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










<証明>

次の定積分の等式を被積分関数に代入します。
\begin{alignat}{2}
&(A) \displaystyle\int_0^p x^ada=\left[\frac{x^a}{\log x}\right]_0^p=\frac{x^p-1}{\log x}\\
&(B) \displaystyle\int_q^p x^ada=\left[\frac{x^a}{\log x}\right]_q^p=\frac{x^p-x^q}{\log x}\\
&(C) \displaystyle\int_{q-1}^{p-1} x^ada=\left[\frac{x^a}{\log x}\right]_{q-1}^{p-1}=\frac{x^{p-1}-x^{q-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_0^p x^ada\right)(x^q-1)dx=\displaystyle\int_0^p \displaystyle\int_0^1 x^a(x^q-1)dxda\\
&=\displaystyle\int_0^p \displaystyle\int_0^1 (x^{a+q}-x^a)dxda=\displaystyle\int_0^p \left[\frac{x^{a+q+1}}{a+q+1}-\frac{x^{a+1}}{a+1}\right]_0^1da\\
&=\displaystyle\int_0^p \left(\frac{1}{a+q+1}-\frac{1}{a+1}\right)da=\left[\log (a+q+1)-\log (a+1)\right]_0^p\\
&=\left[\log \frac{a+q+1}{a+1}\right]_0^p=\log \frac{p+q+1}{p+1} \cdot \frac{1}{q+1}=\log \frac{p+q+1}{(p+1)(q+1)}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{(x^p-1)(x^q-1)}{\log x}dx=\log \frac{p+q+1}{(p+1)(q+1)}$$







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







\begin{alignat}{2}
&(3) \displaystyle\int_0^1 \frac{x^{p-1}-x^{q-1}}{(1+x^r)\log x}dx\\
&=\displaystyle\int_0^1 \left(\displaystyle\int_{q-1}^{p-1} x^ada\right)\frac{1}{1+x^r}dx=\displaystyle\int_{q-1}^{p-1} \displaystyle\int_0^1 \frac{x^a}{1+x^r}dxda\\
&=\displaystyle\int_{q-1}^{p-1} \displaystyle\int_0^1 \frac{(1-x^r)x^a}{(1-x^r)(1+x^r)}dxda=\displaystyle\int_{q-1}^{p-1} \displaystyle\int_0^1 \frac{x^a-x^{a+r}}{1-x^{2r}}dxda\\
\end{alignat}\(x^{2r}=t\) と置きます。\((2rx^{2r-1}dx=dt)\)
\begin{alignat}{2}
&=\displaystyle\int_{q-1}^{p-1} \displaystyle\int_0^1 \frac{t^{\frac{a}{2r}}-t^{\frac{a+r}{2r}}}{1-t} \cdot \frac{1}{2rt^{\frac{2r-1}{2r}}}dtda\\
&=\frac{1}{2r}\displaystyle\int_{q-1}^{p-1} \displaystyle\int_0^1 \frac{t^{\frac{a+1}{2r}-1}-t^{\frac{a+r+1}{2r}-1}}{1-t}dtda\\
&=\frac{1}{2r}\displaystyle\int_{q-1}^{p-1} \left\{ψ\left(\frac{a+r+1}{2r}\right)-ψ\left(\frac{a+1}{2r}\right)\right\}da\\
&=\left[\log Γ\left(\frac{a+r+1}{2r}\right)-\log Γ\left(\frac{a+1}{2r}\right)\right]_{q-1}^{p-1}\\
&=\left[\log \frac{Γ\left(\frac{a+r+1}{2r}\right)}{Γ\left(\frac{a+1}{2r}\right)}\right]_{q-1}^{p-1}=\log \frac{Γ\left(\frac{p+r}{2r}\right)}{Γ\left(\frac{p}{2r}\right)} \cdot \frac{Γ\left(\frac{q}{2r}\right)}{Γ\left(\frac{q+r}{2r}\right)}=\log \frac{Γ\left(\frac{p+r}{2r}\right)Γ\left(\frac{q}{2r}\right)}{Γ\left(\frac{q+r}{2r}\right)Γ\left(\frac{p}{2r}\right)}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{x^{p-1}-x^{q-1}}{(1+x^r)\log x}dx=\log \frac{Γ\left(\frac{p+r}{2r}\right)Γ\left(\frac{q}{2r}\right)}{Γ\left(\frac{q+r}{2r}\right)Γ\left(\frac{p}{2r}\right)}$$



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