x^{μ-1}logx/(e^x+1)^n[0,∞]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^{\infty} \frac{x^{μ-1}\log x}{e^x+1}dx=Γ(μ)\displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^μ}\{ψ(μ)-\log k\}\\
&(2)  \displaystyle\int_0^{\infty} \frac{x^{μ-1}\log x}{(e^x+1)^2}dx=Γ(μ)\displaystyle\sum_{k=2}^{\infty} \frac{(-1)^{k}(k-1)}{k^μ}\{ψ(μ)-\log k\}\\
&(3)  \displaystyle\int_0^{\infty} \frac{x^{μ-1}\log x}{(e^x+1)^n}dx=(-1)^n\frac{Γ(μ)}{(n-1)!}\displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k}(k-1)!}{(n-k)!k^μ}\{ψ(μ)-\log k\}\\
\end{alignat}ただし、全て \(μ \gt 0,\,n \in \mathrm{N}\)










<証明>

\begin{alignat}{2}
(1)  \displaystyle\int_0^{\infty} \frac{x^{μ-1}\log x}{e^x+1}dx&=\displaystyle\int_0^{\infty} \frac{x^{μ-1}e^{-x}\log x}{1+e^{-x}}dx\\
&=\displaystyle\int_0^{\infty} x^{μ-1}e^{-x}\log x\left\{\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1} e^{-(k-1)x}\right\}dx\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1} \displaystyle\int_0^{\infty} x^{μ-1} e^{-kx}\log xdx\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1} \cdot \frac{d}{dμ} \displaystyle\int_0^{\infty} x^{μ-1}e^{-kx}dx\\
\end{alignat}\(kx=t\) と置きます。\((kdx=dt)\)
\begin{alignat}{2}
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1} \cdot \frac{d}{dμ}\displaystyle\int_0^{\infty} \left(\frac{t}{k}\right)^{μ-1} e^{-t} \cdot \frac{1}{k}dt\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1} \cdot \frac{d}{dμ} \cdot \frac{1}{k^μ} \displaystyle\int_0^{\infty} t^{μ-1} e^{-t}dt\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1} \cdot \frac{d}{dμ} \cdot \frac{Γ(μ)}{k^μ}\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1} \cdot \frac{Γ’(μ)\cdot k^μ-Γ(μ)(\log k)k^μ}{k^{2μ}}\\
&=Γ(μ)\displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^μ}\{ψ(μ)-\log k\}
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{x^{μ-1}\log x}{e^x+1}dx=Γ(μ)\displaystyle\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^μ}\{ψ(μ)-\log k\}$$







\begin{alignat}{2}
(2)  \displaystyle\int_0^{\infty} \frac{x^{μ-1}\log x}{(e^x+1)^2}dx&=\displaystyle\int_0^{\infty} \frac{x^{μ-1}e^{-2x}\log x}{(1+e^{-x})^2}dx\\
&=\displaystyle\int_0^{\infty} x^{μ-1}e^{-2x}\log x\left\{\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1} ke^{-(k-1)x}\right\}dx\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}k \displaystyle\int_0^{\infty} x^{μ-1} e^{-(k+1)x}\log xdx\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}k \cdot \frac{d}{dμ} \displaystyle\int_0^{\infty} x^{μ-1}e^{-(k+1)x}dx\\
\end{alignat}\((k+1)x=t\) と置きます。\([(k+1)dx=dt]\)
\begin{alignat}{2}
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}k \cdot \frac{d}{dμ}\displaystyle\int_0^{\infty} \left(\frac{t}{k+1}\right)^{μ-1} e^{-t} \cdot \frac{1}{k+1}dt\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}k \cdot \frac{d}{dμ} \cdot \frac{1}{(k+1)^μ} \displaystyle\int_0^{\infty} t^{μ-1} e^{-t}dt\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}k \cdot \frac{d}{dμ} \cdot \frac{Γ(μ)}{(k+1)^μ}\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}k \cdot \frac{Γ’(μ)\cdot (k+1)^μ-Γ(μ)\{\log (k+1)\}(k+1)^μ}{(k+1)^{2μ}}\\
&=Γ(μ)\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}k \cdot \frac{ψ(μ)-\log (k+1)}{(k+1)^μ}\\
&=Γ(μ)\displaystyle\sum_{k=2}^{\infty} \frac{(-1)^{k}(k-1)}{k^μ}\{ψ(μ)-\log k\}
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{x^{μ-1}\log x}{(e^x+1)^2}dx=Γ(μ)\displaystyle\sum_{k=2}^{\infty} \frac{(-1)^{k}(k-1)}{k^μ}\{ψ(μ)-\log k\}$$








\begin{alignat}{2}
(3)  \displaystyle\int_0^{\infty} \frac{x^{μ-1}\log x}{(e^x+1)^n}dx&=\displaystyle\int_0^{\infty} \frac{x^{μ-1}e^{-nx}\log x}{(1+e^{-x})^n}dx\\
&=\displaystyle\int_0^{\infty} x^{μ-1}e^{-nx}\log x\left\{\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1} {}_{n+k-2}\mathrm{C}_{n-1}e^{-(k-1)x}\right\}dx\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}{}_{n+k-2}\mathrm{C}_{n-1} \displaystyle\int_0^{\infty} x^{μ-1} e^{-(n+k-1)x}\log xdx\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}{}_{n+k-2}\mathrm{C}_{n-1} \cdot \frac{d}{dμ} \displaystyle\int_0^{\infty} x^{μ-1}e^{-(n+k-1)x}dx\\
\end{alignat}\((n+k-1)x=t\) と置きます。\([(n+k-1)dx=dt]\)
\begin{alignat}{2}
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}{}_{n+k-2}\mathrm{C}_{n-1} \cdot \frac{d}{dμ}\displaystyle\int_0^{\infty} \left(\frac{t}{n+k-1}\right)^{μ-1} e^{-t} \cdot \frac{1}{n+k-1}dt\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}{}_{n+k-2}\mathrm{C}_{n-1} \cdot \frac{d}{dμ} \cdot \frac{1}{(n+k-1)^μ} \displaystyle\int_0^{\infty} t^{μ-1} e^{-t}dt\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}{}_{n+k-2}\mathrm{C}_{n-1} \cdot \frac{d}{dμ} \cdot \frac{Γ(μ)}{(n+k-1)^μ}\\
&=\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}{}_{n+k-2}\mathrm{C}_{n-1}\cdot \frac{Γ’(μ)\cdot (n+k-1)^μ-Γ(μ)\{\log (n+k-1)\}(n+k-1)^μ}{(n+k-1)^{2μ}}\\
&=Γ(μ)\displaystyle\sum_{k=1}^{\infty} (-1)^{k-1}{}_{n+k-2}\mathrm{C}_{n-1} \cdot \frac{ψ(μ)-\log (n+k-1)}{(n+k-1)^μ}\\
&=Γ(μ)\displaystyle\sum_{k=n}^{\infty} (-1)^{k-n-2}{}_{k-1}\mathrm{C}_{n-1} \cdot \frac{ψ(μ)-\log k}{k^μ}\\
&=(-1)^nΓ(μ)\displaystyle\sum_{k=n}^{\infty} (-1)^k \cdot \frac{(k-1)!}{(n-1)!(n-k)!}\cdot \frac{ψ(μ)-\log k}{k^μ}\\
&=(-1)^n\frac{Γ(μ)}{(n-1)!}\displaystyle\sum_{k=n}^{\infty} \frac{(-1)^{k}(k-1)!}{(n-k)!k^μ}\{ψ(μ)-\log k\}
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{x^{μ-1}\log x}{(e^x+1)^n}dx=(-1)^n\frac{Γ(μ)}{(n-1)!}\displaystyle\sum_{k=n}^{\infty} \frac{(-1)^{k}(k-1)!}{(n-k)!k^μ}\{ψ(μ)-\log k\}$$

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