# Hankel变换

$F_{\nu}(k)=\int_{0}^{\infty}f(r)J_{\nu}(kr)rdr$

$f(r)=\int_{0}^{\infty}F_{\nu}(k)J_{\nu}(kr)kdk$

## 正交性

$\int_0^\infty J_\nu(kr)J_\nu(k'r)r~\operatorname{d}r = \frac{\delta (k-k')}{k}$

## 与其他函数变换的关系

### 傅立叶变换

$F(\boldsymbol{k})=\iint f(\boldsymbol{r})e^{i\boldsymbol{k}\cdot\boldsymbol{r}}d\boldsymbol{r}$

$F(\boldsymbol{k})=\int_{0}^{\infty}\int_{0}^{2\pi}f(r,\theta)e^{ikr\cos\theta}rdrd\theta$

$F(\boldsymbol{k})=F(k)=2\pi\int_{0}^{\infty}f(r)J_{0}(kr)rdr$

## 常见汉克尔变换函数对

$f(r)\,$ $F_0(k)\,$
$1\,$ $\delta(k)/k\,$
$1/r\,$ $1/k\,$
$r\,$ $-1/k^3\,$
$r^3\,$ $9/k^5\,$
$r^{m}\,$ $\frac{2^{m+1}\Gamma(m/2+1)}{k^{m+2}\Gamma(-m/2)}\,$ for -2<Re(m)<-1/2
$\frac{1}{\sqrt{r^2+z^2}}\,$ $\frac{e^{-k|z|}}{k}=\sqrt{\frac{2|z|}{\pi k}}K_{-1/2}(k|z|)\,$
$\frac{1}{r^2+z^2}\,$ $K_0(kz)\,$, $z$可为复数
$e^{iar}/r\,$ $i/\sqrt{ a^2 - k^2} \quad (a>0, k
$\,$ $1/\sqrt{ k^2 - a^2} \quad (a>0, k>a) \,$
$e^{-a^2r^2/2}\,$ $\frac{e^{-k^2/2a^2}}{a^2}$
$-r^2 f(r)\,$ $\frac{\operatorname{d}^2 F_0}{\operatorname{d}k^2}+\frac{1}{k}\frac{\operatorname{d}F_0}{\operatorname{d}k}$
$f(r)\,$ $F_{\nu}(k)\,$
$r^s\,$ $\frac{\Gamma\left(\frac 1 2 (2+\nu+s)\right)}{\Gamma(\tfrac 1 2 (\nu-s))} \frac{2^{s+1}}{k^{s+2}} \,$
$r^{\nu-2s}\Gamma\left(s,r^2 h\right)\,$ $\frac12 \left(\frac k 2\right)^{2s-\nu-2}\gamma\left(1-s+\nu,\frac{k^2}{4h}\right)\,$
$e^{-r^2}r^\nu U\left(a,b,r^2\right)\,$ $\frac{\Gamma(2+\nu-b)}{2\Gamma(2+\nu-b+a)}\left(\frac k 2\right)^\nu e^{-\frac{k^2}4}\,_1F_1\left(a,2+a-b+\nu,\frac{k^2}4\right)$
$-r^2 f(r)\,$ $\frac{\operatorname{d}^2 F_\nu}{\operatorname{d}k^2}+\frac{1}{k}\frac{\operatorname{d}F_\nu}{\operatorname{d}k}-\frac{\nu^2}{k^2}F_\nu$