一道有难度的定积分

数学星空

\[\int_0^1 \frac{\arctan(x)}{1+x}dx=\int_0^1 {(\sum_{k=0}^{\infty}\frac{(-1)^{k+1}x^{2k+1}}{2k+1})(\sum_{k=0}^{\infty} (-1)^{k+1} x^k}) dx\]

\[=\int_0^1 ((x-x^2)+(1-\frac{1}{3})(x^3-x^4)+(1-\frac{1}{3}+\frac{1}{5})(x^5-x^6)+\ldots)dx\]

\[=(\frac{1}{2}-\frac{1}{3})+(\frac{1}{4}-\frac{1}{5})(1-\frac{1}{3})+(\frac{1}{6}-\frac{1}{7})(1-\frac{1}{3}+\frac{1}{5})+\ldots\]

谁能给出精确的数值解？

kastin

$$\begin{align*}\int_0^1 \frac{\arctan x}{1+x}\dif x &=\arctan x \ln(1+x)|_0^1-\int_0^1\frac{\ln(1+x)}{1+x^2}\dif x\\
&=\frac{\pi}{4}\ln2-\int_0^{\frac{\pi}{4}}\ln(1+\tan t)\dif t\quad\quad(令x =\tan t)\\
&=\frac{\pi}{4}\ln2-v\quad\quad
\end{align*}$$
代换 `t =\D\frac{\pi}{4} -u`，则$$\int_0^{\frac{\pi}{4}}\ln(1+\tan t)\dif t=\int_0^{\frac{\pi}{4}}\ln2-\ln(1+\tan u)\dif u$$即 `v=\D \int_0^{\frac{\pi}{4}}\ln(1+\tan t)\dif t=\frac{\pi}{8}\ln2`，从而$$\int_0^1 \frac{\arctan x}{1+x}\dif x =\frac{\pi}{8}\ln2$$

葡萄糖

\[\begin{array}{l}
\int_0^1 {\frac{{\arctan x}}{{1 + x}}dx}  = \int_0^1 {\arctan xd\left[ {\ln \left( {1 + x} \right)} \right]} \\
= \left. {\left[ {\arctan x\ln \left( {1 + x} \right)} \right]} \right|_0^1 - \int_0^1 {\frac{{\ln \left( {1 + x} \right)}}{{1 + {x^2}}}} dx \\
= \frac{\pi }{4}\ln 2 - \int_0^1 {\ln \left( {1 + x} \right)} d\left( {\arctan x} \right)\\
\mathop  = \limits^{x = \tan t} \frac{\pi }{4}\ln 2 - \int_0^{\frac{\pi }{4}} {\ln \left( {1 + \tan t} \right)} dt\\
\mathop  = \limits^{t = \frac{\pi }{4} - \theta } \frac{\pi }{4}\ln 2 - \int_{ - \frac{\pi }{4}}^0 {\ln \left( {1 + \frac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right)d\left( {\frac{\pi }{4} - \theta } \right)} \\
= \frac{\pi }{4}\ln 2 - \int_0^{\frac{\pi }{4}} {\ln \left( {\frac{2}{{1 + \tan \theta }}} \right)d\theta } \\
= \frac{\pi }{4}\ln 2 - \int_0^{\frac{\pi }{4}} {\ln 2d\theta }  + \int_0^{\frac{\pi }{4}} {\ln \left( {1 + \tan \theta } \right)d\theta } \\
= \int_0^{\frac{\pi }{4}} {\ln \left( {1 + \tan \theta } \right)d\theta } \\
= \frac{\pi }{8}\ln 2
\end{array}\]

数学星空

进一步我们来讨论
\[\sum_{n=1}^{\infty} \left(\left(\frac{1}{2n}-\frac{1}{2n+1}\right)\sum_{k=1}^n\frac{(-1)^{k+1}}{2k-1}\right)=\left(\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n}\right)\left(\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}\right)=\frac{\ln2}{2}\left(\frac{\pi}{4}\right)=\frac{\pi\ln2}{8}\]

数学星空

\[\sum_{n=1}^{\infty}(\frac{(-1)^{n+1}x^{2n+1}}{2n+1}\sum_{k=1}^{2n}\frac{1}{k})=\frac{\ln(1+x^2)}{2}\arctan(x)\]

yinhow

数学星空 发表于 2016-1-26 20:57
\[\sum_{n=1}^{\infty}(\frac{(-1)^{n+1}x^{2n+1}}{2n+1}\sum_{k=1}^{2n}\frac{1}{k})=\frac{\ln(1+x^2)}{2 ...

log(1+ix)的虚部