梅林变换

之所以伽马函数与积分变换的理论联系密切，是因为伽马函数同时是指数函数的拉普拉斯变换和幂函数的梅林变换，这也展示了两种积分变换之间的联系。

双边拉普拉斯变换

双边拉普拉斯变换可以用梅林变换来表示，如下式

${\displaystyle \left\{{\mathcal {B}}f\right\}(s)=\left\{{\mathcal {M}}f(-\ln x)\right\}(s)}$ 

梅林变换也可以用双边拉普拉斯变换来表示，如下式

${\displaystyle \left\{{\mathcal {M}}f\right\}(s)=\left\{{\mathcal {B}}f(e^{-x})\right\}(s)}$ 

傅立叶变换

傅立叶变换可以用梅林变换来表示，如下式

${\displaystyle \left\{{\mathcal {F}}f\right\}(-s)=\left\{{\mathcal {B}}f\right\}(-is)=\left\{{\mathcal {M}}f(-\ln x)\right\}(-is)\ }$ 

梅林变换变换也可以用傅立叶来表示，如下式

${\displaystyle \left\{{\mathcal {M}}f\right\}(s)=\left\{{\mathcal {B}}f(e^{-x})\right\}(s)=\left\{{\mathcal {F}}f(e^{-x})\right\}(-is)\ }$ 

Cahen–Mellin 积分

对于 ${\displaystyle c>0}$ ，${\displaystyle \Re (y)>0}$ ，且 ${\displaystyle y^{-s}}$ 在主要分支（principal branch）上，我们有

${\displaystyle e^{-y}={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }\Gamma (s)y^{-s}\;ds}$ 

其中 ${\displaystyle \Gamma (s)}$ 为 Γ函数。

数论

假设

${\displaystyle \Re (s+a)<0}$ 

我们有

${\displaystyle f(x)={\begin{cases}0&x<1\\x^{a}&x>1\end{cases}}}$ 

其中

${\displaystyle {\mathcal {M}}f(s)=-{\frac {1}{s+a}}}$ 

在任何维度的圆柱坐标系中，拉普拉斯算子总是会包含下式

${\displaystyle {\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)}$ 

例如，拉普拉斯算子在二维空间的极坐标表示法

${\displaystyle \nabla ^{2}f={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \theta ^{2}}}}$ 

或是在三维空间的柱坐标表示法

${\displaystyle \nabla ^{2}f={\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \varphi ^{2}}}+{\frac {\partial ^{2}f}{\partial z^{2}}}}$ 

而利用梅林变换可以很简单的处理此项

${\displaystyle {\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)=f_{rr}+{\frac {f_{r}}{r}}}$ 

${\displaystyle {\mathcal {M}}\left(r^{2}f_{rr}+rf_{r},r\to s\right)=s^{2}{\mathcal {M}}\left(f,r\to s\right)=s^{2}F}$ 

举例来说，二维拉普拉斯方程的极坐标表示法具有以下形式

${\displaystyle r^{2}f_{rr}+rf_{r}+f_{\theta \theta }=0}$ 

或是

${\displaystyle {\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \theta ^{2}}}=0}$ 

利用梅林变换，可以转换成一个简谐振子的形式

${\displaystyle F_{\theta \theta }+s^{2}F=0}$ 

通解为

${\displaystyle F(s,\theta )=C_{1}(s)\cos(s\theta )+C_{2}(s)\sin(s\theta )}$ 

若给定边界条件

${\displaystyle f(r,-\theta _{0})=a(r),\quad f(r,\theta _{0})=b(r)}$ 

其梅林变换为

${\displaystyle F(s,-\theta _{0})=A(s),\quad F(s,\theta _{0})=B(s)}$ 

则通解可以写成

${\displaystyle F(s,\theta )=A(s){\frac {\sin(s(\theta _{0}-\theta ))}{\sin(2\theta _{0}s)}}+B(s){\frac {\sin(s(\theta _{0}+\theta ))}{\sin(2\theta _{0}s)}}}$ 

最后利用逆变换以及卷积定理

${\displaystyle {\mathcal {M}}^{-1}\left({\frac {\sin(s\varphi )}{\sin(2\theta _{0}s)}};s\to r\right)={\frac {1}{2\theta _{0}}}{\frac {r^{m}\sin(m\varphi )}{1+2r^{m}\cos(m\varphi )+r^{2m}}}}$ 

其中

${\displaystyle m={\frac {\pi }{2\theta _{0}}}}$ 

可以得到

${\displaystyle f(r,\theta )={\frac {r^{m}\cos(m\theta )}{2\theta _{0}}}\int _{0}^{\infty }\left\{{\frac {a(x)}{x^{2m}+2r^{m}x^{m}\sin(m\theta )+r^{2m}}}+{\frac {b(x)}{x^{2m}-2r^{m}x^{m}\sin(m\theta )+r^{2m}}}\right\}x^{m-1}\,dx}$ 

• Galambos, Janos; Simonelli, Italo. Products of random variables: applications to problems of physics and to arithmetical functions. Marcel Dekker, Inc. 2004. ISBN 0-8247-5402-6.