More on Integration Methods
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
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.