Exercise:
Compute, $\int_{0}^{1}{\frac{\arcsin x}{1+x}dx}$
Solution: Let $x=\sin \theta $ so $\theta =\arcsin x$ and $dx=\cos \theta d\theta $
Hence $\int_{0}^{1}{\frac{\arcsin x}{1+x}dx=\int_{0}^{\frac{\pi }{2}}{\frac{\theta }{1+\sin \theta }}\cos \theta d\theta }$
Let $u=\theta \,\,\And \,\,dv=\frac{\cos \theta }{1+\sin \theta }d\theta \Leftrightarrow du=d\theta \,\,\,\And \,\,v=\ln \left( 1+\sin \theta \right)+c$
So $\int_{0}^{\frac{\pi }{2}}{\theta \frac{\cos \theta }{1+\sin \theta }d\theta }=\left[ \theta \ln \left( 1+\sin \theta \right) \right]_{0}^{\frac{\pi }{2}}-\int_{0}^{\frac{\pi }{2}}{\ln \left( 1+\sin \theta \right)d\theta }=\frac{\pi }{2}\ln 2-\int_{0}^{\frac{\pi }{2}}{\ln \left( 1+\sin \theta \right)d\theta }$
Put $I=\int_{0}^{\frac{\pi }{2}}{\ln \left( 1+\sin \theta \right)d\theta }$
By using the formula $\int_{0}^{a}{f\left( x \right)dx=\int_{0}^{a}{f\left( a-x \right)dx}}$ we get
$I=\int_{0}^{\frac{\pi }{2}}{\ln \left( 1+\sin \theta \right)d\theta =}\int_{0}^{\frac{\pi }{2}}{\ln \left( 1+\sin \left( \frac{\pi }{2}-\theta \right) \right)d\theta }=\int_{0}^{\frac{\pi }{2}}{\ln \left( 1+\cos \theta \right)d\theta }$
But $\cos \theta =\frac{{{e}^{i\theta }}+{{e}^{-i\theta }}}{2}\Leftrightarrow 1+\cos \theta =\frac{2+{{e}^{i\theta }}+{{e}^{-i\theta }}}{2}$
so$\ln \left( 1+\cos \theta \right)=\ln \left( \frac{2+{{e}^{i\theta }}+{{e}^{-i\theta }}}{2} \right)=\ln \left( 2+{{e}^{i\theta }}+{{e}^{-i\theta }} \right)-\ln 2$
but $2+{{e}^{i\theta }}+{{e}^{-i\theta }}=1+{{e}^{i\theta }}+{{e}^{i\theta }}{{e}^{-i\theta }}+{{e}^{-i\theta }}=1+{{e}^{i\theta }}+{{e}^{-i\theta }}\left( 1+{{e}^{i\theta }} \right)=\left( 1+{{e}^{i\theta }} \right)\left( 1+{{e}^{-i\theta }} \right)$
hence $\ln \left( 1+\cos \theta \right)=\ln \left( 1+{{e}^{i\theta }} \right)\left( 1+{{e}^{-i\theta }} \right)-\ln 2=\ln \left( 1+{{e}^{i\theta }} \right)+\ln \left( 1+{{e}^{-i\theta }} \right)-\ln 2$
Put $x={{e}^{i\theta }}$ so $\ln \left( 1+x \right)=x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+...=\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k+1}}{{x}^{k}}}{k}}$
Hence $\ln \left( 1+{{e}^{i\theta }} \right)={{e}^{i\theta }}-\frac{{{e}^{i2\theta }}}{2}+\frac{{{e}^{i3\theta }}}{3}-\frac{{{e}^{i4\theta }}}{4}+...=\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k-1}}{{e}^{ik\theta }}}{k}}$
And $\ln \left( 1+{{e}^{-i\theta }} \right)={{e}^{-i\theta }}-\frac{{{e}^{-i2\theta }}}{2}+\frac{{{e}^{-i3\theta }}}{3}-\frac{{{e}^{-i4\theta }}}{4}+...=\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k-1}}{{e}^{-ik\theta }}}{k}}$
So $\ln \left( 1+{{e}^{i\theta }} \right)+\ln \left( 1+{{e}^{-i\theta }} \right)=\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k-1}}\left( {{e}^{ik\theta }}+{{e}^{-ik\theta }} \right)}{k}}$
But $\cos kx=\frac{{{e}^{ikx}}+{{e}^{-ikx}}}{2}\Leftrightarrow {{e}^{ikx}}+{{e}^{-ikx}}=2\cos kx$
So $\ln \left( 1+{{e}^{i\theta }} \right)+\ln \left( 1+{{e}^{-i\theta }} \right)=2\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k-1}}\cos k\theta }{k}}$
So $\int_{0}^{\frac{\pi }{2}}{\ln \left( 1+\sin \theta \right)d\theta }=2\int_{0}^{\frac{\pi }{2}}{\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k-1}}}{k}\cos k\theta }}\,d\theta -\int_{0}^{\frac{\pi }{2}}{\ln \left( 2 \right)d\theta }$
$=2\int_{0}^{\frac{\pi }{2}}{\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k-1}}}{k}\cos k\theta \,d\theta -\frac{\pi }{2}\ln 2}}$
$=2\sum\limits_{k=1}^{\infty }{\int_{0}^{\frac{\pi }{2}}{\frac{{{\left( -1 \right)}^{k-1}}}{k}\cos k\theta \,d\theta -\frac{\pi }{2}\ln 2}}$
But $\int_{0}^{\frac{\pi }{2}}{\cos \left( k\theta \right)d\theta }=\left[ \frac{1}{k}\sin \left( k\theta \right) \right]_{0}^{\frac{\pi }{2}}=\frac{1}{k}\sin \left( \frac{k\pi }{2} \right)$
Thus $\int_{0}^{\frac{\pi }{2}}{\ln \left( 1+\sin \theta \right)d\theta }=2\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k-1}}}{{{k}^{2}}}\sin \left( \frac{k\pi }{2} \right)-\frac{\pi }{2}\ln 2}$
But $\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k-1}}}{{{k}^{2}}}}\sin \left( \frac{k\pi }{2} \right)=\sin \left( \frac{\pi }{2} \right)-\frac{1}{4}\sin \left( \pi \right)+\frac{1}{9}\sin \left( \frac{3\pi }{2} \right)-\frac{1}{16}\sin \left( 2\pi \right)+\frac{1}{25}\sin \left( \frac{5\pi }{2} \right)+.....$
$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k-1}}}{{{k}^{2}}}}\sin \left( \frac{k\pi }{2} \right)=1+0-\frac{1}{9}+0+\frac{1}{25}+....=\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{{{\left( 2k+1 \right)}^{2}}}}$
Hence $\int_{0}^{\frac{\pi }{2}}{\ln \left( 1+\sin \theta \right)d\theta =2\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{{{\left( 2k+1 \right)}^{2}}}-\frac{\pi }{2}\ln 2}}$
Thus $\int_{0}^{\frac{\pi }{2}}{\frac{\theta \cos \theta }{1+\sin \theta }d\theta =\frac{\pi }{2}\ln 2-2\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{{{\left( 2k+1 \right)}^{2}}}+\frac{\pi }{2}\ln 2}=\pi \ln 2-2\sum\limits_{k=0}^{\infty }{\frac{{{\left( -1 \right)}^{k}}}{{{\left( 2k+1 \right)}^{2}}}=\pi \ln 2-2G}}$
Where $G$ is called Catalan's constant
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