#9. ニセベータ関数(?)
便利メモ/数学/懐石メニュー  Share on Twitter

\(\ln z\)の偏角は,\(-\pi <\arg z\leq \pi \)とします。
$$ \begin {aligned} \int _{0}^{1}(1+x)^{p-1}x^{q-1}dx &=\frac {1}{\Gamma (p+q)}\int _{0}^{\infty }x^{q-1}e^x\Gamma (p,2x)dx \end {aligned} $$
$$ I=\int _{0}^{1}(1+x)^{p-1}x^{q-1}dx $$ とおきます。\(x=\sinh^2\varphi \)で置換すると、
$$ \begin {aligned} I&=2\int _{0}^{\sinh ^{-1} (1)}\cosh ^{2p-1}\varphi \sinh ^{2q-1}\varphi d\varphi\\ 2I\int _{0}^{\infty }r^{2(p+q)-1}e^{-r^{2}}dr&=4\int _{0}^{\sinh ^{-1} }\int _{0}^{\infty }(r\cosh \varphi)^{2p-1}(r\sinh \varphi)^{2q-1}e^{-r^2}rdrd\varphi\\ I\Gamma (p+q)&=4\iint _{0\leq y\leq \frac {x}{\sqrt 2}}x^{2p-1}y^{2q-1}e^{-x^2+y^2}dxdy\\ I&=\frac {1}{\Gamma (p+q)}\iint _{0\leq y\leq \frac {x}2}x^{p-1}y^{q-1}e^{-x+y}dxdy\\ &=\frac {1}{\Gamma (p+q)}\int _{0}^{\infty }\int _{0}^{\infty }(s+2t)^{p-1}t^{q-1}e^{-(s+2t)+t}dsdt\\ &=\textcolor{blue}{\frac {1}{\Gamma (p+q)}\int _{0}^{\infty }t^{q-1}e^t\Gamma (p,2t)dt}. \end {aligned} $$

倍角公式(?)

$$ \begin {aligned} I&=\int _{0}^{\frac {\pi }4}\cos ^{-2p+2}x\tan ^{2q-2}x\cdot 2\tan x\cos ^{-2}xdx\\ &=2\int _{0}^{\frac {\pi }4}\cos ^{-2(p+q)+1}x\sin ^{2q-1}xdx\\ 1=p+2qのとき\\ &=2\int _{0}^{\frac {\pi }4}\sin ^{-p}x\cos ^{-p}xdx\\ &=2^{p}\int _{0}^{\frac {\pi }2}\sin ^{-2p}xdx\\ &=2^{p-1}B\left (\frac {1}2,-p+\frac {1}2\right ) \end {aligned} $$ $$ \begin {aligned} \int _{0}^{1}\frac {x^{s-1}}{(1+x)^{2s}}dx &=\frac {1}2\int _{0}^{\infty }\frac {x^{s-1}}{(1+x)^{2s}}dx\\ &=\int _{0}^{\frac {\pi }2}\sin ^{2s-1}\theta \cos ^{2s-1}\theta d\theta \quad (x=\tan ^{2}\theta )\\ &=\frac {1}2B(s,s)\\ &=\frac {\Gamma ^{2}(s)}{2\Gamma (2s)} \end {aligned} $$