Definition of the Laplace Transform

The Laplace transform of $f (t)$ is the improper integral $\int_{0}^{\infty} e^{- s t} f (t) d t$ , whenever it converges. This section computes it for the standard functions, establishes that $ℒ$ is a linear operator, and extends the power rule from integer exponents to arbitrary real ones using the gamma function.

Quick Reference

$f (t)$	$F (s) = ℒ {f (t)}$	Valid for
$e^{a t}$	$\frac{1}{s - a}$	$s > a$
$\sin k t$	$\frac{k}{s^{2} + k^{2}}$	$s > 0$
$\cos k t$	$\frac{s}{s^{2} + k^{2}}$	$s > 0$
$\sinh k t$	$\frac{k}{s^{2} - k^{2}}$	$s > \| k \|$
$\cosh k t$	$\frac{s}{s^{2} - k^{2}}$	$s > \| k \|$
$1$	$\frac{1}{s}$	$s > 0$
$t^{n}$ , $n = 0, 1, 2, \dots$	$\frac{n!}{s^{n + 1}}$	$s > 0$
$t^{α}$ , $α > - 1$	$\frac{Γ (α + 1)}{s^{α + 1}}$	$s > 0$

Property	Statement
Linearity	$ℒ {a f + b g} = a ℒ {f} + b ℒ {g}$
Gamma recursion	$Γ (p + 1) = p Γ (p)$ for $p > 0$
Gamma at integers	$Γ (n + 1) = n!$
A key value	$Γ (\frac{1}{2}) = \sqrt{π}$

Definition

Suppose we are given a function $f (t)$ for $t > 0$ . The Laplace transform of $f (t)$ is defined by

ℒ {f (t)} = \int_{0}^{\infty} e^{- s t} f (t) d t,

provided the integral exists, that is, has a finite value. The result is a function of the parameter $s$ , often denoted $F (s)$ .

Notice that the original function depends on $t$ and its transform is a function of $s$ . We denote the original by a lowercase letter and its transform by the same letter in capital.

Since the upper limit is $\infty$ , the defining integral is improper. It exists, or converges, when

lim_{R \to \infty} (\int_{0}^{R} e^{- s t} f (t) d t)

has a finite value. The symbol $ℒ$ is called the Laplace transform operator.

It can be shown that if the Laplace transform of a function exists for $s = a$ , then it also exists for every $s \geq a$ . So the region of convergence is always a half-line $s > a$ , and the number $a$ is what the theory of Section 7.2 is designed to identify.

Engineering interpretation. If $t$ represents time, $s$ is called the (complex) frequency variable. If $t$ is measured in seconds, the units of $s$ are ${sec}^{- 1}$ , so the product $s t$ in the exponent is dimensionless, as it must be.

In this book we take $s$ to be real, and to assure convergence $s$ must be positive. In general $s$ may be complex, provided its real part is large enough, $Re (s) > 0$ in the simplest cases.

There is nothing special about the letter $s$ , and many textbooks use $p$ instead. To indicate the parameter explicitly one may write $ℒ {f (t); s}$ . When transforming a multivariable function such as $f (t, x)$ , some texts indicate the integration variable, writing

ℒ {f (t, x); t \to s} = \int_{0}^{\infty} f (t, x) e^{- s t} d t

and

ℒ {f (t, x); x \to p} = \int_{0}^{\infty} f (t, x) e^{- p x} d x .

The Basic Transforms

Find the Laplace transform of $f (t) = e^{a t}$ .

Solution

\begin{aligned} \mathscr{L}\left\{e^{at}\right\} = F(s) &= \int_0^{\infty}e^{-st}\cdot e^{at}\,dt \\ &= \int_0^{\infty}e^{-(s-a)t}\,dt \\ &= \lim_{R\to\infty}\int_0^{R}e^{-(s-a)t}\,dt \\ &= \lim_{R\to\infty}\left[-\frac{1}{s-a}e^{-(s-a)t}\right]_0^{R} \\ &= \lim_{R\to\infty}\left[-\frac{1}{s-a}e^{-(s-a)R} + \frac{1}{s-a}\right] \\ &= \lim_{R\to\infty}\left[\frac{1}{s-a}\left(1 - e^{-(s-a)R}\right)\right] \end{aligned}

If $s > a$ , then $e^{- (s - a) R} \to 0$ as $R \to \infty$ , and thus

ℒ {e^{a t}} = \frac{1}{s - a} if s > a .

Notice that if $s < a$ , then $e^{- (s - a) R} \to \infty$ as $R \to \infty$ and the integral diverges. The condition $s > a$ is not decoration; it is the entire content of convergence here.

What if

s

is complex?

Writing $s = σ + i ω$ ,

\begin{aligned} e^{-(s-a)R} &= e^{-\left((\sigma-a) + i\omega\right)R} \\ &= e^{-(\sigma-a)R}\,e^{-i\omega R} \\ &= e^{-(\sigma-a)R}\left(\cos\omega R - i\sin\omega R\right) \end{aligned}

using Euler's formula. The trigonometric factor stays bounded, so the behavior is decided entirely by $e^{- (σ - a) R}$ . If $σ = Re (s) > a$ , this tends to zero and

ℒ {e^{a t}} = \frac{1}{s - a} if Re (s) > a .

If $k$ is a real number, show that $ℒ {e^{i k t}} = \frac{1}{s - i k}$ for $s > 0$ .

Solution

\begin{aligned} \mathscr{L}\left\{e^{ikt}\right\} &= \int_0^{\infty}e^{-st}e^{ikt}\,dt \\ &= \lim_{R\to\infty}\int_0^{R}e^{-(s-ik)t}\,dt \\ &= \lim_{R\to\infty}\left[\frac{-1}{s-ik}e^{-(s-ik)t}\right]_0^{R} \\ &= \lim_{R\to\infty}\left[-\frac{1}{s-ik}e^{-(s-ik)R}\right] + \frac{1}{s-ik} \\ &= \lim_{R\to\infty}\left[-\frac{1}{s-ik}e^{-sR}\underbrace{\left(\cos kR + i\sin kR\right)}_{e^{ikR}}\right] + \frac{1}{s-ik} \end{aligned}

The bracketed trigonometric factor has modulus $1$ , so the whole first term is bounded by $\frac{e^{- s R}}{| s - i k |}$ . If $s > 0$ , this tends to zero as $R \to \infty$ and therefore

ℒ {e^{i k t}} = \frac{1}{s - i k} .

More generally, for a complex constant $a$ the same computation gives $ℒ {e^{a t}} = \frac{1}{s - a}$ provided $Re (s) > Re (a)$ .

Show that $ℒ {1} = \frac{1}{s}$ for $s > 0$ .

Solution

Method (a). This is the special case $a = 0$ of $ℒ {e^{a t}} = \frac{1}{s - a}$ , valid for $s > a$ . Setting $a = 0$ gives $ℒ {1} = \frac{1}{s}$ for $s > 0$ .

Method (b). Directly from the definition,

\begin{aligned} \mathscr{L}\left\{1\right\} &= \int_0^{\infty}e^{-st}\,dt = \lim_{R\to\infty}\int_0^{R}e^{-st}\,dt \\ &= \lim_{R\to\infty}\left[-\frac{1}{s}e^{-st}\right]_{t=0}^{t=R} = \lim_{R\to\infty}\left(-\frac{1}{s}e^{-sR}\right) + \frac{1}{s} \end{aligned}

If $s > 0$ then $e^{- s R} \to 0$ and the integral converges to $\frac{1}{s}$ .

Show that

ℒ {\sin k t} = \frac{k}{s^{2} + k^{2}}, ℒ {\cos k t} = \frac{s}{s^{2} + k^{2}}

where $k$ is a constant.

Solution

From the previous example, $ℒ {e^{i k t}} = \frac{1}{s - i k}$ . Since $e^{i k t} = \cos k t + i \sin k t$ ,

\int_{0}^{\infty} e^{- s t} (\cos k t + i \sin k t) d t = \frac{1}{s - i k}

that is,

\underset{ℒ {\cos k t}}{\underset{⏟}{\int_{0}^{\infty} e^{- s t} \cos k t d t}} + i \underset{ℒ {\sin k t}}{\underset{⏟}{\int_{0}^{\infty} e^{- s t} \sin k t d t}} = \frac{1}{s - i k} .

Both transforms are real, so

ℒ {\cos k t} = Re (\frac{1}{s - i k}), ℒ {\sin k t} = Im (\frac{1}{s - i k}) .

Rationalizing the denominator,

\frac{1}{s - i k} = \frac{1}{s - i k} \cdot \frac{s + i k}{s + i k} = \frac{s + i k}{s^{2} + k^{2}}

using $(s - i k) (s + i k) = s^{2} - (i k)^{2} = s^{2} + k^{2}$ . Therefore

ℒ {\cos k t} = \frac{s}{s^{2} + k^{2}}, ℒ {\sin k t} = \frac{k}{s^{2} + k^{2}} .

These can also be computed directly from the defining integral using integration by parts twice, but extracting them from a single complex transform is considerably faster.

An amusing byproduct: since $lim_{s \to 0} ℒ {\sin k t} = lim_{s \to 0} \frac{k}{s^{2} + k^{2}} = \frac{1}{k}$ , the Laplace transform assigns the value $\frac{1}{k}$ to the divergent integral $\int_{0}^{\infty} \sin k t d t$ . The integral does not converge in the ordinary sense, since $\int_{0}^{R} \sin k t d t = \frac{1 - \cos k R}{k}$ oscillates forever, but the transform assigns it the average value.

The Linear Property

If $f (t)$ and $g (t)$ have Laplace transforms and $a$ and $b$ are any constants, then

ℒ {a f (t) + b g (t)} = a ℒ {f (t)} + b ℒ {g (t)} .

The theorem says the Laplace transform of any linear combination of two functions is the same linear combination of their transforms.

Proof

The result holds because integration is a linear operation:

\begin{aligned} \mathscr{L}\left\{af(t) + bg(t)\right\} &= \int_0^{\infty}e^{-st}\left(af(t) + bg(t)\right)dt \\ &= \lim_{R\to\infty}\left[\int_0^{R}e^{-st}\left(af(t) + bg(t)\right)dt\right] \\ &= \lim_{R\to\infty}\left[a\int_0^{R}e^{-st}f(t)\,dt + b\int_0^{R}e^{-st}g(t)\,dt\right] \\ &= a\lim_{R\to\infty}\int_0^{R}e^{-st}f(t)\,dt + b\lim_{R\to\infty}\int_0^{R}e^{-st}g(t)\,dt \\ &= a\int_0^{\infty}e^{-st}f(t)\,dt + b\int_0^{\infty}e^{-st}g(t)\,dt \\ &= a\mathscr{L}\left\{f(t)\right\} + b\mathscr{L}\left\{g(t)\right\} . \end{aligned}

Splitting the limit of a sum into a sum of limits is valid because both limits are assumed to exist.

Find $ℒ {\sinh k t}$ and $ℒ {\cosh k t}$ , where $k$ is a constant.

Solution

(a) Since $\sinh k t = \frac{e^{k t} - e^{- k t}}{2}$ and $ℒ {e^{k t}} = \frac{1}{s - k}$ for $s > k$ ,

\begin{aligned} \mathscr{L}\left\{\sinh kt\right\} &= \frac{1}{2}\left[\mathscr{L}\left\{e^{kt}\right\} - \mathscr{L}\left\{e^{-kt}\right\}\right] \\ &= \frac{1}{2}\left[\frac{1}{s-k} - \frac{1}{s+k}\right] \\ &= \frac{1}{2}\cdot\frac{(s+k) - (s-k)}{(s-k)(s+k)} \\ &= \frac{1}{2}\cdot\frac{2k}{s^{2}-k^{2}} = \frac{k}{s^{2}-k^{2}} \end{aligned}

valid when $s > k$ and $s > - k$ , which together are equivalent to $s > | k |$ .

(b) Since $\cosh k t = \frac{e^{k t} + e^{- k t}}{2}$ ,

\begin{aligned} \mathscr{L}\left\{\cosh kt\right\} &= \frac{1}{2}\left[\frac{1}{s-k} + \frac{1}{s+k}\right] \\ &= \frac{1}{2}\cdot\frac{(s+k) + (s-k)}{s^{2}-k^{2}} = \frac{s}{s^{2}-k^{2}} \end{aligned}

for $s > | k |$ .

Note how the hyperbolic transforms differ from the trigonometric ones only in the sign of $k^{2}$ in the denominator, mirroring the identity $\sinh k t = - i \sin (i k t)$ .

Find $ℒ {3 e^{4 t} - 5 \sin 3 t}$ .

Solution

Using $ℒ {e^{a t}} = \frac{1}{s - a}$ for $s > a$ and $ℒ {\sin k t} = \frac{k}{s^{2} + k^{2}}$ for $s > 0$ , together with linearity,

\begin{aligned} \mathscr{L}\left\{3e^{4t} - 5\sin 3t\right\} &= 3\mathscr{L}\left\{e^{4t}\right\} - 5\mathscr{L}\left\{\sin 3t\right\} \\ &= \frac{3}{s-4} - 5\left(\frac{3}{s^{2}+9}\right) \\ &= \frac{3}{s-4} - \frac{15}{s^{2}+9} \end{aligned}

valid for $s > 4$ , the more restrictive of the two conditions.

Power Functions and the Gamma Function

ℒ {t^{n}} = \frac{n!}{s^{n + 1}} (n = 0, 1, 2, \dots)

Derivation for integer

n

Integrate by parts with $u = t^{n}$ and $d v = e^{- s t} d t$ , so $v = - \frac{1}{s} e^{- s t}$ and $d u = n t^{n - 1} d t$ :

\begin{aligned} \mathscr{L}\left\{t^{n}\right\} &= \int_0^{\infty}t^{n}e^{-st}\,dt \\ &= \underbrace{\left.-\frac{1}{s}e^{-st}t^{n}\right|_0^{\infty}}_{uv} - \int_0^{\infty}\underbrace{\left(-\frac{1}{s}e^{-st}\right)}_{v}\underbrace{nt^{n-1}\,dt}_{du} \\ &= 0 + \frac{n}{s}\underbrace{\int_0^{\infty}t^{n-1}e^{-st}\,dt}_{\mathscr{L}\left\{t^{n-1}\right\}} \\ &= \frac{n}{s}\mathscr{L}\left\{t^{n-1}\right\} \end{aligned}

The boundary term vanishes at both ends: at $t = 0$ because $t^{n} = 0$ for $n \geq 1$ , and at $t = \infty$ because the exponential beats the polynomial.

We know $ℒ {1} = \frac{1}{s}$ , so iterating,

\begin{aligned} \mathscr{L}\left\{t\right\} &= \frac{1}{s}\mathscr{L}\left\{1\right\} = \frac{1}{s^{2}} \\ \mathscr{L}\left\{t^{2}\right\} &= \frac{2}{s}\mathscr{L}\left\{t\right\} = \frac{2\cdot 1}{s^{3}} \\ \mathscr{L}\left\{t^{3}\right\} &= \frac{3}{s}\mathscr{L}\left\{t^{2}\right\} = \frac{3\cdot 2\cdot 1}{s^{4}} \\ \mathscr{L}\left\{t^{4}\right\} &= \frac{4}{s}\mathscr{L}\left\{t^{3}\right\} = \frac{4\cdot 3\cdot 2\cdot 1}{s^{5}} \\ &\ \ \vdots \end{aligned}

and in general $ℒ {t^{n}} = \frac{n!}{s^{n + 1}}$ .

To obtain the transform of non-integral powers of $t$ we need a special function called the gamma function.

The gamma function is defined by

Γ (p) = \int_{0}^{\infty} x^{p - 1} e^{- x} d x, (p > 0) .

Note the convention: $p$ on the left and $p - 1$ in the integrand.

The recursion

Γ (p + 1) = p Γ (p)

Substituting $p + 1$ for $p$ gives

Γ (p + 1) = \int_{0}^{\infty} x^{p} e^{- x} d x .

Integrating by parts with $u = x^{p}$ and $d v = e^{- x} d x$ ,

Γ (p + 1) = {- x^{p} e^{- x} |}_{0}^{\infty} + \int_{0}^{\infty} p x^{p - 1} e^{- x} d x .

Since $p > 0$ , we have $x^{p} \to 0$ as $x \to 0$ , and $x^{p} e^{- x} \to 0$ as $x \to \infty$ , so the boundary term vanishes:

Γ (p + 1) = p \int_{0}^{\infty} x^{p - 1} e^{- x} d x .

The remaining integral is $Γ (p)$ , proving

Γ (p + 1) = p Γ (p) (p > 0) .

If $p = n$ is a positive integer, iterating gives

\begin{aligned} \Gamma(n+1) &= n\Gamma(n) = n(n-1)\Gamma(n-1) = n(n-1)(n-2)\Gamma(n-2) \\ &\ \ \vdots \\ &= n(n-1)(n-2)\cdots 2\cdot 1\cdot\Gamma(1) = n!\,\Gamma(1) . \end{aligned}

And

Γ (1) = \int_{0}^{\infty} e^{- x} x^{0} d x = {- e^{- x} |}_{0}^{\infty} = 1,

Γ (n + 1) = n! .

The gamma function therefore interpolates the factorial to all positive real arguments, which is exactly what is needed for non-integer powers of $t$ .

The value

Γ (\frac{1}{2}) = \sqrt{π}

Γ (1 / 2) = \int_{0}^{\infty} x^{- 1 / 2} e^{- x} d x .

Let $u = \sqrt{x}$ , so $x = u^{2}$ and $d u = \frac{d x}{2 \sqrt{x}}$ , giving $x^{- 1 / 2} d x = 2 d u$ :

Γ (1 / 2) = 2 \int_{0}^{\infty} e^{- u^{2}} d u .

So it suffices to show $\int_{0}^{\infty} e^{- u^{2}} d u = \frac{\sqrt{π}}{2}$ . Let $I$ denote that integral. Then

\begin{aligned} I^{2} &= \left(\int_0^{\infty}e^{-u^{2}}\,du\right)\left(\int_0^{\infty}e^{-v^{2}}\,dv\right) \\ &= \int_0^{\infty}\!\!\int_0^{\infty}e^{-\left(u^{2}+v^{2}\right)}\,du\,dv \end{aligned}

Converting to polar coordinates with $u = r \cos θ$ and $v = r \sin θ$ , so that $d u d v = r d r d θ$ and the first quadrant corresponds to $0 \leq θ \leq π / 2$ ,

\begin{aligned} I^{2} &= \int_0^{\pi/2}\left(\int_0^{\infty}e^{-r^{2}}r\,dr\right)d\theta \\ &= \int_0^{\pi/2}\left[-\frac{1}{2}e^{-r^{2}}\right]_0^{\infty}d\theta \\ &= \int_0^{\pi/2}\frac{1}{2}\,d\theta = \frac{\pi}{4} \end{aligned}

Therefore $I = \frac{\sqrt{π}}{2}$ and

Γ (1 / 2) = 2 \cdot \frac{\sqrt{π}}{2} = \sqrt{π} .

The polar-coordinate trick is worth remembering: the single integral is not elementary, but its square is.

Graph of the gamma function for positive arguments, falling steeply from infinity near p = 0, reaching a minimum near p = 1.46, then rising steeply. — The gamma function $Γ (p)$ for $p > 0$ . It blows up as $p \to 0^{+}$ , dips to a minimum slightly below $1$ near $p \approx 1.46$ , and then grows faster than any exponential. The values $Γ (1) = Γ (2) = 1$ bracket that minimum.

Now we can compute the transform of $t^{α}$ . Setting $s t = x$ , so $s d t = d x$ and $t = x / s$ ,

\begin{aligned} \mathscr{L}\left\{t^{\alpha}\right\} &= \int_0^{\infty}e^{-st}t^{\alpha}\,dt \\ &= \int_0^{\infty}e^{-x}\left(\frac{x}{s}\right)^{\alpha}\frac{dx}{s} \\ &= \frac{1}{s^{\alpha+1}}\int_0^{\infty}e^{-x}x^{\alpha}\,dx \end{aligned}

The last integral is $Γ (α + 1)$ . Therefore

ℒ {t^{α}} = \frac{Γ (α + 1)}{s^{α + 1}} (α > - 1)

When $α = n$ is an integer, $Γ (n + 1) = n!$ recovers $ℒ {t^{n}} = \frac{n!}{s^{n + 1}}$ as before.

Find $ℒ {t^{5 / 2}}$ .

Solution

ℒ {t^{5 / 2}} = \frac{Γ (\frac{5}{2} + 1)}{s^{5 / 2 + 1}} = \frac{Γ (\frac{7}{2})}{s^{7 / 2}} .

Using $Γ (p + 1) = p Γ (p)$ repeatedly to walk down to $Γ (\frac{1}{2})$ ,

\begin{aligned} \Gamma\left(\frac{7}{2}\right) &= \frac{5}{2}\cdot\Gamma\left(\frac{5}{2}\right) \\ &= \frac{5}{2}\cdot\frac{3}{2}\cdot\Gamma\left(\frac{3}{2}\right) \\ &= \frac{5}{2}\cdot\frac{3}{2}\cdot\frac{1}{2}\cdot\Gamma\left(\frac{1}{2}\right) \\ &= \frac{15}{8}\sqrt{\pi} \end{aligned}

Therefore

ℒ {t^{5 / 2}} = \frac{15 \sqrt{π}}{8 s^{7 / 2}} = \frac{15}{8 s^{3}} \sqrt{\frac{π}{s}} .

Table of Laplace Transforms of Basic Functions

$f (t)$	$F (s)$
$e^{a t}$	$\frac{1}{s - a}$
$\sin k t$	$\frac{k}{s^{2} + k^{2}}$
$\cos k t$	$\frac{s}{s^{2} + k^{2}}$
$\sinh k t$	$\frac{k}{s^{2} - k^{2}}$
$\cosh k t$	$\frac{s}{s^{2} - k^{2}}$
$1$	$\frac{1}{s}$
$t^{n}$ ( $n = 0, 1, \dots$ )	$\frac{n!}{s^{n + 1}}$
$t^{α}$ ( $α > - 1$ )	$\frac{Γ (α + 1)}{s^{α + 1}}$

Exercises

Find $ℒ {4 t^{3} - 2 e^{- 5 t} + \cos 2 t}$ .

Solution

By linearity and the table,

4 \cdot \frac{3!}{s^{4}} - 2 \cdot \frac{1}{s + 5} + \frac{s}{s^{2} + 4} = \frac{24}{s^{4}} - \frac{2}{s + 5} + \frac{s}{s^{2} + 4},

valid for $s > 0$ .

Find $ℒ {\sin^{2} t}$ using a trigonometric identity.

Solution

Since $\sin^{2} t = \frac{1 - \cos 2 t}{2}$ ,

ℒ {\sin^{2} t} = \frac{1}{2} [\frac{1}{s} - \frac{s}{s^{2} + 4}] = \frac{1}{2} \cdot \frac{s^{2} + 4 - s^{2}}{s (s^{2} + 4)} = \frac{2}{s (s^{2} + 4)} .

Note that the transform is linear, not multiplicative: $ℒ {\sin^{2} t}$ is not the square of $ℒ {\sin t}$ . Rewriting a product as a sum first is the standard move.

Find $ℒ {t^{- 1 / 2}}$ and explain why the answer is finite despite the integrand blowing up at $t = 0$ .

Solution

ℒ {t^{- 1 / 2}} = \frac{Γ (\frac{1}{2})}{s^{1 / 2}} = \sqrt{\frac{π}{s}} .

The exponent $α = - \frac{1}{2}$ satisfies $α > - 1$ , which is exactly the condition making $\int_{0}^{1} t^{α} d t$ converge. The singularity at the origin is integrable, so the transform exists even though the function is unbounded there. This example returns in Section 7.2.

Frequently Asked Questions

What does the condition

s > a

mean?

It is the range of $s$ for which the defining integral converges. Below it the integral diverges and $F (s)$ simply does not exist. When combining several transforms, the valid range is the intersection, that is, the largest of the individual lower bounds.

Why is

ℒ

called an operator?

Because it takes a function as input and returns a function as output, just like the differential operator $L$ of Chapter 3. And like $L$ , it is linear, which is what allows a complicated function to be transformed piece by piece.

Is the transform of a product the product of the transforms?

No, and this is the most common misconception. $ℒ {f g} \neq ℒ {f} ℒ {g}$ . Products in the $t$ domain correspond to convolutions in the $s$ domain and vice versa. For a product like $\sin^{2} t$ , use a trigonometric identity to rewrite it as a sum first.

Why is the gamma function needed?

Because $ℒ {t^{n}} = \frac{n!}{s^{n + 1}}$ requires a meaning for $n!$ when $n$ is not an integer. The gamma function supplies exactly that, satisfying $Γ (n + 1) = n!$ at integers and defined for every $p > 0$ .

Why the restriction

α > - 1

Because $\int_{0}^{1} t^{α} d t$ converges only when $α > - 1$ . For $α \leq - 1$ the singularity at the origin is too severe and the defining integral diverges no matter how large $s$ is.

Can I use complex

s

Yes, and the general theory does. All the transforms here remain valid with the convergence condition restated in terms of $Re (s)$ . This book keeps $s$ real for simplicity, but the complex viewpoint is what makes the inversion formula and the theory of poles possible.

Quick Reference

Definition

The Basic Transforms

The Linear Property

Power Functions and the Gamma Function

Table of Laplace Transforms of Basic Functions

Exercises

Frequently Asked Questions

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