Difference between revisions of "Laplace transform"
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|$\dfrac{(\sqrt{z^2+a^2}-z)^{\nu}}{\sqrt{z^2+a^2}};\nu>-1$ | |$\dfrac{(\sqrt{z^2+a^2}-z)^{\nu}}{\sqrt{z^2+a^2}};\nu>-1$ | ||
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| − | |$ | + | |$\mathrm{abs}(\sin(kt))$ |
|$\dfrac{k}{z^2+k^2}\coth\left( \dfrac{\pi z}{2k} \right)$ | |$\dfrac{k}{z^2+k^2}\coth\left( \dfrac{\pi z}{2k} \right)$ | ||
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=See also= | =See also= | ||
[[Two-dimensional Laplace transform]] | [[Two-dimensional Laplace transform]] | ||
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| + | =External links= | ||
| + | [http://math.stackexchange.com/questions/150390/laplace-transform-identity Laplace transform identity]<br /> | ||
=References= | =References= | ||
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=findme|next=Two-dimensional Laplace transform}}: $29.1.1$ | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=findme|next=Two-dimensional Laplace transform}}: $29.1.1$ | ||
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[[Category:SpecialFunction]] | [[Category:SpecialFunction]] | ||
Latest revision as of 03:37, 21 December 2017
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.
Properties
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
- 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 $s=\dfrac{\pi}{2} - i\log(\pi + \sqrt{\pi^2-1})$.
Videos
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
Two-dimensional Laplace transform
External links
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $29.1.1$
