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