Laplace transform

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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[edit]

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


  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 $s=\dfrac{\pi}{2} - i\log(\pi + \sqrt{\pi^2-1})$.


Laplace transform of power function with real exponent
Laplace transform of $\sin(\sqrt{t})$
Laplace transform of impulse function
Laplace transform of sine integral
Laplace transform of cosine integral
Laplace transform of exponential integral

See also[edit]

Two-dimensional Laplace transform

External links[edit]

Laplace transform identity