{(1-x^p)/(1-x)-p}1/logx[0,1]などの定積分

\begin{alignat}{2}
&(1) \displaystyle\int_0^1 \frac{x^{p-1}-x^{q-1}}{\log x}dx=\log \frac{p}{q}\\
&(2) \displaystyle\int_0^1 \frac{x^{p-1}-x^{q-1}}{(1+x)\log x}dx=\log \frac{Γ\left(\frac{q}{2}\right)Γ\left(\frac{p+1}{2}\right)}{Γ\left(\frac{p}{2}\right)Γ\left(\frac{q+1}{2}\right)}\\
&(3) \displaystyle\int_0^1 \left(\frac{1-x^p}{1-x}-p\right)\frac{1}{\log x}dx=\log Γ(p+1)
\end{alignat}ただし、全て \(p,q \gt 0\)








<証明>

次の定積分における等式を用います。
\begin{alignat}{2}
&(A) \displaystyle\int_q^p x^{a-1}da=\left[\frac{x^{a-1}}{\log x}\right]_q^p=\frac{x^{p-1}-x^{q-1}}{\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}





\begin{alignat}{2}
&(1) \displaystyle\int_0^1 \frac{x^{p-1}-x^{q-1}}{\log x}dx\\
&=\displaystyle\int_0^1 \left(\displaystyle\int_q^p x^{a-1}da\right)dx=\displaystyle\int_q^p \displaystyle\int_0^1 x^{a-1}dxda\\
&=\displaystyle\int_q^p \left[\frac{x^a}{a}\right]_0^1 da=\displaystyle\int_q^p \frac{1}{a}da=[\log a]_q^p=\log \frac{p}{q}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{x^{p-1}-x^{q-1}}{\log x}dx=\log \frac{p}{q}$$







\begin{alignat}{2}
&(2) \displaystyle\int_0^1 \frac{x^{p-1}-x^{q-1}}{(1+x)\log x}dx\\
&=\displaystyle\int_0^1 \frac{1}{1+x}\left(\displaystyle\int_q^p x^{a-1}da\right)dx=\displaystyle\int_q^p \displaystyle\int_0^1 \frac{x^{a-1}}{1+x}dxda\\
&=\displaystyle\int_q^p \displaystyle\int_0^1 \frac{(1-x)x^{a-1}}{(1-x)(1+x)}dxda=\displaystyle\int_q^p \displaystyle\int_0^1 \frac{x^{a-1}-x^a}{1-x^2}dxda\\
\end{alignat}\(x^2=t\) と置きます。\((2xdx=dt)\)
\begin{alignat}{2}
&=\displaystyle\int_q^p \displaystyle\int_0^1 \frac{t^{\frac{a-1}{2}}-t^{\frac{a}{2}}}{1-t} \cdot \frac{1}{2\sqrt{t}}dtda=\frac{1}{2}\displaystyle\int_q^p \displaystyle\int_0^1 \frac{t^{\frac{a}{2}-1}-t^{\frac{a-1}{2}}}{1-t}dtda\\
&=\frac{1}{2}\displaystyle\int_q^p \left\{ψ\left(\frac{a+1}{2}\right)-ψ\left(\frac{a}{2}\right)\right\}\\
&=\left[\log Γ\left(\frac{a+1}{2}\right)-\log Γ\left(\frac{a}{2}\right)\right]_q^p=\log \frac{Γ\left(\frac{p+1}{2}\right)Γ\left(\frac{q}{2}\right)}{Γ\left(\frac{p}{2}\right)Γ\left(\frac{q+1}{2}\right)}\\
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{x^{p-1}-x^{q-1}}{(1+x)\log x}dx=\log \frac{Γ\left(\frac{q}{2}\right)Γ\left(\frac{p+1}{2}\right)}{Γ\left(\frac{p}{2}\right)Γ\left(\frac{q+1}{2}\right)}$$







\((3)\) 定積分を次のように分けます。$$\displaystyle\int_0^1 \left(\frac{1-x^p}{1-x}-p\right)\frac{1}{\log x}dx=\displaystyle\int_0^1 \frac{1-x^p}{(1-x)\log x}dx-p\displaystyle\int_0^1 \frac{1}{\log x}dx$$左の積分について
\begin{alignat}{2}
&\displaystyle\int_0^1 \frac{1-x^p}{(1-x)\log x}dx=\displaystyle\int_0^1 \frac{1}{x-1}\left(\displaystyle\int_0^p x^ada\right)dx=\displaystyle\int_0^p \displaystyle\int_0^1 \frac{x^a}{x-1}dxda\\
&                =-\displaystyle\int_0^p \displaystyle\int_0^1 \frac{x^a}{1-x}dxda=-\displaystyle\int_0^p \displaystyle\int_0^1 x^a \displaystyle\sum_{n=0}^{\infty} x^n dxda\\
&                =-\displaystyle\sum_{n=0}^{\infty} \displaystyle\int_0^p \displaystyle\int_0^1 x^{a+n}dxda=-\displaystyle\sum_{n=0}^{\infty} \displaystyle\int_0^p \left[\frac{x^{a+n+1}}{a+n+1}\right]_0^1da\\
&                =-\displaystyle\sum_{n=0}^{\infty}\displaystyle\int_0^p \frac{1}{a+n+1}da=-\displaystyle\sum_{n=0}^{\infty}[\log (a+n+1)]_0^p\\
&                =-\displaystyle\sum_{n=0}^{\infty} \log \frac{p+n+1}{n+1}=\displaystyle\sum_{n=0}^{\infty} \log \frac{n+1}{p+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{1-x^p}{1-x}-p\right)\frac{1}{\log x}dx=\displaystyle\int_0^1 \frac{1-x^p}{(1-x)\log x}dx-p\displaystyle\int_0^1 \frac{1}{\log x}dx\\
&                      =\displaystyle\sum_{n=0}^{\infty} \log \frac{n+1}{p+n+1}+p\displaystyle\sum_{n=0}^{\infty} \log \frac{n+2}{n+1}\\
&                      =\displaystyle\sum_{n=0}^{\infty} \log \frac{n+1}{p+n+1} \left(\frac{n+2}{n+1}\right)^p=\displaystyle\sum_{n=1}^{\infty} \log \frac{n}{p+n} \left(\frac{n+1}{n}\right)^p\\
&                      =\displaystyle\sum_{n=1}^{\infty} \log \left(1+\frac{1}{n}\right)^p \left(1+\frac{p}{n}\right)^{-1}=\log \displaystyle\prod_{n=1}^{\infty}\left(1+\frac{1}{n}\right)^p \left(1+\frac{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$$Γ(z)=\frac{1}{z}\displaystyle\prod_{n=1}^{\infty}\left\{\left(1+\frac{1}{n}\right)^z\left(1+\frac{z}{n}\right)^{-1}\right\}$$\(z=p\) として両辺を \(p\) 倍すれば$$pΓ(p)=\displaystyle\prod_{n=1}^{\infty}\left\{\left(1+\frac{1}{n}\right)^p\left(1+\frac{p}{n}\right)^{-1}\right\}$$となるので$$\displaystyle\int_0^1 \left(\frac{1-x^p}{1-x}-p\right)\frac{1}{\log x}dx=\log \displaystyle\prod_{n=1}^{\infty}\left(1+\frac{1}{n}\right)^p \left(1+\frac{p}{n}\right)^{-1}=\log pΓ(p)=\log Γ(p+1)$$以上より$$\displaystyle\int_0^1 \left(\frac{1-x^p}{1-x}-p\right)\frac{1}{\log x}dx=\log Γ(p+1)$$

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