# Laplace transform

Let $f \colon \mathbb{R} \rightarrow \mathbb{C}$ be a function, then the Laplace transform of $f$ is the function defined by $$\mathscr{L}\{f\}(z) = \displaystyle\int_0^{\infty} e^{-zt}f(t) \mathrm{d}t,$$ provided this integral exists.

# Table of Laplace Transforms

 Original function $f(t)$ Laplace transform $\mathscr{L}\{f\}(z)$ $1$ $\dfrac{1}{z}$ $t$ $\dfrac{1}{z^2}$ $t^{n}$ $\dfrac{\Gamma(n+1)}{z^{n+1}}$ $\dfrac{1}{\sqrt{\pi t}}$ $\dfrac{1}{\sqrt{z}}$ $\dfrac{2^nt^{n-\frac{1}{2}}}{1 \cdot 3 \cdot 5 \ldots (2n-1)\sqrt{\pi}}$ $\dfrac{1}{z^{n+\frac{1}{2}}}$ $e^{-at}$ $\dfrac{1}{z+a}$ $te^{-at}$ $\dfrac{1}{(z+a)^2}$ $\dfrac{1}{a}\sin(at)$ $\dfrac{1}{z^2+a^2}$ $\cos(at)$ $\dfrac{z}{z^2+a^2}$ $J_0(at)$ $\dfrac{1}{\sqrt{z^2+a^2}}$ $a^{\nu}J_{\nu}(a t)$ $\dfrac{(\sqrt{z^2+a^2}-z)^{\nu}}{\sqrt{z^2+a^2}};\nu>-1$ $\mathrm{abs}(\sin(kt))$ $\dfrac{k}{z^2+k^2}\coth\left( \dfrac{\pi z}{2k} \right)$

# Examples

1. Find a function $f$ such that $\mathscr{L}\{f\}(z)=f(z)$.

Solution: We know that if $g(t)=t^{\ell}$ and $\mathrm{Re}(\ell) > 1$, then $\mathscr{L}\{g\}(z)=\dfrac{\Gamma(\ell+1)}{z^{\ell+1}}$, where $\Gamma$ denotes the gamma function. We will seek a function $f$ with the property that $\mathscr{L}\{f\}(z)=f(z)$. Let $s \in \mathbb{C}$ with $0<\mathrm{Re}(s)<1$ be as of yet undetermined. Suppose $f$ has the form $$f(t)=\sqrt{\Gamma(s)}t^{-s}+\sqrt{\Gamma(1-s)}t^{s-1}.$$ Then we may compute $$\begin{array}{ll} \mathscr{L}\{f\}(z)&=\sqrt{\Gamma(s)} \dfrac{\Gamma(1-s)}{z^{-s+1}} + \sqrt{\Gamma(1-s)} \dfrac{\Gamma(s)}{z^s} \\ &= \sqrt{\Gamma(s)\Gamma(1-s)} \left[ \dfrac{\sqrt{\Gamma(s)}\Gamma(1-s)}{z^{-s+1}\sqrt{\Gamma(s)\Gamma(1-s)}} + \dfrac{\sqrt{\Gamma(1-s)}\Gamma(s)}{z^s\sqrt{\Gamma(s)\Gamma(1-s)}} \right] \\ &= \sqrt{\Gamma(s)\Gamma(1-s)} \left[ \dfrac{\sqrt{\Gamma(1-s)}}{z^{-s+1}} + \dfrac{\sqrt{\Gamma(s)}}{z^s} \right] \\ &= \sqrt{\Gamma(s)\Gamma(1-s)} \left[ \sqrt{\Gamma(1-s)}z^{s-1} + \sqrt{\Gamma(s)}z^{-s} \right] \\ &= \sqrt{\Gamma(s)\Gamma(1-s)} f(z). \end{array}.$$ If we pick $s$ so that $\sqrt{\Gamma(1-s)}z^{s-1} + \sqrt{\Gamma(s)}=1$ we will have found the function $f$. By properties of the gamma function we know that $\Gamma(1-z)\Gamma(z)=\dfrac{\pi}{\sin(\pi z)}$ and so we see that we must pick $s$ so that $\dfrac{\pi}{\sin(\pi s)}=1,$ i.e., $s = \dfrac{\arcsin(\pi)}{\pi},$ which yields $s=\dfrac{\pi}{2} - i\log(\pi + \sqrt{\pi^2-1})$.