log(1+x)/(ax+b)^2[0,1]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^1 \frac{\log (1+x)}{(ax+b)^2}dx=
\begin{cases}
\displaystyle \frac{1}{a(a-b)}\log \frac{a+b}{b}+\frac{2 \log 2}{b^2-a^2}  (a≠b)\\
\displaystyle \frac{1}{2a^2}(1- \log 2)  (a=b)\\
\end{cases}\\
&(2)  \displaystyle\int_0^{\infty} \frac{\log (1+x)}{(ax+b)^2}dx=\frac{1}{a(a-b)}\log \frac{a}{b}  (a≠b)\\
\end{alignat}ただし 全て \(a,b \gt 0\)









<証明>

全て。部分積分を行い計算を進めます。

\((1)\) \((A)\)  \(a≠b\) のとき
\begin{alignat}{2}
&  \displaystyle\int_0^1 \frac{\log (1+x)}{(ax+b)^2}dx\\
&=\left[-\frac{\log (1+x)}{a(ax+b)}\right]_0^1+\frac{1}{a}\displaystyle\int_0^1 \frac{1}{(x+1)(ax+b)}dx\\
&=-\frac{\log 2}{a(a+b)}+\frac{1}{a^2}\displaystyle\int_0^1 \frac{1}{(x+1)\left(x+\frac{b}{a}\right)}dx\\
&=-\frac{\log 2}{a(a+b)}+\frac{1}{a^2} \cdot \frac{1}{\frac{b}{a}-1} \displaystyle\int_0^1 \left(\frac{1}{x+1}-\frac{1}{x+\frac{b}{a}}\right)dx\\
&=-\frac{\log 2}{a(a+b)}+\frac{1}{a(b-a)}\left[\log \left|\frac{x+1}{x+\frac{b}{a}}\right|\right]_0^1 \\
&=-\frac{\log 2}{a(a+b)}-\frac{1}{a(a-b)}\left(\log \frac{2}{1+\frac{b}{a}}-\log \frac{a}{b}\right)\\
&=-\frac{\log 2}{a(a+b)}-\frac{1}{a(a-b)}\log \left(\frac{2a}{a+b} \cdot \frac{b}{a}\right)\\
&=-\frac{\log 2}{a(a+b)}-\frac{1}{a(a-b)}\log \frac{2b}{a+b}\\
&=-\frac{\log 2}{a(a+b)}-\frac{1}{a(a-b)}\left(\log 2-\log \frac{a+b}{b}\right)\\
&=\frac{1}{a(a-b)}\log \frac{a+b}{b}-\frac{\log 2}{a(a+b)}-\frac{\log 2}{a(a-b)}\\
&=\frac{1}{a(a-b)}\log \frac{a+b}{b}-\frac{\log 2}{a}\left(\frac{1}{a+b}+\frac{1}{a-b}\right)\\
&=\frac{1}{a(a-b)}\log \frac{a+b}{b}-\frac{\log 2}{a} \cdot \frac{2a}{a^2-b^2}\\
&=\displaystyle \frac{1}{a(a-b)}\log \frac{a+b}{b}+\frac{2 \log 2}{b^2-a^2}
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{\log (1+x)}{(ax+b)^2}dx=\displaystyle \frac{1}{a(a-b)}\log \frac{a+b}{b}+\frac{2 \log 2}{b^2-a^2}   (a≠b)$$





\((B)\)  \(a=b\) のとき
\begin{alignat}{2}
\displaystyle\int_0^1 \frac{\log (1+x)}{(ax+b)^2}dx&=\displaystyle\int_0^1 \frac{\log (1+x)}{(ax+a)^2}dx=\frac{1}{a^2}\displaystyle\int_0^1 \frac{\log (1+x)}{(x+1)^2}dx\\
&=\frac{1}{a^2}\left\{\left[-\frac{\log (1+x)}{1+x}\right]_0^1 +\displaystyle\int_0^1 \frac{1}{(1+x)^2}dx\right\}\\
&=\frac{1}{a^2}\left(-\frac{\log 2}{2}+\left[-\frac{1}{1+x}\right]_0^1\right)\\
&=\frac{1}{a^2}\left(-\frac{\log 2}{2}-\frac{1}{2}+1\right)=\frac{1}{2a^2}(1-\log 2)\\
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{\log (1+x)}{(ax+b)^2}dx=\displaystyle \frac{1}{2a^2}(1- \log 2)   (a=b)$$







\begin{alignat}{2}
(2)   \displaystyle\int_0^{\infty} \frac{\log (1+x)}{(ax+b)^2}dx&=\left[-\frac{\log (1+x)}{a(ax+b)}\right]_0^{\infty}+\frac{1}{a}\displaystyle\int_0^{\infty} \frac{1}{(x+1)(ax+b)}dx\\
&=\frac{1}{a^2}\displaystyle\int_0^{\infty} \frac{1}{(x+1)\left(x+\frac{b}{a}\right)}dx\\
&=\frac{1}{a^2} \cdot \frac{1}{\frac{b}{a}-1} \displaystyle\int_0^{\infty}\left(\frac{1}{x+1}-\frac{1}{x+\frac{b}{a}}\right)dx\\
&=\frac{1}{a(b-a)}\left[\log \left|\frac{x+1}{x+\frac{b}{a}}\right|\right]_0^{\infty} \\
&=\frac{1}{a(b-a)}\left(-\log \frac{a}{b}\right)=\frac{1}{a(a-b)}\log \frac{a}{b}
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{\log (1+x)}{(ax+b)^2}dx=\frac{1}{a(a-b)}\log \frac{a}{b}$$

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