Laplace 變換
Laplace transform$ {\cal L}\lbrack f\rbrack
$ {\cal L}[f(t)](s):=\int^\infty_0 f(t)e^{-st}{\rm d}t ,$ t\in\R,s\in\Complex
※$ {\cal F}[f(x)](\xi):=\int^\infty_{-\infty}f(x)e^{-ix\xi}{\rm d}x
$ \sin x=\sinh(ix)
逆 Laplace 變換$ {\cal L}^{-1}[F(s)](t):=\lim_{p\to\infty}\frac 1{2\pi i}\int_{c-ip}^{c+ip}F(s)e^{st}{\rm d}s
※$ {\cal F}^{-1}[F(\xi)](x):=\int^\infty_{-\infty}F(x)e^{ix\xi}{\rm d}\xi
$ {\cal L}^{-1}[F(s)](t)=\lim_{k\to\infty}\frac{-(-\frac k t)^{k+1}\frac{\partial^k F(\frac k t)}{\partial s^k}}{k!}
Z 變換 (Z-transform。兩側 Z 變換) $ {\cal Z}[x_n](z)=X(z):=\sum_{n=-\infty}^\infty x_n z^{-n} ,$ n\in\Z,z\in\Complex
※離散 Fourier 變換は$ F(k)=\sum_{n=0}^{N-1}f(x)e^{-i\frac{2\pi kx_n}{x_N-x_0}} 逆 Z 變換$ x_n={\cal Z}^{-1}[X(z)]=\frac 1{2\pi i}\oint_C X(z)z^{n-1}{\rm d}z
片側 Z 變換$ {\cal Z}[x_n]:=\sum_{n=0}^\infty x_n z^{-n}
兩側 Laplace 變換$ {\cal B}\lbrack f\rbrack
$ {\cal B}\lbrack f\rbrack(t):=\int_{-\infty}^\infty f(t)e^{-st}dt.
$ {\cal L}\lbrack f(t)\rbrack={\cal B}\lbrack f(t)H(t)\rbrack.
$ {\cal B}\lbrack f\rbrack(s)={\cal L}\lbrack f(t)\rbrack(s)+{\cal L}\lbrack f(-t)\rbrack(-s).
$ {\cal F}\lbrack f\rbrack=\frac 1{\sqrt{2\pi}}{\cal B}\lbrack f\rbrack(s).
$ M_X(s)={\cal B}\lbrack f\rbrack(-s)
$ {\cal M}\lbrack f\rbrack(s):=\int_0^\infty f(x)x^{s-1}dx.
$ {\cal M}\lbrack f\rbrack(s)={\cal B}\lbrack f(e^{-x})\rbrack(s).
$ {\cal B}\lbrack f\rbrack(s)={\cal M}\lbrack f(-\log x)\rbrack(s).