# 分数微积分

${\displaystyle \,f^{2}(x)=f(f(x))}$

${\displaystyle {\sqrt {D}}=D^{\frac {1}{2}}}$

${\displaystyle D^{n}}$

## 试探法

${\displaystyle H^{2}f(x)=Df(x)={\frac {d}{dx}}f(x)=f'(x)}$

${\displaystyle (P^{a}f)(x)=f'(x)}$

${\displaystyle n!=\Gamma (n+1)}$

${\displaystyle (Jf)(x)=\int _{0}^{x}f(t)\;dt}$

${\displaystyle (J^{2}f)(x)=\int _{0}^{x}(Jf)(t)dt=\int _{0}^{x}\left(\int _{0}^{t}f(s)\;ds\right)\;dt}$,

${\displaystyle (J^{n}f)(x)={\frac {1}{(n-1)!}}\int _{0}^{x}(x-t)^{n-1}f(t)\;dt}$

${\displaystyle (J^{\alpha }f)(x)={\frac {1}{\Gamma (\alpha )}}\int _{0}^{x}(x-t)^{\alpha -1}f(t)\;dt}$

${\displaystyle (J^{\alpha })(J^{\beta })f=(J^{\beta })(J^{\alpha })f=(J^{\alpha +\beta })f={\frac {1}{\Gamma (\alpha +\beta )}}\int _{0}^{x}(x-t)^{\alpha +\beta -1}f(t)\;dt}$

## 分数微分在一个简单函数上的应用

${\displaystyle f(x)=x^{k}\;}$。它的一阶导数一般是：
${\displaystyle f'(x)={\dfrac {d}{dx}}f(x)=kx^{k-1}\;}$。重复这一过程，得到更一般的结果：
${\displaystyle {\dfrac {d^{a}}{dx^{a}}}x^{k}={\dfrac {k!}{(k-a)!}}x^{k-a}\;}$，将阶乘伽玛函数替换，可得：
${\displaystyle {\dfrac {d^{a}}{dx^{a}}}x^{k}={\dfrac {\Gamma (k+1)}{\Gamma (k-a+1)}}x^{k-a}\;}$。当k = 1,并且a = 1/2时我们可以得到函数${\displaystyle x}$的半导数：
${\displaystyle {\dfrac {d^{\frac {1}{2}}}{dx^{\frac {1}{2}}}}x={\dfrac {\Gamma (1+1)}{\Gamma (1-{\frac {1}{2}}+1)}}x^{1-{\frac {1}{2}}}={\dfrac {1!}{\Gamma ({\frac {3}{2}})}}x^{\frac {1}{2}}={\dfrac {2x^{\frac {1}{2}}}{\sqrt {\pi }}}}$。重复这一过程，得：
${\displaystyle {\dfrac {d^{\frac {1}{2}}}{dx^{\frac {1}{2}}}}2\pi ^{-{\frac {1}{2}}}x^{\frac {1}{2}}=2\pi ^{-{\frac {1}{2}}}{\dfrac {\Gamma (1+{\frac {1}{2}})}{\Gamma ({\frac {1}{2}}-{\frac {1}{2}}+1)}}x^{{\frac {1}{2}}-{\frac {1}{2}}}=2\pi ^{-{\frac {1}{2}}}{\dfrac {\Gamma ({\frac {3}{2}})}{\Gamma (1)}}x^{0}={\dfrac {2{\sqrt {\pi }}x^{0}}{2{\sqrt {\pi }}0!}}=1}$，这正是期望的结果：
${\displaystyle \left({\dfrac {d^{\frac {1}{2}}}{dx^{\frac {1}{2}}}}{\dfrac {d^{\frac {1}{2}}}{dx^{\frac {1}{2}}}}\right)x={\dfrac {d}{dx}}x=1}$

${\displaystyle D^{\alpha }f(x)={\frac {1}{\Gamma (1-\alpha )}}{\frac {d}{dx}}\int _{0}^{x}{\frac {f(t)}{(x-t)^{\alpha }}}dt}$

${\displaystyle D^{\frac {3}{2}}f(x)=D^{\frac {1}{2}}D^{1}f(x)=D^{\frac {1}{2}}{\frac {d}{dx}}f(x)}$

## 拉普拉斯變換

${\displaystyle {\mathcal {L}}\left\{Jf\right\}(s)={\mathcal {L}}\left\{\int _{0}^{t}f(\tau )\,d\tau \right\}(s)={\frac {1}{s}}{\bigl (}{\mathcal {L}}\left\{f\right\}{\bigr )}(s)}$

${\displaystyle {\mathcal {L}}\left\{J^{2}f\right\}={\frac {1}{s}}{\bigl (}{\mathcal {L}}\left\{Jf\right\}{\bigr )}(s)={\frac {1}{s^{2}}}{\bigl (}{\mathcal {L}}\left\{f\right\}{\bigr )}(s)}$

${\displaystyle J^{\alpha }f={\mathcal {L}}^{-1}\left\{s^{-\alpha }{\bigl (}{\mathcal {L}}\{f\}{\bigr )}(s)\right\}}$

${\displaystyle J^{\alpha }(t^{k})={\mathcal {L}}^{-1}\left\{{\frac {\Gamma (k+1)}{s^{\alpha +k+1}}}\right\}={\frac {\Gamma (k+1)}{\Gamma (\alpha +k+1)}}t^{\alpha +k}}$

${\displaystyle {\mathcal {L}}\{f*g\}={\bigl (}{\mathcal {L}}\{f\}{\bigr )}{\bigl (}{\mathcal {L}}\{g\}{\bigr )}}$

{\displaystyle {\begin{aligned}\left(J^{\alpha }f\right)(t)&={\frac {1}{\Gamma (\alpha )}}{\mathcal {L}}^{-1}\left\{{\bigl (}{\mathcal {L}}\{p\}{\bigr )}{\bigl (}{\mathcal {L}}\{f\}{\bigr )}\right\}\\&={\frac {1}{\Gamma (\alpha )}}(p*f)\\&={\frac {1}{\Gamma (\alpha )}}\int _{0}^{t}p(t-\tau )f(\tau )\,d\tau \\&={\frac {1}{\Gamma (\alpha )}}\int _{0}^{t}\left(t-\tau \right)^{\alpha -1}f(\tau )\,d\tau \\\end{aligned}}}

## 应用

WKB近似

${\displaystyle V^{-1}(x)=2{\sqrt {\pi }}{\frac {d^{\frac {1}{2}}}{dx^{\frac {1}{2}}}}n(x)}$

## 参考文献

1. ^ Fractional Calculus. An Introduction for Physicists, by Richard Herrmann. Hardcover. Publisher: World Scientific, Singapore;（2014）ISBN 978-981-4551-09-0http://www.worldscientific.com/worldscibooks/10.1142/8934）页面存档备份，存于互联网档案馆