对数与反正切的定积分问题

青青子衿

\begin{align*}
\int_{0}^{1}\arctan\left(x\right)\ln\left(1+x\right){\mathrm{d}}x=\frac{3\pi\ln2}{8}+\frac{\ln2}{2}-\frac{\pi^{2}}{96}-\frac{\ln^2{2}}{8}-\frac{\pi}{4}
\end{align*}

abababa

回复 1# 青青子衿
令$\ln(1+x) = [\ln(1+x)+x\ln(1+x)-x]' = [u(x)]'$，然后
\[
\begin{align*}
\int_{0}^{1}\arctan x\ln(1+x)dx
&= [\arctan xu(x)]_{0}^{1}-\int_{0}^{1}\frac{u(x)}{1+x^2}dx\\
&= \frac{\pi}{4}(2\ln2-1)-\int_{0}^{1}\frac{u(x)}{1+x^2}dx
\end{align*}
\]
\[
\begin{align*}
\int_{0}^{1}\frac{\ln(1+x)+x\ln(1+x)-x}{1+x^2}dx
&= \int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx+\frac{x\ln(1+x)}{1+x^2}dx-[\frac{\ln(1+x^2)}{2}]_{0}^{1}\\
&= \int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx+\frac{x\ln(1+x)}{1+x^2}dx-\frac{\ln2}{2}
\end{align*}
\]
令$I_1=\int_{0}^{1}\frac{x\ln(1+x)}{1+x^2}, I_2=\int_{0}^{1}\frac{x\ln(1-x)}{1+x^2}$
就有$I_1+I_2=\int_{0}^{1}\frac{x\ln(1-x^2)}{1+x^2}dx$，换元$t=\frac{1-x^2}{1+x^2}$，即$x^2=\frac{1-t}{1+t}$，就得到
\[I_1+I_2=\frac{1}{2}\int_{0}^{1}\frac{\ln\frac{2t}{1+t}}{1+t}dt=\frac{1}{2}\left(\int_{0}^{1}\frac{\ln2}{1+t}dt+\int_{0}^{1}\frac{\ln t}{1+t}dt-\int_{0}^{1}\frac{\ln(1+t)}{1+t}dt\right)\]
还有$I_1-I_2=\int_{0}^{1}\frac{x\ln\frac{1+x}{1-x}}{1+x^2}dx$，换元$t=\frac{1+x}{1-x}$，就得到
\[I_1-I_2=\int_{1}^{\infty}\frac{(t-1)\ln t}{(1+t)(1+t^2)}dt = \int_{1}^{\infty}\frac{t\ln t}{1+t^2}dt-\int_{1}^{\infty}\frac{\ln t}{1+t}dt= \int_{0}^{1}\frac{\ln t}{1+t}dt-\int_{0}^{1}\frac{t\ln t}{1+t^2}dt\]

然后
\[
\int_{0}^{1}\frac{\ln x}{1+x}dx=\sum_{k=0}^{\infty}(-1)^k\int_{0}^{1}x^k\ln xdx=\sum_{k=0}^{\infty}(-1)^{k}\cdot-\frac{1}{(k+1)^2}=-\frac{\pi^2}{12}
\]
\[
\int_{0}^{1}\frac{x\ln x}{1+x^2}dx=\sum_{k=0}^{\infty}(-1)^k\int_{0}^{1}x^{2k+1}\ln xdx=\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{4(k+1)^{2}} = -\frac{\pi^2}{48}
\]
这样就能算出$I_1$了，之后还剩下一个$\int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx$，换元$x=\frac{1-t}{1+t}$，就得到
\[
\begin{align*}
\int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx&=\int_{0}^{1}\frac{\ln\frac{2}{1+t}}{1+t^2}dt\\
&=\int_{0}^{1}\frac{\ln2}{1+t^2}dt-\int_{0}^{1}\frac{\ln(1+t)}{1+t^2}dt\\
&=\frac{\pi}{4}-\int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx
\end{align*}
\]
就都算出来了。

hbghlyj

abababa 发表于 2021-4-21 12:12
\begin{align*}
&=\int_{0}^{1}\frac{\ln2}{1+t^2}dt-\int_{0}^{1}\frac{\ln(1+t)}{1+t^2}dt\\
&=\color{#f00}{\frac{\pi}{4}}-\int_{0}^{1}\frac{\ln(1+x)}{1+x^2}dx
\end{align*}

这里好像漏了$\ln2$