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

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









<証明>

次の定積分における等式を被積分関数に代入します。$$\displaystyle\int_0^p x^a da=\left[\frac{x^a}{\log x}\right]_0^p=\frac{x^p-1}{\log x}$$




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







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







\begin{alignat}{2}
&(3) \displaystyle\int_0^1 \frac{(1-x^p)(1-x^q)(1-x^r)x^{s-1}}{\log x}dx\\
&=-\displaystyle\int_0^1 x^{s-1}(x^q-1)(x^r-1)\left(\displaystyle\int_0^p x^a da\right)dx=-\displaystyle\int_0^p \displaystyle\int_0^1 x^{a+s-1}(x^q-1)(x^r-1)dxda\\
&=-\displaystyle\int_0^p \displaystyle\int_0^1 x^{a+s-1}(x^{q+r}-x^q-x^r+1)dxda=-\displaystyle\int_0^p \displaystyle\int_0^1 (x^{a+q+r+s-1}-x^{a+q+s-1}-x^{a+r+s-1}+x^{a+s-1})dxda\\

&=-\displaystyle\int_0^p \left[\frac{x^{a+q+r+s}}{a+q+r+s}-\frac{x^{a+q+s}}{a+q+s}-\frac{x^{a+r+s}}{a+r+s}+\frac{x^{a+s}}{a+s}\right]_0^1 da\\

&=-\displaystyle\int_0^p \left(\frac{1}{a+q+r+s}-\frac{1}{a+q+s}-\frac{1}{a+r+s}+\frac{1}{a+s}\right)da\\
&=-\left[\log \frac{(a+q+r+s)(a+s)}{(a+q+s)(a+r+s)}\right]_0^p\\
&=\log \frac{(p+q+s)(p+r+s)}{(p+q+r+s)(p+s)} \cdot \frac{(q+r+s)s}{(q+s)(r+s)}\\
&=\log \frac{(p+q+s)(p+r+s)(q+r+s)s}{(p+q+r+s)(p+s)(q+s)(r+s)}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{(1-x^p)(1-x^q)(1-x^r)x^{s-1}}{\log x}dx=\log \frac{(p+q+s)(p+r+s)(q+r+s)s}{(p+q+r+s)(p+s)(q+s)(r+s)}$$

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