Eufisky - The lost book

## 向老师的题目

$\int_0^1{\frac{1}{1+a^2x^2}\left[\left(1-\frac{x}{2}\right)\ln\frac{1+x}{1-x}+\frac{\pi^2x^2}{4}\right]^{-1}\textrm{d}x}.$

\begin{align*}&\hspace{0.5cm}\int_0^1{\frac{1}{1+a^2x^2}\left[\left(1-\frac{x}{2}\right)\ln\frac{1+x}{1-x}+\frac{\pi^2x^2}{4}\right]^{-1}\textrm{d}x}\\&=\int_0^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\left(\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2x^2}{4}}\right)\textrm{d}x}\\&=\frac{1}{2}\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\left(\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2x^2}{4}}\right)\textrm{d}x}\end{align*}

\begin{align*}&\int_{ - \infty  - \frac{\pi }{2}{\rm{i}}}^{\infty  - \frac{\pi }{2}{\rm{i}}} {f\left( z \right){\rm{d}}z}  - \int_{ - \infty  + \frac{\pi }{2}{\rm{i}}}^{\infty  + \frac{\pi }{2}{\rm{i}}} {f\left( z \right){\rm{d}}z} \\&= 2\pi {\rm{i}}\left( {{\rm{Res}}\left[ {f\left( z \right),z = 0} \right] + \left( {{\rm{Res}}\left[ {f\left( z \right),{\rm{i}} \cdot \arctan \left( a \right)} \right] + {\rm{Res}}\left[ {f\left( z \right),z =  - {\rm{i}} \cdot \arctan \left( a \right)} \right]} \right)} \right)\\&+ \pi {\rm{i}}\left( {{\rm{Res}}\left[ {f\left( z \right),z = \frac{\pi }{2}{\rm{i}}} \right] + {\rm{Res}}\left[ {f\left( z \right),z =  - \frac{\pi }{2}{\rm{i}}} \right]} \right)\\&= 2\pi {\rm{i}}\left( {\frac{3}{{{a^2}}} - \frac{a}{{2\left( {a - \arctan \left( a \right)} \right)}} - \frac{a}{{2\left( {a - \arctan \left( a \right)} \right)}}} \right) + \pi {\rm{i}}\left( {1 + 1} \right)\\&= 2\pi {\rm{i}}\left( {\frac{3}{{{a^2}}} - \frac{{\arctan \left( a \right)}}{{a - \arctan \left( a \right)}}} \right).\end{align*}

\begin{align*}&\int_{-\infty -\frac{\pi}{2}\textrm{i}}^{\infty -\frac{\pi}{2}\textrm{i}}{f\left(z\right)\textrm{d}z}-\int_{-\infty +\frac{\pi}{2}\textrm{i}}^{\infty +\frac{\pi}{2}\textrm{i}}{f\left(z\right)\textrm{d}z}=\int_{-\infty}^{\infty}{f\left(x-\frac{\pi}{2}\textrm{i}\right)\textrm{d}x}-\int_{-\infty}^{\infty}{f\left(x+\frac{\pi}{2}\textrm{i}\right)\textrm{d}x}\\&=\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\coth x-\left(x-\frac{\pi}{2}\textrm{i}\right)^2}\textrm{d}x}-\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\coth x-\left(x+\frac{\pi}{2}\textrm{i}\right)^2}\textrm{d}x}\\&=-\pi\textrm{i}\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2}{4}}\textrm{d}x}=2\pi\textrm{i}\left(\frac{3}{a^2}-\frac{\arctan\left(a\right)}{a-\arctan\left(a\right)}\right).\end{align*}

\begin{align*}\int_0^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2}{4}}\textrm{d}x}&=\frac{1}{2}\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2}{4}}\textrm{d}x}\\&=\frac{\arctan\left(a\right)}{a-\arctan\left(a\right)}-\frac{3}{a^2}\end{align*}

$\int_0^1{x^{20}\left[ \left( 1-\frac{x}{2}\ln \frac{1+x}{1-x} \right) ^2+\frac{\pi ^2x^2}{4} \right] ^{-1}\text{d}x}=\frac{5588512716806912356}{374010621408251953125}$

$\sum_{n=1}^\infty\frac{\mathrm{Si}(n\pi)}{n^3}$

\begin{align*}\text{Si}\left( n\pi \right) &=\int_0^{n\pi}{\frac{\sin t}{t}\text{d}t}=\int_0^{\pi}{\frac{\sin nx}{x}\text{d}x}=\int_0^{\pi}{\sin nx\text{d}\left( \ln x \right)}=-n\int_0^{\pi}{\cos nx\ln x\text{d}x}\\&=-n\int_0^{\pi}{\cos nx\text{d}\left( x\ln x-x \right)}=n\left[ \left( -1 \right) ^{n-1}\left( \pi \ln \pi -\pi \right) -n\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x} \right]\end{align*}

$\sum_{n=1}^{\infty}{\frac{\text{Si}(n\pi)}{n^3}}=\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}}{n^2}\left( \pi \ln \pi -\pi \right)}-\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}$

\begin{align*}\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}&=\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\text{d}\left( \frac{1}{2}x^2\ln x-\frac{3}{4}x^2 \right)}}\\&=\sum_{n=1}^{\infty}{\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \cos nx\text{d}x}}.\end{align*}

$\widetilde{f}\left( x \right) =\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n\cos nx}$

$\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n}=f(0)=0$

$\sum_{n=1}^{\infty}{\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \text{d}x}}=\frac{\pi}{2}\sum_{n=1}^{\infty}{a_n}=-\frac{\pi}{4}a_0=\frac{\pi ^3}{12}\ln \pi -\frac{11}{72}\pi ^3$

$\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^3}}=\frac{\pi ^2}{12}\left( \pi \ln \pi -\pi \right) -\left( \frac{\pi ^3}{12}\ln \pi -\frac{11}{72}\pi ^3 \right) =\frac{5\pi ^3}{72}$

$\sum_{n=1}^{\infty}{\left( -1 \right) ^n\frac{\text{Si}\left( n\pi \right)}{n^3}}=-\frac{\pi ^2}{6}\left( \pi \ln \pi -\pi \right) -\left( \frac{2\pi ^3}{9}-\frac{\pi ^3}{6}\ln \pi \right) =-\frac{\pi ^3}{18}$

$\sum_{n=1}^{\infty}{\left( \frac{\text{Si}\left( n\pi \right)}{n} \right) ^2}$

$\text{Si}\left( n\pi \right) =-n\int_0^{\pi}{\cos nx\ln x\text{d}x}$

$\sum_{n=1}^{\infty}{\left( \frac{\text{Si}\left( n\pi \right)}{n} \right) ^2}=\sum_{n=1}^{\infty}{\left( \int_0^{\pi}{\cos nx\ln x\text{d}x} \right) ^2}$

$\widetilde{f}\left( x \right) =\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n\cos nx}$

$\frac{a_{0}^{2}}{2}+\sum_{n=1}^{\infty}{a_{n}^{2}}=\frac{2}{\pi}\int_0^{\pi}{f^2\left( x \right) \text{d}x}=\frac{2}{\pi}\int_0^{\pi}{\ln ^2x\text{d}x}=4-4\ln \pi +2\ln ^2\pi$

$\sum_{n=1}^{\infty}{\left( \frac{\text{Si}\left( n\pi \right)}{n} \right) ^2}=\sum_{n=1}^{\infty}{\left( \int_0^{\pi}{\cos nx\ln x\text{d}x} \right) ^2}=\frac{\pi^2}2$

$\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^5}}=\frac{269}{43200}\pi ^5,\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}\text{Si}\left( n\pi \right)}{n^5}}=\frac{4}{675}\pi ^5$

$\sum_{n=1}^{\infty}{\frac{\text{Si}(n\pi)}{n^5}}=\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}}{n^4}\left( \pi \ln \pi -\pi \right)}-\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}$

\begin{align*}\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}&=\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\text{d}\left( \frac{1}{2}x^2\ln x-\frac{3}{4}x^2 \right)}}\\&=\sum_{n=1}^{\infty}\frac1{n^2}{\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \cos nx\text{d}x}}\\&=\sum_{n=1}^{\infty}{\frac{1}{n^2}\int_0^{\pi}{\cos nx\text{d}\left( \frac{11}{36}x^3-\frac{1}{6}x^3\ln x \right)}}\\&=\left( \frac{11}{36}\pi ^3-\frac{\pi ^3}{6}\ln \pi \right) \sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n^2}}+\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\left( \frac{11}{36}x^3-\frac{x^3}{6}\ln x \right) \sin nx\text{d}x}}\\&=-\left( \frac{11}{36}\pi ^3-\frac{\pi ^3}{6}\ln \pi \right) \frac{\pi ^2}{12}+\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\left( \frac{11}{36}x^3-\frac{x^3}{6}\ln x \right) \sin nx\text{d}x}}.\end{align*}
\begin{align*}\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\left( \frac{11}{36}x^3-\frac{x^3}{6}\ln x \right) \sin nx\text{d}x}}&=\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\text{d}\left( \frac{25}{288}x^4-\frac{1}{24}x^4\ln x \right)}}\\&=\sum_{n=1}^{\infty}{\int_0^{\pi}{\left( \frac{25}{288}x^4-\frac{1}{24}x^4\ln x \right) \cos nx\text{d}x}}.\end{align*}

$\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^5}}=\frac{7\pi ^4}{720}\left( \pi \ln \pi -\pi \right) +\frac{\pi ^2}{12}\left( \frac{11\pi ^3}{36}-\frac{\pi ^3}{6}\ln \pi \right) -\frac{\pi}{2}\cdot \frac{\pi ^4\left( 137-60\ln \pi \right)}{7200}=\frac{269}{43200}\pi ^5$

$\sum_{n=1}^{\infty}{\frac{\text{Si}^2\left( n\pi \right)}{n^4}}=\frac{\pi ^4}{27}$

$\frac{\text{Si}\left( n\pi \right)}{n^2}=\frac{\left( -1 \right) ^{n-1}}{n}\left( \pi \ln \pi -\pi \right) -\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}$

\begin{align*}\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^2}\sin nx}&=\left( \pi \ln \pi -\pi \right) \sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}}{n}\sin nx}-\sum_{n=1}^{\infty}{\sin nx\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}\\&=\frac{\left( \pi \ln \pi -\pi \right) x}{2}-\frac{\pi \left( x\ln x-x \right)}{2}.\end{align*}

$\sum_{n=1}^{\infty}{\frac{\text{Si}^2\left( n\pi \right)}{n^4}}=\frac{2}{\pi}\int_0^{\pi}{\left( \frac{\left( \pi \ln \pi -\pi \right) x}{2}-\frac{\pi \left( x\ln x-x \right)}{2} \right) ^2\text{d}x}=\frac{\pi ^4}{27}$

\begin{align*}\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^3}}&=\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\frac{\sin nx}{x}\text{d}x}}=\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\text{d}x}\int_0^{+\infty}{\text{e}^{-xy}\text{d}y}}\\&=\int_0^{+\infty}{\text{d}y}\int_0^{\pi}{\text{e}^{-xy}}\frac{x^3-3\pi x^2+2\pi ^2x}{12}\text{d}y=\frac{5}{72}\pi ^3.\end{align*}