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

\begin{alignat}{2}
&(1) \displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^n(1+x^q)^mx^{p-1}dx=n! \displaystyle\sum_{k=0}^m \frac{{}_m \mathrm{C}_k}{(p+kq)^{n+1}}\\
&(2) \displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^n(1-x^q)^mx^{p-1}dx=n! \displaystyle\sum_{k=0}^m \frac{(-1)^k{}_m \mathrm{C}_k}{(p+kq)^{n+1}}\\
&(3) \displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^{r-1}\frac{x^{p-1}}{(1+x^q)^m}dx=Γ(r) \displaystyle\sum_{k=0}^{\infty} \frac{{}_{-m} \mathrm{C}_k}{(p+kq)^r}
\end{alignat}ただし、全て \(p,q,r \gt 0,n,m \in \mathrm{N}\)







<証明>

被積分関数に二項定理、及び負の二項定理の等式を代入します。(負の二項定理の詳細はこちらです。)$$(1+x)^{-n} =\displaystyle\sum_{k=0}^{\infty} {}_{-n} \mathrm{C}_k x^k$$

また全て、途中 \(\displaystyle \log \frac{1}{x}=t\) と置きます。$$\log x=-t,  x=e^{-t},  dx=-e^{-t}dt$$その後 \((p+kq)t=s\) と置きます。\([(p+kq)dt=ds]\)




\begin{alignat}{2}
&(1) \displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^n(1+x^q)^mx^{p-1}dx\\
&=\displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^n \left(\displaystyle\sum_{k=0}^m {}_m \mathrm{C}_k x^{kq} \right)x^{p-1}dx\\
&=\displaystyle\sum_{k=0}^m {}_m \mathrm{C}_k \displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^n x^{p+kq-1}dx\\
&=\displaystyle\sum_{k=0}^m {}_m \mathrm{C}_k \displaystyle\int_{\infty}^0 t^n e^{-(p+kq-1)t}(-e^{-t})dt\\
&=\displaystyle\sum_{k=0}^m {}_m \mathrm{C}_k \displaystyle\int_0^{\infty} t^n e^{-(p+kq)t}dt\\
&=\displaystyle\sum_{k=0}^m {}_m \mathrm{C}_k \displaystyle\int_0^{\infty} \left(\frac{s}{p+kq}\right)^n e^{-s} \cdot \frac{1}{p+kq}ds\\
&=\displaystyle\sum_{k=0}^m \frac{{}_m \mathrm{C}_k}{(p+kq)^{n+1}}\displaystyle\int_0^{\infty} s^n e^{-s}ds\\
&=\displaystyle\sum_{k=0}^m \frac{{}_m \mathrm{C}_k}{(p+kq)^{n+1}} \cdot Γ(n+1)=n! \displaystyle\sum_{k=0}^m \frac{{}_m \mathrm{C}_k}{(p+kq)^{n+1}}
\end{alignat}以上より$$\displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^n(1+x^q)^mx^{p-1}dx=n! \displaystyle\sum_{k=0}^m \frac{{}_m \mathrm{C}_k}{(p+kq)^{n+1}}$$







\begin{alignat}{2}
&(2) \displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^n(1-x^q)^mx^{p-1}dx\\
&=\displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^n \left\{\displaystyle\sum_{k=0}^m {}_m \mathrm{C}_k (-x^q)^k \right\}x^{p-1}dx\\
&=\displaystyle\sum_{k=0}^m (-1)^k {}_m \mathrm{C}_k \displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^n x^{p+kq-1}dx\\
&=\displaystyle\sum_{k=0}^m (-1)^k {}_m \mathrm{C}_k \displaystyle\int_{\infty}^0 t^n e^{-(p+kq-1)t}(-e^{-t})dt\\
&=\displaystyle\sum_{k=0}^m (-1)^k {}_m \mathrm{C}_k \displaystyle\int_0^{\infty} t^n e^{-(p+kq)t}dt\\
&=\displaystyle\sum_{k=0}^m (-1)^k {}_m \mathrm{C}_k \displaystyle\int_0^{\infty} \left(\frac{s}{p+kq}\right)^n e^{-s} \cdot \frac{1}{p+kq}ds\\
&=\displaystyle\sum_{k=0}^m \frac{(-1)^k {}_m \mathrm{C}_k}{(p+kq)^{n+1}}\displaystyle\int_0^{\infty} s^n e^{-s}ds\\
&=\displaystyle\sum_{k=0}^m \frac{(-1)^k {}_m \mathrm{C}_k}{(p+kq)^{n+1}} \cdot Γ(n+1)=n! \displaystyle\sum_{k=0}^m \frac{{}_m \mathrm{C}_k}{(p+kq)^{n+1}}
\end{alignat}以上より$$\displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^n(1-x^q)^mx^{p-1}dx=n! \displaystyle\sum_{k=0}^m \frac{(-1)^k{}_m \mathrm{C}_k}{(p+kq)^{n+1}}$$







\begin{alignat}{2}
&(3) \displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^{r-1}\frac{x^{p-1}}{(1+x^q)^m}dx\\
&=\displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^{r-1} \left(\displaystyle\sum_{k=0}^{\infty} {}_{-m} \mathrm{C}_k x^{kq} \right)x^{p-1}dx\\
&=\displaystyle\sum_{k=0}^{\infty} {}_{-m} \mathrm{C}_k \displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^{r-1} x^{p+kq-1}dx\\
&=\displaystyle\sum_{k=0}^{\infty} {}_{-m} \mathrm{C}_k \displaystyle\int_{\infty}^0 t^{r-1} e^{-(p+kq-1)t}(-e^{-t})dt\\
&=\displaystyle\sum_{k=0}^{\infty} {}_{-m} \mathrm{C}_k \displaystyle\int_0^{\infty} t^{r-1} e^{-(p+kq)t}dt\\
&=\displaystyle\sum_{k=0}^{\infty} {}_{-m} \mathrm{C}_k \displaystyle\int_0^{\infty} \left(\frac{s}{p+kq}\right)^{r-1} e^{-s} \cdot \frac{1}{p+kq}ds\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{{}_{-m} \mathrm{C}_k}{(p+kq)^r}\displaystyle\int_0^{\infty} s^{r-1} e^{-s}ds\\
&=\displaystyle\sum_{k=0}^{\infty} \frac{{}_{-m} \mathrm{C}_k}{(p+kq)^r} \cdot Γ(r)=Γ(r) \displaystyle\sum_{k=0}^{\infty} \frac{{}_{-m} \mathrm{C}_k}{(p+kq)^r}
\end{alignat}以上より$$\displaystyle\int_0^1 \left(\log \frac{1}{x}\right)^{r-1}\frac{x^{p-1}}{(1+x^q)^m}dx=Γ(r) \displaystyle\sum_{k=0}^{\infty} \frac{{}_{-m} \mathrm{C}_k}{(p+kq)^r}$$


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