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[数学分析] 请教一道定积分的题目

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发表于 2018-4-22 13:42:26 | 显示全部楼层 |阅读模式
下面这道定积分的题,我感觉求不出具体的值,我只用级数表示了,还望高手赐教!


$$\int_0^1\frac{\ln ^2(1+x^2)}{1+x}\operatorname dx$$





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 楼主| 发表于 2018-5-14 16:08:32 | 显示全部楼层
原问题链接请教一道定积分的题目

原解答链接Qn56.
(by Renascence_5)
\begin{align*}
&\int_0^1\frac{\ln^2(1+x^2)}{1+x^2}\,\mathrm{d} x\\
=&\int_0^1\frac{\ln^2(1-y^2)}{1+\mathrm{i}y}\mathrm{i}\,\mathrm{d} y-\int_0^{\frac{\pi}{2}}\frac{\ln^2(1+e^{\mathrm{i}2\theta})}{1+e^{\mathrm{i}\theta}}\,\mathrm{d}\theta\\
=&\underbrace{\int_0^1\frac{y\ln^2(1-y^2)}{1+y^2}\,\mathrm{d}y}_{I_1}+\underbrace{\frac{1}{2}\int_0^{\frac{\pi}{2}}\tan\left(\frac{\theta}{2}\right)\ln^2(2\cos\theta)\,\mathrm{d}\theta}_{I_2}+\underbrace{\int_0^{\frac{\pi}{2}}\theta\ln(2\cos\theta)\,\mathrm{d}\theta}_{I_3}-\underbrace{\frac{1}{2}\int_0^{\frac{\pi}{2}}\theta^2\tan\left(\frac{\theta}{2}\right)\,\mathrm{d}\theta}_{I_4}
\end{align*}
Evaluation of $I_1$:
\begin{align*}
I_1&=\frac{1}{2}\int_0^1\frac{\ln^2(1-y)}{1+y}\,\mathrm{d}y=\frac{1}{4}\int_0^1\frac{\ln^2y}{1-y/2}\,\mathrm{d}y\\
&=\frac{1}{4}\sum_{n=0}^{\infty}\frac{1}{2^n}\int_0^1y^2\ln^2y\,\mathrm{d}y=\sum_{n=0}^{\infty}\frac{1}{2^{n+1}(n+1)^3}\\
&=\mathrm{Li}_3\left(\frac{1}{2}\right)=\frac{7}{8}\zeta(3)-\frac{\pi^2}{12}\ln2+\frac{1}{6}\ln^32
\end{align*}
Evaluation of $I_2$:
\begin{align*}
I_2&=\int_0^1\frac{x}{1+x^2}\ln^2\left(2\frac{1-x^2}{1+x^2}\right)\,\mathrm{d}x\\
&=\frac{1}{2}\int_0^1\frac{1}{1+x}\ln^2\left(2\frac{1-x}{1+x}\right)\,\mathrm{d}x\\
&=\frac{1}{2}\int_0^1\frac{\ln^2x}{1+x}\,\mathrm{d}x+\ln2\int_0^1\frac{\ln x}{1+x}\,\mathrm{d}x+\frac{1}{2}\ln^22\int_0^1\frac{1}{1+x}\,\mathrm{d}x\\
&=\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n\int_0^1x^n\ln^2x\,\mathrm{d}x+\ln2\sum_{n=0}^{\infty}(-1)^n\int_0^1x^n\ln x\,\mathrm{d}\frac{1}{2}\ln^32\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^3}-\ln2\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^2}+\frac{1}{2}\ln^32\\
&=\frac{3}{4}\zeta(3)-\frac{\pi^2}{12}\ln2+\frac{1}{2}\ln^32       
\end{align*}
Evaluation of $I_3$:
\begin{align*}
I_3&=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\int_0^{\frac{\pi}{2}}\theta\cos(2n\theta)\,\mathrm{d}\theta\\
&=-\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^3}-\frac{1}{4}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3}\\
&=-\frac{7}{16}\zeta(3)
\end{align*}
Evaluation of $I_4$:
\begin{align*}
I_4&=\left[\theta^2\ln\left(\cos\frac{\theta}{2}\right)\right]_0^{\frac{\pi}{2}}-2\int_0^{\frac{\pi}{2}}\theta\ln\left(\cos\frac{\theta}{2}\right)\,\mathrm{d}\theta\\\
&=-\frac{\pi^2}{8}\ln2+2\ln2\int_0^{\frac{\pi}{2}}\theta\,\mathrm{d}\theta-2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\int_0^{\frac{\pi}{2}}\theta\cos(n\theta)\,\mathrm{d}\theta\\
&=\frac{\pi^2}{8}\ln2+2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3}-\pi\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\sin(n\pi/2)}{n^2}-2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\cos(n\pi/2)}{n^2}\\
&=\frac{21}{16}\zeta(3)-\pi\mathrm{\mathbf{G}}+\frac{\pi^2}{8}\ln2
\end{align*}
Result
\begin{align*}
\int_0^1\frac{\ln^2(1+x^2)}{1+x^2}\,\mathrm{d}x
&=\left(\frac{7}{8}+\frac{3}{4}-\frac{7}{16}+\frac{21}{16}\right)\zeta(3)-\pi\mathrm{\mathbf{G}}+\left(-\frac{\pi^2}{12}-\frac{\pi^2}{12}+\frac{\pi^2}{8}\right)\ln2+\left(\frac{1}{6}+\frac{1}{2}\right)\ln^32\\
&=\frac{5}{2}\zeta(3)-\pi\mathrm{\mathbf{G}}-\frac{\pi^2}{24}\ln2+\frac{2}{3}\ln^32
\end{align*}


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感谢s19870810提供解答,奖励5威望。

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