log(1+x^2)/(1+x^2)[0,1]などの定積分

\begin{alignat}{2}
&(1)  \displaystyle\int_0^1 \frac{\log (1+x)}{1+x^2}dx=\frac{π}{8}\log 2\\
&(2)  \displaystyle\int_0^{\infty} \frac{\log (1+x)}{1+x^2}dx=\frac{π}{4}\log 2+G\\
&(3)  \displaystyle\int_0^1 \frac{\log (1+x^2)}{1+x^2}dx=\frac{π}{2}\log 2-G\\
&(4)  \displaystyle\int_1^{\infty} \frac{\log (1+x^2)}{1+x^2}dx=\frac{π}{2}\log 2+G\\
&(5)  \displaystyle\int_0^{\infty} \frac{\log (1+x^2)}{1+x^2}dx=π\log 2\\
&(6)  \displaystyle\int_0^1 \frac{\log (1-x)}{1+x^2}dx=\frac{π}{8}\log 2-G\\
&(7)  \displaystyle\int_1^∞ \frac{\log (x-1)}{1+x^2}dx=\frac{π}{8}\log 2\\
&(8)  \displaystyle\int_0^∞ \frac{\log |x-1|}{1+x^2}dx=\frac{π}{4}\log 2-G\\
&(9)  \displaystyle\int_0^1 \frac{\log (1-x^2)}{1+x^2}dx=\frac{π}{4}\log 2-G\\
&(10)  \displaystyle\int_1^{\infty} \frac{\log (x^2-1)}{1+x^2}dx=\frac{π}{4}\log 2+G\\
&(11)  \displaystyle\int_0^{\infty} \frac{\log |x^2-1|}{1+x^2}dx=\frac{π}{2}\log 2\\
\end{alignat}








<証明>

次の定積分の結果を用います。[詳細はこちらです。(A)(B)(C)(D)]
\begin{alignat}{2}
&(A)  \displaystyle\int_0^{\frac{π}{4}} \log ( \cos x) dx=-\frac{π}{4}\log 2 +\frac{1}{2}G\\
&(B)  \displaystyle\int_0^{\frac{π}{4}}\log (1+\tan x)dx=\frac{π}{8}\log 2\\
&(C)  \displaystyle\int_0^{\frac{π}{4}}\log (1-\tan x)dx=\frac{π}{8}\log 2-G\\
&(D)  \displaystyle\int_0^{\frac{π}{2}}\log (1+\tan x)dx=\frac{π}{4}\log 2+G
\end{alignat}




\((1)(2)(3)(4)(6)(9)\) では \(x=\tan t\) と置きます。\(\displaystyle \left(dx=\frac{1}{\cos^2 t}dt\right)\)

\begin{alignat}{2}
(1)  \displaystyle\int_0^1 \frac{\log (1+x)}{1+x^2}dx&=\displaystyle\int_0^{\frac{π}{4}} \frac{\log (1+\tan t)}{1+\tan^2 t} \cdot \frac{1}{\cos^2 t}dt\\
&=\displaystyle\int_0^{\frac{π}{4}}\log (1+\tan t)dt=\frac{π}{8}\log 2
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{\log (1+x)}{1+x^2}dx=\frac{π}{8}\log 2$$









\begin{alignat}{2}
(2)  \displaystyle\int_0^{\infty} \frac{\log (1+x)}{1+x^2}dx&=\displaystyle\int_0^{\frac{π}{2}} \frac{\log (1+\tan t)}{1+\tan^2 t} \cdot \frac{1}{\cos^2 t}dt\\
&=\displaystyle\int_0^{\frac{π}{2}}\log (1+\tan t)dt=\frac{π}{4}\log 2+G
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{\log (1+x)}{1+x^2}dx=\frac{π}{4}\log 2+G$$







\begin{alignat}{2}
(3)  \displaystyle\int_0^1 \frac{\log (1+x^2)}{1+x^2}dx&=\displaystyle\int_0^{\frac{π}{4}} \frac{\log (1+\tan^2 t)}{1+\tan^2 t} \cdot \frac{1}{\cos^2 t}dt\\
&=\displaystyle\int_0^{\frac{π}{4}}\log (1+\tan^2 t)dt\\
&=\displaystyle\int_0^{\frac{π}{4}} \log \frac{1}{\cos^2 t}dt=-2 \displaystyle\int_0^{\frac{π}{4}} \log (\cos t)dt\\
&=-2 \left(-\frac{π}{4}\log 2+\frac{1}{2}G\right)=\frac{π}{2}\log 2-G\\
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{\log (1+x^2)}{1+x^2}dx=\frac{π}{2}\log 2-G$$







\begin{alignat}{2}
(4)  \displaystyle\int_1^{\infty} \frac{\log (1+x^2)}{1+x^2}dx&=\displaystyle\int_{\frac{π}{4}}^{\frac{π}{2}} \frac{\log (1+\tan^2 t)}{1+\tan^2 t} \cdot \frac{1}{\cos^2 t}dt\\
&=\displaystyle\int_{\frac{π}{4}}^{\frac{π}{2}}\log (1+\tan^2 t)dt\\
&=\displaystyle\int_{\frac{π}{4}}^{\frac{π}{2}} \log \frac{1}{\cos^2 t}dt=-2 \displaystyle\int_{\frac{π}{4}}^{\frac{π}{2}} \log (\cos t)dt\\
&=-2\left\{\displaystyle\int_0^{\frac{π}{2}}\log (\cos t)dt-\displaystyle\int_0^{\frac{π}{4}}\log (\cos t)dt\right\}\\
&=-2\left\{-\frac{π}{2}\log 2-\left(-\frac{π}{4}\log 2+\frac{1}{2}G\right)\right\}\\
&=-2\left(-\frac{π}{4}\log 2-\frac{1}{2}G\right)=\frac{π}{2}\log 2+G
\end{alignat}以上より$$\displaystyle\int_1^{\infty} \frac{\log (1+x^2)}{1+x^2}dx=\frac{π}{2}\log 2+G$$







\((5)\) \((3)(4)\) の積分の和であるので
\begin{alignat}{2}
\displaystyle\int_0^{\infty} \frac{\log (1+x^2)}{1+x^2}dx&=\displaystyle\int_0^1 \frac{\log (1+x^2)}{1+x^2}dx+\displaystyle\int_1^{\infty} \frac{\log (1+x^2)}{1+x^2}dx\\
&=\frac{π}{2}\log 2-G+\frac{π}{2}\log 2+G=π\log 2\\
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{\log (1+x^2)}{1+x^2}dx=π\log 2$$








\begin{alignat}{2}
(6)  \displaystyle\int_1^{\infty} \frac{\log (1-x)}{1+x^2}dx&=\displaystyle\int_0^{\frac{π}{4}} \frac{\log (1-\tan t)}{1+\tan^2 t} \cdot \frac{1}{\cos^2 t}dt\\
&=\displaystyle\int_0^{\frac{π}{4}} \log (1-\tan t)dt=\frac{π}{8}\log 2-G\\
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{\log (1-x)}{1+x^2}dx=\frac{π}{8}\log 2-G$$






\((7)\) \(\displaystyle x=\frac{1}{t}\) と置きます。\(\displaystyle\left(dx=-\frac{1}{t^2}dt\right)\)
\begin{alignat}{2}
\displaystyle\int_1^∞ \frac{\log (x-1)}{1+x^2}dx&=\displaystyle\int_1^0 \frac{\log \left(\frac{1}{t}-1\right)}{1+\frac{1}{t^2}}\cdot \left(-\frac{1}{t^2}\right)dt\\
&=\displaystyle\int_0^1 \frac{\log (1-t)-\log t}{t^2+1}dt\\
&=\displaystyle\int_0^1 \frac{\log (1-t)}{t^2+1}dt-\displaystyle\int_0^1 \frac{\log t}{t^2+1}dt\\
&=\frac{π}{8}\log 2-G-(-G)=\frac{π}{8}\log 2\\
\end{alignat}以上より$$\displaystyle\int_1^∞ \frac{\log (x-1)}{1+x^2}dx=\frac{π}{8}\log 2$$








\((8)\) \((6)(7)\) の式を用います。
\begin{alignat}{2}
\displaystyle\int_0^∞ \frac{\log |x-1|}{1+x^2}dx&=\displaystyle\int_0^1 \frac{\log (1-x)}{1+x^2}dx+\displaystyle\int_1^∞ \frac{\log (x-1)}{1+x^2}dx\\
&=\frac{π}{8}\log 2-G+\frac{π}{8}\log 2=\frac{π}{4}\log 2-G
\end{alignat}以上より$$\displaystyle\int_0^∞ \frac{\log |x-1|}{1+x^2}dx=\frac{π}{4}\log 2-G$$








\begin{alignat}{2}
(9)  \displaystyle\int_1^{\infty} \frac{\log (1-x^2)}{1+x^2}dx&=\displaystyle\int_0^{\frac{π}{4}} \frac{\log (1-\tan^2 t)}{1+\tan^2 t} \cdot \frac{1}{\cos^2 t}dt\\
&=\displaystyle\int_0^{\frac{π}{4}}\log (1-\tan^2 t)dt\\
&=\displaystyle\int_0^{\frac{π}{4}}\log (1-\tan t)(1+\tan t)dt\\
&=\displaystyle\int_0^{\frac{π}{4}}\log (1-\tan t)dt+\displaystyle\int_0^{\frac{π}{4}}\log (1+\tan t)dt\\
&=\frac{π}{8}\log 2-G+\frac{π}{8}\log 2=\frac{π}{4}\log 2-G
\end{alignat}以上より$$\displaystyle\int_0^1 \frac{\log (1-x^2)}{1+x^2}dx=\frac{π}{4}\log 2-G$$







\((10)\) \(\displaystyle x=\frac{1}{t}\) と置きます。\(\displaystyle\left(dx=-\frac{1}{t^2}dt\right)\)
\begin{alignat}{2}
\displaystyle\int_1^{\infty} \frac{\log (x^2-1)}{1+x^2}dx&=\displaystyle\int_1^0 \frac{\log \left(\frac{1}{t^2}-1\right)}{1+\frac{1}{t^2}}\left(-\frac{1}{t^2}\right)dt\\
&=\displaystyle\int_0^1 \frac{\log (1-t^2)-2 \log 2}{1+t^2}dt\\
&=\displaystyle\int_0^1 \frac{\log (1-t^2)}{1+t^2}dt-2 \displaystyle\int_0^1 \frac{\log t}{1+t^2}dt\\
&=\frac{π}{4}\log 2-G+2G=\frac{π}{4}\log 2+G
\end{alignat}以上より$$\displaystyle\int_1^{\infty} \frac{\log (x^2-1)}{1+x^2}dx=\frac{π}{4}\log 2+G$$







\((11)\) \((9)(10)\) の式を用います。
\begin{alignat}{2}
\displaystyle\int_0^{\infty} \frac{\log |x^2-1|}{1+x^2}dx&=\displaystyle\int_0^1 \frac{\log (1-x^2)}{1+x^2}dx+\displaystyle\int_1^{\infty} \frac{\log (x^2-1)}{1+x^2}dx\\
&=\frac{π}{4}\log 2-G+\frac{π}{4}\log 2+G=\frac{π}{2}\log 2\\
\end{alignat}以上より$$\displaystyle\int_0^{\infty} \frac{\log |x^2-1|}{1+x^2}dx=\frac{π}{2}\log 2$$

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