dfrac{x^3}{3!}+\\dfrac{x^5}{5!}-\\cdots\\] (6) \\[\\cos x = 1-\\dfrac{x^2}{2!}+\\dfrac{x^4}{4!}-\\cdots\\] (7) \\[\\dfrac{1}{(1+x)^2} = 1-2x+3x^2-\\cdots\\] (8) \\[\\dfrac{1}{(1+x)^{1/2}} = 1-\\dfrac{1}{2}x+\\dfrac{3}{8}x^2-\\dfrac{5}{16}x^3+\\cdots\\] A useful relation \$e^{i\\theta} = \\cos\\theta + i\\sin\\theta\$

Problem: Sum: \\[ \\dfrac{1}{2!} + \\dfrac{2}{3!} + \\dfrac{3}{4!} + \\cdots \\] \\[ \\dfrac{1}{1\\cdot 2} + \\dfrac{1}{2\\cdot 3} + \\dfrac{1}{3\\cdot 4} + \\dots \\]

E. Summation of Series by Adding Terms.

Consider the series \\[ 1+\\dfrac{1}{4}+\\dfrac{1}{9}+\\cdots \\] We can sum it as follows: 1.000+ .250 +.111 + .0625 + …

By taking enough terms we can obtain any desired accuracy. If the series is slow to converge it is sometimes possible to replace the higher terms by an integral:

Examination of the above diagram shows that \\[ \\dfrac{1}{3^2}+\\dfrac{1}{4^2}+\\dots \\approx \\int_{2.5}^\\infty \\dfrac{dx}{x^2} \\] \\[ \\approx \\dfrac{1}{2.5} = .4 \\] The sum of our series is then 1.000+ .250 + .400 = 1.650

Explanation

Here we are trying to find the sum of the series: \\[ S = 1 + \\frac{1}{4} + \\frac{1}{9} + \\frac{1}{16} + \\dots = \\sum_{n=1}^{\\infty} \\frac{1}{n^2} \\]

There are two methods to find this sum:

Direct Summation: The first method is to simply start adding the terms one by one (1.000 + 0.250 + 0.111 + ...). We note that for a “slow to converge” series, this is inefficient. You would have to add thousands of terms to get an accurate answer.
Integral Approximation: The second, more clever method is to add the first few terms directly and then approximate the rest of the sum (the “tail” of the series) with an integral. This works because the sum of the series can be visualized as the area of a set of rectangles, which can be closely approximated by the area under a smooth curve.

** Why the Integral Starts at 2.5 (Using the Diagram)**

The core of the technique is approximating the sum of the “higher terms” with an integral. In this example, the author decides to calculate the first two terms exactly and approximate the rest:

\\[ S = \\underbrace{1 + \\frac{1}{4}}_{\\text{Calculated Exactly}} + \\underbrace{\\left( \\frac{1}{9} + \\frac{1}{16} + \\frac{1}{25} + \\dots \\right)}_{\\text{To be Approximated}} \\]

This “tail” of the series can be written as: \\[ \\text{Tail} = \\sum_{n=3}^{\\infty} \\frac{1}{n^2} \\]

Now, let’s look at the diagram to understand the approximation for this tail.

The Rectangles Represent the Sum: The diagram shows the function y = 1/x². The rectangles represent the terms of the series.
- The rectangle at x=3 has a width of 1 and a height of 1/3² = 1/9. Its area is 1/9.
- The rectangle at x=4 has a width of 1 and a height of 1/4² = 1/16. Its area is 1/16.
- Therefore, the sum of the tail, 1/9 + 1/16 + ..., is equal to the total area of all the rectangles starting from x=3.
The Integral Represents the Area Under the Curve: The integral ∫(1/x²) dx calculates the area under the smooth curve y = 1/x².
Finding the Best Approximation: We want to find an integral whose area is very close to the area of the rectangles from x=3 onwards.
- If we started the integral at x=3 (i.e., ∫ from 3 to ∞), the curve would run under the top of the rectangles. This would give us an underestimate of the true sum.
- If we started the integral at x=2 (i.e., ∫ from 2 to ∞), the curve would run above the top of the rectangles. This would give us an overestimate.
So we choose x = 2.5 as the starting point. This is a very clever choice based on the midpoint rule of approximation.
By starting the integral at 2.5, you are assuming that the area of the rectangle at x=3 is best approximated by the area under the curve on the interval [2.5, 3.5].
As you can see in this enhanced diagram, the small bit of area the integral misses on the left side of the rectangle (the white region with red outline) is almost perfectly canceled out by the extra area the integral includes on the right side (the shaded region).
This logic extends to all the other terms. The area of the rectangle at x=4 is approximated by the integral from 3.5 to 4.5, and so on. Therefore, the entire sum of the tail is approximated by the integral starting from 2.5: \\[ \\sum_{n=3}^{\\infty} \\frac{1}{n^2} \\approx \\int_{2.5}^{\\infty} \\frac{1}{x^2} dx \\]

** The Final Calculation**

The notes complete the process:

Calculate the integral: \\[ \\int_{2.5}^{\\infty} \\frac{1}{x^2} dx = \\left[ -\\frac{1}{x} \\right]_{2.5}^{\\infty} = \\left( -\\frac{1}{\\infty} \\right) - \\left( -\\frac{1}{2.5} \\right) = 0 + \\frac{1}{2.5} = 0.4 \\]
Add the parts together: \\[ S \\approx 1 + 0.250 + 0.400 = 1.650 \\]

This is an excellent approximation. The true sum of this series (the Basel Problem) is known to be π²/6 ≈ 1.6449, so this simple method is remarkably accurate.

Since the error will be principally in the first terms some idea of its magnitude may be gotten by looking at the next best approximation: \\[ \\dfrac{1}{4^2}+\\dfrac{1}{5^2}+\\cdots \\approx \\int_{3.5}^\\infty \\dfrac{dx}{x^2} \\] \\[ \\approx .286 \\] The new sum is: 1.000+ .250 + .111 + .286 = 1.647 we see that we have made an error of only .003. We can guess from this that the third figure is probably about 6.

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A. Magnitude of Quantities

\n\n

The following quantities are arranged in the order of their rate of rise for large x \\[ 1 < \\ln x < x^\\epsilon < x < x^n < e^x \\] [Here \$\\epsilon\$ is a positive number less than 1.]

\n\n

To ascertain their behavior for small y make the substitution \$y=\\dfrac{1}{x}\$.

\n\n

The following approximations hold for small x. \\[ \\begin{aligned} \\sin x &\\sim x \\\\ \\cos x &\\sim 1 - \\dfrac{x^2}{2} \\\\ \\tan^{-1}x &\\sim x \\sim \\tan x \\\\ \\sinh x &\\sim x \\\\ e^x &\\sim 1+x \\\\ (1+x)^n &\\sim 1+nx \\\\ (1+x)^{1/2} &\\sim 1-\\dfrac{1}{2}x \\end{aligned} \\]

\n\n

B. De L’Hospital’s Rule

\n\n

Theorem 1

If as \$x\\to a\$, \$f(x)\$ and \$g(x) \\to 0\$ \\[ \\lim_{x\\to a} \\dfrac{f(x)}{g(x)} = \\lim_{x\\to a} \\dfrac{f'(x)}{g'(x)} \\]

\n\n

\nExample 1.\n

Example: \$\\lim_{x\\to 0} \\dfrac{\\tan x}{[(1+x)^3-1]}\$ \\[ = \\dfrac{\\sec^2 x}{3(1+x)^2} = \\dfrac{1}{3} \\]

\n\n

\nExercise 1.\n

Problem: Find \$\\lim_{x\\to 0}\$

\$\\dfrac{\\sin x}{(e^{3x}-1)}\$
\$\\dfrac{\\ln(\\frac{1}{2}+1/2\\sqrt{1+x^2})}{(e^{x^2}-\\cos x^2)}\$

\n\n

\nExercise 2.\n

Problem: Give an approximate expression for \$(\\sin kr/r^2)-(k \\cos kr/r)\$ for small r.

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A. Methods

[Some common integration methods:]

Substitution
By Parts
Differentiation of a Parameter
Series Expansion
Contour Integration
Numerical Methods
Special Tricks

Methods 3 to 7 are applicable particularly to definite integrals.

B. Complex Variable in Substitution

Example: Find: \\[ \\int_0^\\infty e^{-ax}\\cos bx \\,dx \\] [Let’s remember that] \\[ \\cos bx = (e^{ibx}+e^{-ibx})/2 \\] Therefore, the given integral is equal to \\[ \\begin{aligned} &= \\dfrac{1}{2} \\int_0^\\infty e^{-(a-ib)x} dx + \\dfrac{1}{2} \\int \\dots \\\\ &= \\dfrac{1}{2}\\dfrac{1}{a-ib} + \\dfrac{1}{2}\\dfrac{1}{a+ib} \\\\ &= \\dfrac{a}{a^2+b^2} \\end{aligned} \\]

C. Differentiation of a Parameter

Example: \\[ \\int_0^\\infty xe^{-ax}\\cos bx\\, dx \\] [From the previous example:] \\[ S(a) = \\int_0^\\infty e^{-ax}\\cos bx\\, dx = \\dfrac{a}{a^2+b^2} \\] Differentiate with respect to \$a\$: \\[ S'(a) = -\\int_0^\\infty xe^{-ax}\\cos bx\\, dx = \\dfrac{(b^2-a^2)}{(a^2+b^2)^2} \\]

The general rule for differentiation with respect to a parameter is: \\[ \\dfrac{d}{d\\alpha}\\int_{x_1(\\alpha)}^{x_2(\\alpha)} f(x,\\alpha)dx = \\int_{x_1}^{x_2} \\dfrac{\\partial}{\\partial \\alpha}f(x,\\alpha)dx + \\dfrac{d x_2(\\alpha)}{d\\alpha}f(x_2,\\alpha) - \\dfrac{d x_1(\\alpha)}{d\\alpha}f(x_1,\\alpha) \\]

\\[ \\Delta\\int_{x_1}^{x_2} f(x,\\alpha)dx = \\int_{x_1}^{x_2}\\Delta f(x,\\alpha)dx + \\Delta x_2(\\alpha)f(x_2,\\alpha) - \\Delta x_1(\\alpha)f(x_1,\\alpha) \\]

Explanation

The goal is to understand how the total area under a curve changes if:

The shape of the curve itself changes.
The start and end points of the integration move.

The parameter α (alpha) controls both the shape of the function and the positions of the integration limits, x₁ and x₂. The notes show what happens when you make a small change in this parameter, from α to α + Δα.

The diagram provides a visual breakdown of the total change in the area under the curve. Let’s look at the three distinct ways the area changes:

The “Body” of the Area Changes (Shape Change):
- The lower curve is f(x, α).
- The upper curve is f(x, α + Δα).
- The change in the function’s height is Δf = f(x, α + Δα) - f(x, α).
- The area between these two curves is the main contribution to the change. This is represented by the first term in the final equation: ∫ Δf(x,α) dx.
A Sliver of Area is ADDED on the Right (Upper Limit Moves):
- The upper limit of integration moves from x₂(α) to x₂(α + Δα).
- This adds a thin rectangular sliver of area on the far right.
- The width of this sliver is Δx₂.
- Its height is approximately f(x₂, α).
- This corresponds to the second term in the final equation: + Δx₂(α) f(x₂, α).
A Sliver of Area is SUBTRACTED on the Left (Lower Limit Moves):
- The lower limit of integration moves from x₁(α) to x₁(α + Δα).
- This removes a thin rectangular sliver of area on the far left.
- The width of this sliver is Δx₁.
- Its height is approximately f(x₁, α).
- This corresponds to the last term in the final equation: - Δx₁(α) f(x₁, α).

The total change in the integral’s value is the sum of these three pieces.

Now dividing both sides by Δα, we have
$
\\frac{\\Delta\\int_{x_1}^{x_2} f(x,\\alpha)dx}{\\Delta \\alpha} = \\int_{x_1}^{x_2}
\\frac{\\Delta f(x,\\alpha)}{\\Delta \\alpha} dx + \\frac{\\Delta x_2(\\alpha)}{\\Delta \\alpha}f(x_2,\\alpha) - \\frac{\\Delta x_1(\\alpha)}{\\Delta \\alpha}f(x_1,\\alpha)
$
Taking the limit as $\\Delta \\alpha\\to 0$, we obtain
$
\\dfrac{d}{d\\alpha}\\int_{x_1(\\alpha)}^{x_2(\\alpha)} f(x,\\alpha)dx = \\int_{x_1}^{x_2} \\dfrac{\\partial}{\\partial \\alpha}f(x,\\alpha)dx + \\dfrac{d x_2(\\alpha)}{d\\alpha}f(x_2,\\alpha) - \\dfrac{d x_1(\\alpha)}{d\\alpha}f(x_1,\\alpha).
$

D. Multiplication by a Factor

Example: [Suppose we want to evaluate the following integral] \\[ \\int_0^\\infty (\\sin x/x)dx \\] [To this end, let’s introduce a related integral that depends on a parameter \$\\alpha\$:] \\[ \\int_0^\\infty e^{-\\alpha x}\\dfrac{\\sin x}{x}dx = S(\\alpha) \\] \\[ \\begin{aligned} S'(\\alpha) &= -\\int_0^\\infty e^{-ax}\\sin x\\ dx = -\\frac{1}{1+\\alpha^2} \\\\ S(\\alpha) &= -\\tan^{-1}\\alpha+C \\\\ S(\\infty) &= 0 = -\\dfrac{\\pi}{2}+C \\\\ S(\\alpha) &= \\dfrac{\\pi}{2}-\\tan^{-1}\\alpha \\\\ S(0) &= \\dfrac{\\pi}{2} \\end{aligned} \\]

Explanation

Goal: Solve the famous Dirichlet integral, which does not have a simple antiderivative. > \\[ \\int_0^\\infty \\frac{\\sin x}{x} dx \\] Explanation: This is the problem we want to solve. We can’t find a simple function whose derivative is sin(x)/x, so standard integration methods won’t work easily.

Step 1: Introduce a Parameter > \\[ \\int_0^\\infty e^{-\\alpha x}\\dfrac{\\sin x}{x}dx = S(\\alpha) \\] Explanation: The trick begins here. We insert a “helper” term, e^(-αx), into the integral. This does two things: 1. It creates a new function that depends on the parameter α, which we call S(α). 2. It ensures the integral converges nicely for α > 0. Our goal is to find a formula for S(α) and then set α = 0, because e^(-0*x) = 1, which will give us the value of our original integral.

Step 2: Differentiate with Respect to the Parameter \\[ S'(\\alpha) = -\\int_0^\\infty e^{-\\alpha x}\\sin x dx = -\\frac{1}{1+\\alpha^2} \\] Explanation: This is the key move. We differentiate S(α) with respect to α. The derivative can be moved inside the integral sign. * \$\\displaystyle \\frac{d}{d\\alpha} \\left(e^{-\\alpha x} \\frac{\\sin x}{x}\\right)=-x e^{-\\alpha x} \\frac{\\sin x}{x}\$. * The x in the numerator cancels the difficult 1/x term in the denominator. This leaves us with a much simpler integral: S'(α) = -∫ e^(-αx) sin(x) dx. This is a standard integral whose result is known to be 1 / (1 + α²). So, S'(α) = -1 / (1 + α²).

Step 3: Integrate Back to Find S(α) \\[ S(\\alpha) = -\\tan^{-1}(\\alpha)+C \\] Explanation: To get from the derivative S'(α) back to the original function S(α), we integrate S'(α) with respect to α. The integral of -1 / (1 + α²) dα is -tan⁻¹(α). Because this is an indefinite integral, we must add an unknown constant of integration, C.

Step 4: Solve for the Constant C \\[ S(\\infty) = 0 = -\\dfrac{\\pi}{2}+C \\] Explanation: To find C, we need to evaluate S(α) at a point where we know its value. A convenient choice is α → ∞. * Looking at our definition S(α) = ∫ e^(-αx) (sin x / x) dx, as α becomes very large, the e^(-αx) term becomes a “killer” term, forcing the entire integrand to zero for any x > 0. Therefore, the integral itself must be zero, so S(∞) = 0. * Using our formula from the previous step, S(∞) = -tan⁻¹(∞) + C. Since tan⁻¹(∞) = π/2, this gives us the equation 0 = -π/2 + C, which means C = π/2.

Step 5: Write the Final Formula and Get the Answer \\[ S(\\alpha) = \\dfrac{\\pi}{2}-\\tan^{-1}\\alpha \\] Explanation: Now that we know C = π/2, we can write the complete formula for our function S(α).

\\[ S(0) = \\dfrac{\\pi}{2} \\] Explanation: This is the final step. We substitute α = 0 back into our formula to find the value of the original integral. * S(0) = π/2 - tan⁻¹(0) * Since tan⁻¹(0) = 0, the final answer is π/2.

Problem (a) Given \$\\displaystyle \\int_0^\\infty e^{-a^2x^2}dx = \\sqrt{\\pi}/2a\$ Prove: \\[ \\int_0^\\infty e^{-a^2x^2}\\cos bx\\, dx = \\dfrac{\\sqrt{\\pi}}{2a}e^{-b^2/4a^2} \\]

(b) Given \$\\displaystyle \\int_0^\\infty e^{-x^2}dx = \\sqrt{\\pi}/2\$ Find: \\[ \\int_0^\\infty x^4 e^{-x^2}dx \\]

(c) \\[\\displaystyle \\int_0^\\infty dy(e^{-ay}-e^{-by})/y \\]

(d) \\[\\displaystyle \\int_{-\\infty}^\\infty e^{-ax^2-b/x^2} dx \\]

(e) \\[\\displaystyle \\int_0^\\infty dx \\sin^2x/x^2 \\]

E. Differentiation under Integral Sign

I.

Example: [Let’s consider the following difficult integral which depends on the parameter $a$] \\[ \\int_0^\\infty e^{-ax}(1+x^2)^{-1}dx = S(a) \\]

\\[ \\int_0^\\infty x^2 e^{-ax}(1+x^2)^{-1}dx = S''(a) \\] \\[ S(a)+S''(a) = \\int_0^\\infty \\left(\\dfrac{1}{1+x^2}+\\dfrac{x^2}{1+x^2}\\right)e^{-ax}dx \\] \\[ \\dfrac{d^2 S}{da^2}+S = \\dfrac{1}{a} \\] This differential equation may be solved for S(a).

Explanation

This example demonstrates a clever technique for solving a difficult integral by transforming it into a differential equation. The integral is first defined as a function, S(a), which depends on the parameter a. By taking the second derivative of S(a) with respect to a, a new integral for S’‘(a) is found. The key insight is to then add S(a) and S’‘(a) together. This allows the complicated parts inside the integral to combine and simplify algebraically, leaving a much simpler integral on the right-hand side that evaluates to just 1/a. As a result, the original problem is successfully converted into the more standard task of solving the second-order differential equation S’’ + S = 1/a to find the value of S(a).

II. Partial Differentiation

[Now let’s consider the following integral that is a function of two parameters, α and β.] \\[ S(\\alpha,\\beta) = \\int_0^\\infty e^{-\\alpha x}\\sin (\\beta x)\\, dx \\] \\[ \\begin{aligned} \\dfrac{\\partial^2 S}{\\partial \\alpha^2} &= \\int_0^\\infty x^2 e^{-\\alpha x} \\sin (\\beta x)\\, dx \\\\ \\dfrac{\\partial^2 S}{\\partial \\beta^2} &= -\\int_0^\\infty x^2 e^{-ax} \\sin (\\beta x)\\, dx \\\\ \\dfrac{\\partial^2 S}{\\partial \\beta^2} &= -\\dfrac{\\partial^2 S}{\\partial \\alpha^2} \\end{aligned} \\]

Explanation

We differentiate under the integral sign twice. Differentiating with respect to α brings down (-x)² = x². Differentiating with respect to β brings down a factor of x twice from the sin(βx) and a minus sign from the second derivative of sine. We see that the two second partial derivatives are negatives of each other. This gives us a partial differential equation, which is the Laplace equation. Our integral is a solution to this equation.

The form of S may be partly determined by making the substitution \$y=\\beta x\$ \\[ S(\\alpha,\\beta) = \\beta^{-1} \\int_0^\\infty e^{-\\frac{\\alpha}{\\beta}y}\\sin y\\, dy = \\beta^{-1} F\\left(\\dfrac{\\alpha}{\\beta}\\right) \\]

Explanation

This is a dimensional analysis or “scaling” argument. By substituting y = βx, we can show that the integral’s dependence on α and β is not arbitrary; it must be a function of the ratio α/β, with a 1/β factor out front. We call this unknown function F. This step dramatically simplifies the problem by reducing two variables (α, β) to one (α/β).

\\[ \\begin{aligned} \\dfrac{\\partial}{\\partial\\beta}\\left[\\beta^{-1}F\\left(\\dfrac{\\alpha}{\\beta}\\right)\\right] &= -\\frac{\\alpha}{\\beta^3}F'\\left(\\dfrac{\\alpha}{\\beta}\\right)-F\\beta^{-2} \\\\ \\dfrac{\\partial^2}{\\partial\\beta^2}\\left[\\beta^{-1}F\\left(\\dfrac{\\alpha}{\\beta}\\right)\\right] &= \\dfrac{\\alpha^2}{\\beta^5}F''\\left(\\dfrac{\\alpha}{\\beta}\\right)+\\frac{3\\alpha}{\\beta^4}F' \\\\ &\\qquad+F\\frac{2}{\\beta^3}+\\frac{\\alpha}{\\beta^4} F'\\\\ \\frac{\\partial^2 S}{\\partial \\beta^2}&=\\frac{2}{\\beta^3}F+\\frac{4\\alpha}{\\beta^4}F'+\\frac{\\alpha^2}{\\beta^5}F''\\\\ \\dfrac{\\partial^2 S}{\\partial\\alpha^2} &= \\dfrac{1}{\\beta^3}F'' \\end{aligned} \\]

Explanation

Here, the second partial derivatives are re-calculated, but this time by applying the chain rule to the new form S = (1/β) F(α/β).

Multiply by \$\\beta^3\$ and let \$z=\\dfrac{\\alpha}{\\beta}\$ \\[ 2F+4zF'+z^2F'' = -F'' \\]

Explanation

We substitute the new expressions for the derivatives back into our Laplace equation ($S_{\\alpha\\alpha}+S_{\\beta\\beta}=0$) and let z = α/β. This turns the complex PDE into a simpler ordinary differential equation (ODE) for the function F(z).

\\[ \\dfrac{d}{dz}(z^2F') + 2\\dfrac{d}{dz}(zF) = -F'' \\] \\[ z^2F'+2zF = -F'+C_1 \\] \\[ \\dfrac{d}{dz}(z^2F) = -F'+C_1 \\] \\[ z^2F = -F+C_1 z + C_0 \\] \\[ F = (C_0+C_1 z)/(1+z^2) \\]

Explanation

The ODE is now solved. We use a clever trick, noticing that some terms are the result of a product rule (d/dz (z²F) and d/dz (z²F')), which allows for direct integration. This process yields the general solution for F, which contains two unknown constants, C₀ and C₁

\\[ S(\\alpha,\\beta) = \\dfrac{1}{\\beta}F = (C_0\\beta+C_1\\alpha)/(\\alpha^2+\\beta^2) \\]

Explanation

We substitute back z = α/β and S = F/β to get the general form of our original integral. Now, all we need to do is find C₀ and C₁.

To evaluate \$C_0+C_1\$ we observe: \\[ S(\\alpha,0)=0 = \\dfrac{C_1\\alpha}{\\alpha^2}\\quad \\therefore C_1=0. \\]

Explanation

We use a simple physical limit. If β=0, then sin(0*x) = 0, so the integral S(α,0) is clearly zero. Plugging β=0 into our general form gives C₁α / α². For this to be zero, C₁ must be zero.

For small \$\\beta\$, \$\\sin\\beta x \\sim \\beta x\$. This is a good approximation since \$e^{-ax}\$ kills the integrand for large x. \\[ \\int_0^\\infty e^{-ax}\\ \\beta x\\ dx = \\dfrac{\\beta}{\\alpha^2} \\] \\[ \\dfrac{C_0\\beta}{\\alpha^2+\\beta^2} \\approx \\dfrac{C_0\\beta}{\\alpha^2} \\qquad\\therefore C_0 = 1 \\]

Explanation

To find C₀, we use another approximation. For very small β, sin(βx) is approximately βx. We solve this much simpler approximate integral and find that it equals β/α².

We compare our general solution (with C₁=0) to the result from the approximation. For small β, (C₀β) / (α² + β²) ≈ (C₀β) / α². For this to match β/α², the constant C₀ must be 1.

[Therefore:] \\[ S(\\alpha,\\beta) = \\dfrac{\\beta}{\\alpha^2+\\beta^2} \\]

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A. Dirac Delta Function

[Let us introduce a quantity 𝛿(x), which depends on x satisfying the following conditions:] \\[ \\delta(x)=0, x\\neq 0 \\text{, and is in some way infinite for } x=0 \\] with \\[ \\int_{-\\infty}^\\infty \\delta(x)dx = \\int_{-\\epsilon}^\\epsilon \\delta(x)dx=1 \\] [This quantity is called the Dirac Delta function and is quite useful in physics. We can show that] \\[ \\int_{-\\infty}^\\infty f(x)\\delta(x)dx = f(0) \\]

Proof: \\[ \\int_{-\\infty}^{-\\epsilon}f(x)\\delta(x)dx + \\int_{-\\epsilon}^\\epsilon f(x)\\delta(x)dx + \\int_\\epsilon^\\infty f(x)\\delta(x)dx \\] If f(x) varies relatively slowly in the range \$-\\epsilon\$ to \$\\epsilon\$ we can replace it by an average of f(x) over that interval. \\[ \\int_{-\\infty}^\\infty f(x)\\delta(x)dx = \\int_{-\\epsilon}^\\epsilon \\overline{f(x)}\\delta(x)dx \\] Letting \$\\epsilon\\to 0\$, we get \\[ \\begin{aligned} &= \\left.\\overline{f(x)}\\right|_{x\\to 0}\\int_{-\\epsilon}^\\epsilon \\delta(x)dx\\\\ &=f(0) \\end{aligned} \\]

The delta function is useful since it may be operated on as though it were a real function. The only justification we give for this is that it gives the correct answer. \\[ f'(t) = -\\int_{-\\infty}^\\infty f(x)\\delta'(x-t)dx \\] \\[ = -f(x)\\delta(x-t)\\bigg|_{-\\infty}^\\infty + \\int_{-\\infty}^\\infty f'(x)\\delta(x-t)dx \\] \\[ = 0+f'(t) \\] The following useful relations may be easily proved:

\$\\delta(x)=\\delta(-x)\$
\$\\delta'(x)=-\\delta'(-x)\$
\$x\\delta(x)=0\$
\$x\\delta'(x)=-\\delta(x)\$
\$\\delta(ax) = \\dfrac{1}{|a|}\\delta(x)\$
\$f(x)\\delta(x-a)=f(a)\\delta(x-a)\$

The following figures are approximate representations of \$\\delta(x)\$ and \$\\delta'(x)\$.

Problem: (with integrals)

Prove: \$\\delta(ax)=\\dfrac{1}{a}\\delta(x)\$ for \$a>0\$
Find: \$x\\delta''(x)=?\$

B. Step Function

\\[ \\int_{-\\infty}^x \\delta(t)dt = \\begin{cases} 0 & x<0 \\\\ 1 & x>0 \\end{cases} \\]

Explanation

The function being described on the right-hand side is the Heaviside step function, often written as H(x) or θ(x)

Case 1 (x<0) The integral’s upper limit x is negative. The integration range is from −∞ to x. This range does not include t=0 where the delta functon is non-zero. Therefore, you are integrating a function that is zero over the entire interval, and the result is 0.
Case 2 (x>0): The integral’s upper limit x is positive. The integration range from −∞ to xdoes include t=0. Because the entire “spike” of the delta function is included in the range, the integral evaluates to the total area of the delta function, which is 1.

Notice that \$H^\\prime(x)=\\delta(x)\$.

Consider the curve in the following figure. Its derivative may be expressed in terms of the delta function as follows: \\[ f'(x) = f'(x)_{x\\neq a} + c\\delta(x-a) \\]

Explanation

Let’s take any function f(x) that is smooth everywhere except for a single jump discontinuity of height c at x=a.

We can represent this function f(x) as the sum of two parts:

A continuous smooth “base” function, let’s call it g(x).
A step function that adds the jump at the right place.

The formula is: \\[ f(x) = g(x) + c \\cdot H(x-a) \\]

How do we find g(x)? The function g(x) is simply the “pre-jump” part of the function extended across the entire domain.

For x < a, we have H(x-a)=0, so f(x) = g(x). This means g(x) is identical to f(x) before the jump.
For x > a, we have H(x-a)=1, so f(x) = g(x) + c. This correctly describes the function after the jump.

Differentiating the Representation

Now that we have expressed f(x) in a form that is a sum of well-behaved functions (a continuous function and a Heaviside function), we can differentiate it using the sum rule: \\[ f'(x) = \\frac{d}{dx} \\left[ g(x) + c \\cdot H(x-a) \\right] \\] \\[ f'(x) = \\frac{d}{dx} g(x) + c \\cdot \\frac{d}{dx} H(x-a) \\]

Using the relationship \$\\frac{d}{dx}H(x-a) = \\delta(x-a)\$, we get: \\[ f'(x) = g'(x) + c \\cdot \\delta(x-a) \\]

Now, what is g’(x)? Since g(x) is the continuous version of f(x), its derivative g’(x) is exactly the same as the derivative of f(x) away from the jump, i.e., for \$x \\neq a\$.

So, we can replace g’(x) with the notation f’(x)_{x a} to get the final formula: \\[ f'(x) = f'(x)_{x\\neq a} + c \\cdot \\delta(x-a). \\]

C. Delta function and the integral of cos(ßx)

Consider the integral: \$\\displaystyle \\int_{-\\infty}^\\infty \\cos \\beta x\\, dx\$. We shall show that it is equal to \$2\\pi\\delta(\\beta)\$. \\[ \\int_0^{\\infty} \\cos\\beta x\\, dx = \\pi\\delta(\\beta). \\]

Explanation

Notice that \$\\int_0^\\infty\$ cos(ßx) dx does not converge in the usual sense because the cosine function oscillates forever. To give it a meaningful value, we treat it as a distribution by examining its behavior in a limit.

\\[ \\lim_{a\\to 0} \\int_0^\\infty e^{-ax}\\cos\\beta x\\, dx = \\lim_{a\\to 0} \\dfrac{a}{a^2+\\beta^2} \\]

Explanation

It was previously shown that \\[ \\int_0^\\infty e^{-ax}\\cos(\\beta x) dx=\\frac{a}{a^2+\\beta^2} \\]

For small \$a\$ this integral looks approximately as in the following figure.

The area under the curve is: \\[ A = \\int_{-\\infty}^\\infty \\left(\\dfrac{a}{a^2+\\beta^2}\\right)d\\beta = \\int_{-\\infty}^\\infty \\dfrac{a^2}{a^2+a^2z^2}dz \\] \\[ = \\int_{-\\infty}^\\infty \\dfrac{dz}{1+z^2} = \\left. \\tan^{-1}z \\right|_{-\\infty}^\\infty = \\pi \\] Note that the area is independent of the value of a.

Explanation

The function \$\\frac{a}{a^2+\\beta^2}\$ has an area of π for any value of a. As a→0, the function becomes an infinitely thin, infinitely high spike at β=0, while its total area remains π. This is precisely the behavior of πδ(β).

D. Use of the delta function in evaluating a definite integral

Example: Consider \\[ \\int_0^\\pi \\cos(m\\tan\\theta)d\\theta \\] Substitute \$\\tan\\theta = x\$ \\[ \\int_0^\\infty \\cos mx(1+x^2)^{-1}dx = S(m) \\] \\[ -\\int_0^\\infty x^2 \\cos(mx)(1+x^2)^{-1}\\, dx=S''(m) \\] \\[ S(m)-S''(m) = \\int_0^\\infty \\cos mx\\ dx = \\pi\\delta(m) \\] The general solution of such a differential equation is: \\[ S(m) = \\dfrac{e^{-m}}{2}\\int_{-\\infty}^m e^{-t} f(t)dt - \\dfrac{e^m}{2}\\int_{-\\infty}^m e^{t}f(t)dt + Ae^m+Be^{-m} \\] where \$f(t) = \\pi\\delta(t)\$

Explanation

From differential equations, recall that if \$y_1(x)\$ and \$y_2(x)\$ are two solutions the equation \$y''+P(x)y'+Q(x)=0\$, then a particular solution of \$y''+P(x)y'+Q(x)=R(x)\$ is given by \\[ -y_1(x)\\int_{x_0}^x \\frac{y_2(t)R(t)}{W(t)}dt+y_2(x)\\int_{x_0}^x \\frac{y_1(t)R(t)}{W(t)}dt \\] where \$W(t)=W[y_1(t),y_2(t)]\$ is the Wronskian.

For the equation \$S''(m)-S(m)=0\$, we look for solutions of the form \$e^{rm}\$. The characteristic equation is \$r^2-1=0\$, which has roots \$r=\\pm 1\$. This gives us two independent solutions of the corresponding homogeneous equation: \\[ S_1(m)=e^{m},\\qquad S_2(m)=e^{-m}. \\] The homogeneous solution is \\[ S_h(m)=A S_1(m)+B S_2(m)=A e^m+B e^{-m}. \\] The Wronskian is \\[ \\begin{vmatrix} S_1(m) & S_2(m)\\\\ S_1^\\prime(m) & S_2^\\prime(m) \\end{vmatrix}=\\begin{vmatrix} e^m & e^{-m}\\\\ e^m & -e^{-m} \\end{vmatrix}=-1-1=-2. \\] According to the theorem mentioned above, a particular solution is given by \\[ S_p(m)=e^m \\int_{-\\infty}^m \\frac{e^{-t}f(t)}{-2} dt-e^{-m}\\int_{-\\infty}^m \\frac{e^t f(t)}{-2}dt \\] and the total solution is \\[ S(m)=S_h(m)+S_p(m) \\]

Solutions: \\[ \\begin{aligned} m<0 \\quad &S=Ae^m+Be^{-m}\\\\ m>0 \\quad &S=Ae^m+Be^{-m}-\\dfrac{\\pi}{2}e^m+\\dfrac{\\pi}{2}e^{-m} \\end{aligned} \\]

Explanation

Previously, we obtained \\[ S_p(m)=\\frac{\\pi}{2}e^{-m}\\int_{-\\infty}^m \\delta(t) e^{-t}dt-\\frac{\\pi}{2} e^m\\int_{-\\infty}^m \\delta(t)e^{-t} dt \\] If \$m<0\$: \\[ S_p(m)=0 \\]and the total solution is \\[ S(m)=A e^m+B e^{-m} \\] If \$m>0\$ then \\[ S_p(m)=\\frac{\\pi}{2}e^{-m}e^0-\\frac{\\pi}{2}e^m e^0 \\] and the total solution is \\[ S(m)=A e^m+B e^{-m}+\\frac{\\pi}{2} e^{-m}-\\frac{\\pi}{2}e^m \\]

\\[ S(0) = \\dfrac{\\pi}{2} = A+B \\] \\[ S(m)=S(-m)\\quad \\therefore A-B=\\dfrac{\\pi}{2} \\] \\[A=\\dfrac{\\pi}{2}, B=0 \\] \\[m<0 \\quad S=\\dfrac{\\pi}{2}e^m\\] \\[ m>0 \\quad S=\\dfrac{\\pi}{2}e^{-m} \\]

Example: Let us consider the integral \\[ \\int_{-\\infty}^\\infty \\dfrac{\\sin\\beta x}{x}dx = S(\\beta) \\] [If we differentiate with respect to \$\\beta\$, we get the \$\\int_0^\\infty \\cos \\beta x dx\$ where we know is \$2\\pi \\delta(\\beta)\$:] \\[ S'(\\beta) = \\int_{-\\infty}^\\infty x\\dfrac{\\cos\\beta x}{x} dx= 2\\pi\\delta(\\beta) \\] [Solving this differential equation, we get:] \\[ \\begin{aligned} S(\\beta) &= 2\\pi+C \\quad \\beta>0 \\\\ S(\\beta) &= C \\quad \\beta<0 \\end{aligned} \\] We note that \$S(\\beta) = -S(-\\beta)\$ \\[ 2\\pi+C = -C \\] \\[ C=-\\pi \\] \\[ \\begin{aligned} \\beta>0 \\quad S(\\beta) = \\pi \\\\ \\beta<0 \\quad S(\\beta) = -\\pi \\end{aligned} \\] (A graph shows a step function, going from -π for β<0 to +π for β>0)

Consider now the integral \\[ \\int_0^\\infty e^{-\\alpha x}\\dfrac{\\sin\\beta x}{x}dx = \\dfrac{\\pi}{2}-\\tan^{-1}(\\alpha/\\beta) \\]

Explanation

Notice that we previously studied this integral.

A graph of this integral looks as follows:

In the lim \$\\alpha\\to 0\$, this integral reduces to the one considered previous to it (there is a factor 2 to be considered)

Reference: Principals of Quantum Mechanics, Paul Dirac, 4th ed., p. 58–61

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A. Investigation of the Existence of Integrals.

Explanation

Alright, before we start trying to solve a fancy integral, we should ask a basic question: does it even have a finite answer? An integral from zero to infinity can go wrong in two places: at the starting point, \$x=0\$ (or at a finite number), if the function blows up, or at the end, \$x=\\infty\$, if the function doesn’t die off fast enough. We have to check both “danger zones.” If it fails at either end, the whole thing is infinite, and we can stop wasting our time.

Example: Consider: \\[ \\int_0^\\infty e^{-ax}\\dfrac{dx}{x} \\] Near 0 \\[ \\sim \\int \\dfrac{dx}{x} = \\ln 0 + \\text{numbers} \\] The integral is infinite.

Explanation

Let’s look at this guy. We check the first danger zone: \$x\$ near zero. When \$x\$ is a very small number, what is \$e^{-ax}\$? It’s \$e\$ to the power of something tiny, which is practically \$e^0 = 1\$. So, our complicated-looking integrand \$\\frac{e^{-ax}}{x}\$ behaves almost exactly like the much simpler function \$\\frac{1}{x}\$ when we’re close to the origin.

And we all know what happens when you try to integrate \$\\frac{1}{x}\$ starting from zero. The antiderivative is \$\\ln(x)\$. As \$x\$ gets closer and closer to zero, \$\\ln(x)\$ goes to minus infinity. Since the integral blows up at the lower limit, we don’t even need to check what happens at infinity. The whole thing is infinite. It’s no good.

Example: Consider: \\[ \\int_0^\\infty \\dfrac{1-e^{-ax}}{x}dx \\] \$x\\sim 0\$ \\[ \\int_0 a\\, dx= 0+\\dots \\]

\$x\\to\\infty\$ \\[ \\int^\\infty \\dfrac{dx}{x} = \\ln\\infty+\\dots \\] Again the integral is infinite.

Explanation

Now for this one. It looks almost the same, but that little $1 - ...$ in the numerator makes all the difference. Let’s check our danger zones again.

Near zero: As \$x \\to 0\$, we have a problem. The denominator is zero. But wait! The numerator \$1-e^{-ax}\$ is also going to zero, because \$e^{-ax} \\to 1\$. We’ve got a \$0/0\$ situation, which means we have to be more careful. You remember the Taylor series for an exponential, right? For any small thing ‘\$u\$’, \$e^{-u}\$ is approximately \$1 - u\$. Here, our small thing is \$ax\$. So, for small \$x\$, \$e^{-ax} \\approx 1-ax\$.

Let’s plug that approximation into the numerator: \$1 - (1-ax) = ax\$. So our whole function looks like \$\\frac{ax}{x}\$. The \$x\$’s cancel! Near the origin, our integrand behaves just like the constant $a$ . Integrating a constant is no problem at all; it’s perfectly finite. So, it’s safe at zero.

Now for infinity: What happens when \$x\$ gets very, very big? The \$e^{-ax}\$ term (for \$a>0\$) decays exponentially fast. It goes to zero. So for large \$x\$, our integrand looks like \$\\frac{1-0}{x} = \\frac{1}{x}\$. Ah, but we know that integrating \$\\frac{1}{x}\$ all the way out to infinity gives you a logarithm, \$\\ln(\\infty)\$, which blows up.

So this one is the other way around: it’s perfectly fine at the start, but it goes to infinity at the end. The final result is still that the integral is infinite.

The following table of functions may prove useful in determining whether or not a given integral exists.

	\$x\\to\\infty\$	\$x\\to 0\$
\$dx/x^2\$	O.K.	N.G.
\$dx/x\$	N.G.	N.G.
\$dx/x^{1-\\epsilon}\$	N.G.	O.K.
\$dx/x^{1+\\epsilon}\$	O.K.	N.G.
\$dx\$	N.G.	O.K.

Explanation

This table is just a summary of what we’ve been doing. Most of the time, the convergence of an integral at either end depends on how it behaves compared to \$\\int \\frac{dx}{x^p}\$. The function \$\\frac{1}{x}\$ (where \$p=1\$) is the great dividing line.

At infinity, your function must die off faster than \$\\frac{1}{x}\$ to converge. That’s why \$1/x^{1+\\epsilon}\$ is O.K. but \$1/x^{1-\\epsilon}\$ is No Good.

Near zero, the situation is reversed. Your function must blow up slower than \$\\frac{1}{x}\$ to converge. That’s why \$1/x^{1-\\epsilon}\$ is O.K. there, but \$1/x^{1+\\epsilon}\$ blows up too fast and is No Good. It’s a very useful rule of thumb for quickly checking if an integral is worth your time.

B. Special Method of Evaluating a Definite Integral.

(This method is applicable to problems 1 and 3) \\[ A = \\int_{-\\infty}^\\infty e^{-x^2}dx \\]

Explanation

Alright, how do you solve this integral? If you try all the standard high-school methods—integration by parts, substitution—you’ll get absolutely nowhere. The antiderivative of \$e^{-x^2}\$ is not an “elementary function.” You can’t write it down with sines, cosines, logs, and so on. So we have to be clever. We need a trick.

\\[ \\begin{aligned} A^2 &= \\int_{-\\infty}^\\infty e^{-x^2}dx \\int_{-\\infty}^\\infty e^{-y^2}dy \\\\ A^2 &= \\int_{-\\infty}^\\infty \\int_{-\\infty}^\\infty e^{-(x^2+y^2)}dxdy \\end{aligned} \\]

Explanation

The trick is this: instead of calculating \$A\$, let’s try to calculate \$A^2\$. It seems mad to make the problem more complicated, but watch. We write \$A^2\$ as the integral times itself. Since the variable of integration is just a dummy, we can call it \$x\$ in the first integral and \$y\$ in the second. Now we can combine them into a single double integral over the entire \$xy\$-plane.

Substitute \$r^2=x^2+y^2\$ \$x=r\\cos\\theta\$ \$y=r\\sin\\theta\$ \\[ \\begin{aligned} A^2 &= 4\\int_0^{\\pi/2}\\int_0^\\infty r e^{-r^2}drd\\theta \\\\ &= \\pi \\int_0^\\infty 2r e^{-r^2}dr = -\\left. \\pi e^{-r^2} \\right|_0^\\infty \\\\ A &= \\sqrt{\\pi} \\end{aligned} \\]

Explanation

Now, look at what’s inside the integral: \$e^{-(x^2+y^2)}\$. Whenever you see an \$x^2+y^2\$, your brain should light up and scream “Circles! Polar coordinates!” In polar coordinates, \$x^2+y^2\$ is just \$r^2\$. The area element \$dxdy\$ becomes \$r dr d\\theta\$.

This is where the magic happens. Our integral becomes \$\\int e^{-r^2} (r dr d\\theta)\$. We got an extra factor of \$r\$ from the coordinate change, and that’s exactly what we needed to solve the integral! The derivative of the exponent \$(-r^2)\$ is \$-2r\$. We have an \$r\$, so the integral is now easy.

The integral over \$\\theta\$ from \$0\$ to \$2\\pi\$ just gives a factor of \$2\\pi\$ (the notes use symmetry and go from \$0\$ to \$\\pi/2\$, giving a factor of \$4 \\times \\pi/2 = 2\\pi\$ in the end, but the result is \$\\pi\$). The integral \$\\int_0^\\infty 2r e^{-r^2} dr\$ is just \$-e^{-r^2}\$ evaluated from \$0\$ to \$\\infty\$, which is \$- (0 - 1) = 1\$.

So, \$A^2 = \\pi \\times 1 = \\pi\$. And that means our original, impossible-looking integral is simply \$A = \\sqrt{\\pi}\$. It’s a beautiful trick!

C. Series Solution for Evaluating Integrals.

Explanation

We’re going to learn another powerful trick for cracking open tough integrals.

The idea is wonderfully simple. If you have a function that’s too hard to integrate, why not replace it with something easier? Or better yet, replace it with an infinite list of things that are all incredibly easy to integrate. That’s the whole game. We’re going to trade one hard problem for an infinite number of easy ones. As long as we can add up the answers from all the easy problems, we’ve solved the original hard one.

Example: Consider: \\[ \\int_0^\\infty \\dfrac{dx}{e^{mx}+e^{-mx}} = \\dfrac{\\pi}{4m} \\] \\[ \\begin{aligned} \\int_0^\\infty \\dfrac{e^{-mx}dx}{(1+e^{-2mx})} &= \\int_0^\\infty dx(1-e^{-2mx}+e^{-4mx}\\dots) \\\\ &= \\int_0^\\infty dx(e^{-mx}-e^{-3mx}+e^{-5mx}\\dots) \\\\ \\end{aligned} \\]

Explanation

Here’s the plan: if you can’t integrate a function directly, maybe you can turn it into an infinite series of simpler functions that you can integrate.

First, let’s clean up the integrand. We can multiply the top and bottom by \$e^{-mx}\$. This gives us \$\\frac{e^{-mx}}{1+e^{-2mx}}\$. Why is this better? Because as \$x\$ gets large, the term \$e^{-2mx}\$ becomes very small. This means our denominator looks like \$1/(1+u)\$ where \$0<u< 1\$ .

And we know exactly what to do with that! It’s a geometric series: \$\\frac{1}{1+u} = 1 - u + u^2 - u^3 + \\dots\$. So we can replace our complicated fraction with an infinite sum of simple, beautiful exponentials.

\\[ \\begin{aligned} &= \\int_0^\\infty e^{-mx}dx - \\int_0^\\infty e^{-3mx}dx + \\int_0^\\infty e^{-5mx}dx - \\dots \\\\ &= \\left(\\dfrac{1}{m}-\\dfrac{1}{3m}+\\dfrac{1}{5m}-\\dots\\right) = \\dfrac{1}{m}\\left(1-\\dfrac{1}{3}+\\dfrac{1}{5}-\\dots\\right) \\end{aligned} \\]

Explanation

We integrate the series term by term. Each term is just \$\\int_0^\\infty e^{-kx} dx\$, which is one of the easiest integrals in the world. The answer is simply \$1/k\$.

So our integral becomes a sum of numbers: \$\\frac{1}{m} - \\frac{1}{3m} + \\frac{1}{5m} - \\dots\$. We can pull out the common factor of \$1/m\$.

\\[ \\therefore \\int_0^\\infty \\dfrac{dx}{e^{mx}+e^{-mx}} = \\dfrac{1}{m}(\\dfrac{\\pi}{4}) \\]

Explanation

We are left with the famous alternating series \$1 - \\frac{1}{3} + \\frac{1}{5} - \\frac{1}{7} + \\dots\$. This series is well-known; it’s the Leibniz series, and it converges to \$\\pi/4\$. You can derive it from the Taylor series for \$\\tan^{-1}(x)\$ by plugging in \$x=1\$.

So, we substitute \$\\pi/4\$ for the series, and there’s our answer! We’ve traded a single hard integral for an infinite number of easy ones, and it paid off beautifully.

A combination of the methods of integration discussed this far is sufficient to to solve any integral which exists. There are still other methods which may be used and are in many cases easier. We shall discuss some of them later.

Problem 1: The \$\\displaystyle\\int_0^\\infty \\dfrac{1-e^{-ax^2}}{x^2}dx\$ exists. Find its value.

Problem 2: Integrate: \\[ \\int_0^\\infty \\dfrac{\\cos mx}{(1+x^2)^2}dx \\]

Problem 3: Prove: \\[ \\int_0^\\infty \\cos(x^2)dx = \\int_0^\\infty \\sin(x^2)dx = \\dfrac{1}{2}\\sqrt{\\pi/2} \\]

Problem 4: Integrate \\[ \\int_0^\\infty \\dfrac{x dx}{e^{mx}-e^{-mx}} \\]

Problem 5: Let \\[ \\int_0^\\infty \\dfrac{\\sin ax}{e^x - e^{-x}} dx = S(a) \\]

Find S(1) to 3 significant figures
Find approximate expressions for S(a) for large and small a.

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trial \\(x\\)	result \\(x\\)
0	\\(\\to .5\\)
.5	\\(\\to .4\\)
.4	\\(\\to .431\\)
…	…

	\\(x\\to\\infty\\)	\\(x\\to 0\\)
\\(dx/x^2\\)	O.K.	N.G.
\\(dx/x\\)	N.G.	N.G.
\\(dx/x^{1-\\epsilon}\\)	N.G.	O.K.
\\(dx/x^{1+\\epsilon}\\)	O.K.	N.G.
\\(dx\\)	N.G.	O.K.

Mathematical Methods in Physics

By Unknown Author

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