# 分数微积分

$\, f^2 (x) = f(f(x))$

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

$D^n$

## 试探法

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

$(P ^ a f)(x) = f'(x)$

$n! = \Gamma(n+1)$

$(J f )( x ) = \int_0^x f (t) \; dt$

$(J^2 f)(x) = \int_0^x ( J f )(t ) dt = \int_0^x \left( \int_0^t f(s) \; ds \right) \; dt$,

$(J^{n} f)(x) = \frac{1}{(n-1)!}\int_0^x (x-t)^{n-1}f (t) \; dt$

$(J^{\alpha}f)(x)=\frac{1}{\Gamma(\alpha)}\int_0^x (x-t)^{\alpha-1}f (t)\; dt$

$(J^{\alpha})(J^{\beta})f=(J^{\beta})(J^{\alpha})f=(J^{\alpha+\beta})f=\frac{1}{\Gamma(\alpha+\beta)}\int_0^x (x-t)^{\alpha+\beta-1}f (t) \; dt$

## 分数微分在一个简单函数上的应用

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

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

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

## 应用

WKB近似

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

## 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）