{(x^p-x^{p+q})/(1-x)-q}1/logx[0,1]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^1 \left(\frac{x^p-x^{p+q}}{1-x}-q\right)\frac{1}{\log x}dx=\log \frac{Γ(p+q+1)}{Γ(p+1)}\\
&(2) \displaystyle\int_0^1 \left\{\frac{(1-x^p)(1-x^q)}{x(1-x)}-\frac{1-x}{x}\right\}\frac{1}{\log x}dx=\log B(p,q)\\
\end{alignat}ただし、全て \(p,q \gt 0\)








<証明>


\((1)\) 積分を切り離します。$$\displaystyle\int_0^1 \left(\frac{x^p-x^{p+q}}{1-x}-q\right)\frac{1}{\log x}dx=\displaystyle\int_0^1 \frac{x^p-x^{p+q}}{(1-x)\log x}dx-q \displaystyle\int_0^1 \frac{1}{\log x}dx$$

次の定積分における等式を代入します。
\begin{alignat}{2}
&(A) \displaystyle\int_p^{p+q} x^ada=\left[\frac{x^a}{\log x}\right]_p^{p+q}=\frac{x^{p+q}-x^p}{\log x}\\
&(B) \displaystyle\int_0^1 x^ada=\left[\frac{x^a}{\log x}\right]_0^1=\frac{x-1}{\log x}\\
&(C) \displaystyle\int_0^p x^ada=\left[\frac{x^a}{\log x}\right]_0^p=\frac{x^p-1}{\log x}\\
\end{alignat}

左の積分について
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{x^p-x^{p+q}}{(1-x)\log x}dx=-\displaystyle\int_0^1 \frac{1}{1-x}\left(\displaystyle\int_p^{p+q} x^ada\right)dx=-\displaystyle\int_p^{p+q} \displaystyle\int_0^1 \frac{x^a}{1-x}dxda\\
&                =-\displaystyle\int_p^{p+q} \displaystyle\int_0^1 \frac{x^a}{1-x}dxda=-\displaystyle\int_p^{p+q} \displaystyle\int_0^1 x^a \displaystyle\sum_{n=0}^{\infty} x^n dxda\\
&                =-\displaystyle\sum_{n=0}^{\infty} \displaystyle\int_p^{p+q} \displaystyle\int_0^1 x^{a+n}dxda=-\displaystyle\sum_{n=0}^{\infty} \displaystyle\int_p^{p+q} \left[\frac{x^{a+n+1}}{a+n+1}\right]_0^1da\\
&                =-\displaystyle\sum_{n=0}^{\infty}\displaystyle\int_p^{p+q} \frac{1}{a+n+1}da=-\displaystyle\sum_{n=0}^{\infty}[\log (a+n+1)]_p^{p+q}\\
&                =-\displaystyle\sum_{n=0}^{\infty} \log \frac{p+q+n+1}{p+n+1}=\displaystyle\sum_{n=0}^{\infty} \log \frac{p+n+1}{p+q+n+1}\\
\end{alignat}
右の積分について
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{1}{\log x}dx=\displaystyle\int_0^1 \frac{1}{x-1}\left(\displaystyle\int_0^1 x^ada\right)dx=\displaystyle\int_0^1 \displaystyle\int_0^1 \frac{x^a}{x-1}dxda\\
&          =-\displaystyle\int_0^1 \displaystyle\int_0^1 \frac{x^a}{1-x}dxda=-\displaystyle\int_0^1 \displaystyle\int_0^1 x^a \displaystyle\sum_{n=0}^{\infty} x^n dxda\\
&          =-\displaystyle\sum_{n=0}^{\infty} \displaystyle\int_0^1 \displaystyle\int_0^1 x^{a+n}dxda=-\displaystyle\sum_{n=0}^{\infty} \displaystyle\int_0^1 \left[\frac{x^{a+n+1}}{a+n+1}\right]_0^1da\\
&          =-\displaystyle\sum_{n=0}^{\infty}\displaystyle\int_0^1 \frac{1}{a+n+1}da=-\displaystyle\sum_{n=0}^{\infty}[\log (a+n+1)]_0^1=-\displaystyle\sum_{n=0}^{\infty} \log \frac{n+2}{n+1}\\
\end{alignat}よって、元の積分は
\begin{alignat}{2}
&  \displaystyle\int_0^1 \left(\frac{x^p-x^{p+q}}{1-x}-q\right)\frac{1}{\log x}dx=\displaystyle\int_0^1 \frac{x^p-x^{p+q}}{(1-x)\log x}dx-q \displaystyle\int_0^1 \frac{1}{\log x}dx\\
&=\displaystyle\sum_{n=0}^{\infty} \log \frac{p+n+1}{p+q+n+1}+q\displaystyle\sum_{n=0}^{\infty} \log \frac{n+2}{n+1}\\
&=\displaystyle\sum_{n=0}^{\infty} \log \frac{p+n+1}{p+q+n+1}\left(\frac{n+2}{n+1}\right)^q=\displaystyle\sum_{n=1}^{\infty} \log \frac{p+n}{p+q+n}\left(\frac{n+1}{n}\right)^q\\
&=\displaystyle\sum_{n=1}^{\infty} \log \left(1+\frac{1}{n}\right)^q \left(1+\frac{q}{p+n}\right)^{-1}=\log \displaystyle\prod_{n=1}^{\infty} \left(1+\frac{1}{n}\right)^q \left(1+\frac{q}{p+n}\right)^{-1}\\
\end{alignat}
ここで次のガンマ関数のオイラーの表示式より$$Γ(z+1)=\displaystyle\prod_{n=1}^{\infty}\left\{\left(1+\frac{1}{n}\right)^z\left(1+\frac{z}{n}\right)^{-1}\right\}$$これが \(z=p+q\) と \(z=p\) のとき
\begin{alignat}{2}
&Γ(p+q+1)=\displaystyle\prod_{n=1}^{\infty}\left\{\left(1+\frac{1}{n}\right)^{p+q}\left(1+\frac{p+q}{n}\right)^{-1}\right\}\\
&\\
&Γ(p+1)=\displaystyle\prod_{n=1}^{\infty}\left\{\left(1+\frac{1}{n}\right)^p\left(1+\frac{p}{n}\right)^{-1}\right\}\\
\end{alignat}この \(2\) つの式を割ると$$\frac{Γ(p+q+1)}{Γ(p+1)}=\displaystyle\prod_{n=1}^{\infty} \left(1+\frac{1}{n}\right)^q \left(1+\frac{q}{p+n}\right)^{-1}$$となるので、以上より$$\displaystyle\int_0^1 \left(\frac{x^p-x^{p+q}}{1-x}-q\right)\frac{1}{\log x}dx=\log \frac{Γ(p+q+1)}{Γ(p+1)}$$








\((2)\) 積分を切り離します。
\begin{alignat}{2}
&  \displaystyle\int_0^1 \left\{\frac{(1-x^p)(1-x^q)}{x(1-x)}-\frac{1-x}{x}\right\}\frac{1}{\log x}dx\\
&=\displaystyle\int_0^1 \frac{(1-x^p)(1-x^q)}{x(1-x)\log x}dx-\displaystyle\int_0^1 \frac{1-x}{x \log x}dx\\
\end{alignat}
途中、極限を用いて式を表し、後でまとめて計算します。

左の積分について
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{(1-x^p)(1-x^q)}{x(1-x)\log x}dx=\displaystyle\int_0^1 \frac{x^q-1}{x(1-x)}\left(\displaystyle\int_0^p x^ada\right)dx=\displaystyle\int_0^p \displaystyle\int_0^1 \frac{x^{a-1}(x^q-1)}{1-x}dxda\\
&                    =\displaystyle\int_0^p \displaystyle\int_0^1 \frac{x^{a+q-1}-x^{a-1}}{1-x}dxda=\displaystyle\int_0^p \{ψ(a)-ψ(a+q)\}da\\
&                    =\displaystyle\lim_{s \to 0} \displaystyle\int_s^p \{ψ(a)-ψ(a+q)\}da=\displaystyle\lim_{s \to 0} [\log Γ(a)-\log Γ(a+q)]_s^p\\
&                    =\displaystyle\lim_{s \to 0} \log \frac{Γ(p)Γ(s+q)}{Γ(p+q)Γ(s)}\\
\end{alignat}

右の積分について
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{1-x}{x \log x}dx=-\displaystyle\int_0^1 \frac{1}{x}\left(\displaystyle\int_0^1 x^ada\right)dx=-\displaystyle\int_0^1 \displaystyle\int_0^1 x^{a-1}dxda\\
&           =-\displaystyle\int_0^1 \left[\frac{x^a}{a}\right]_0^1 da=-\displaystyle\int_0^1 \frac{1}{a}da=-\displaystyle\lim_{s \to 0} \displaystyle\int_s^1 \frac{1}{a}da\\
&           =-\displaystyle\lim_{s \to 0} [\log a]_s^1=\displaystyle\lim_{s \to 0} \log s\\
\end{alignat}よって
\begin{alignat}{2}
&  \displaystyle\int_0^1 \left\{\frac{(1-x^p)(1-x^q)}{x(1-x)}-\frac{1-x}{x}\right\}\frac{1}{\log x}dx\\
&=\displaystyle\int_0^1 \frac{(1-x^p)(1-x^q)}{x(1-x)\log x}dx-\displaystyle\int_0^1 \frac{1-x}{x \log x}dx\\
&=\displaystyle\lim_{s \to 0}\left\{\log \frac{Γ(p)Γ(s+q)}{Γ(p+q)Γ(s)}- \log s\right\}=\displaystyle\lim_{s \to 0} \log \frac{Γ(p)Γ(s+q)}{Γ(p+q) \cdot sΓ(s)}\\
&=\displaystyle\lim_{s \to 0} \log \frac{Γ(p)Γ(s+q)}{Γ(p+q) \cdot Γ(s+1)}=\log \frac{Γ(p)Γ(q)}{Γ(p+q)}=\log B(p,q)\\
\end{alignat}以上より$$\displaystyle\int_0^1 \left\{\frac{(1-x^p)(1-x^q)}{x(1-x)}-\frac{1-x}{x}\right\}\frac{1}{\log x}dx=\log B(p,q)$$

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