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Δημοσιεύτηκε: **Κυρ Φεβ 07, 2016 10:14 am**

Δείξτε ότι

$\displaystyle{\int_{0}^{1}\frac{(\sin^{-1}(x^{2}))^{2}}{\sqrt{1-x^{2}}}dx=\frac{\pi^{3}}{4}-\frac{3\pi}{4}\log^{2}(2)-2\pi Li_{2}\left(\frac{1}{\sqrt{2}}\right)}$

Δημοσιεύτηκε: **Πέμ Αύγ 24, 2017 9:03 pm**

galactus έγραψε:Δείξτε ότι $\displaystyle{\int_{0}^{1}\frac{(\sin^{-1}(x^{2}))^{2}}{\sqrt{1-x^{2}}}dx=\frac{\pi^{3}}{4}-\frac{3\pi}{4}\log^{2}(2)-2\pi Li_{2}\left(\frac{1}{\sqrt{2}}\right)}$

$\displaystyle{\int\limits_{0}^{1}{\frac{\left( \arcsin x^{2} \right)^{2}}{\sqrt{1-x^{2}}}dx}=\frac{1}{2}\int\limits_{0}^{1}{\frac{\left( \arcsin x \right)^{2}}{\sqrt{x}\sqrt{1-x}}dx}=\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}{\frac{x^{2}\cos x}{\sqrt{\sin x}\sqrt{1-\sin x}}dx}=\frac{\sqrt{2}}{2}\int\limits_{0}^{\frac{\pi }{2}}{\frac{\left( \frac{\pi }{2}-x \right)^{2}\cos \frac{x}{2}}{\sqrt{1-2\sin ^{2}\frac{x}{2}}}dx}}$

$\displaystyle{=\int\limits_{0}^{\frac{\pi }{2}}{\left( \frac{\pi }{2}-2\arcsin \left( \frac{\sin x}{\sqrt{2}} \right) \right)^{2}dx}=\left( \frac{\pi }{2} \right)^{2}\int\limits_{0}^{\frac{\pi }{2}}{dx}-2\pi \int\limits_{0}^{\frac{\pi }{2}}{\arcsin \left( \frac{\sin x}{\sqrt{2}} \right)dx}+4\int\limits_{0}^{\frac{\pi }{2}}{\arcsin ^{2}\left( \frac{\sin x}{\sqrt{2}} \right)dx}}$

$\displaystyle{=\frac{\pi ^{3}}{8}-2\pi \underbrace{\int\limits_{0}^{\frac{\pi }{2}}{\arcsin \left( \frac{\sin x}{\sqrt{2}} \right)dx}}_{I_{1}}+4\underbrace{\int\limits_{0}^{\frac{\pi }{2}}{\arcsin ^{2}\left( \frac{\sin x}{\sqrt{2}} \right)dx}}_{I_{2}}}$

$\displaystyle{I_{1}=\int\limits_{0}^{\frac{\pi }{2}}{\arcsin \left( \frac{\sin x}{\sqrt{2}} \right)dx}=\frac{\sqrt{2}}{2}\sum\limits_{j=0}^{+\infty }{\frac{\left( \begin{matrix} 2j \\ j \\ \end{matrix} \right)}{\left( 2j+1 \right)2^{3j}}\int\limits_{0}^{\frac{\pi }{2}}{sin^{2j+1}xdx}}=\frac{\sqrt{2}}{2}\sum\limits_{j=0}^{+\infty }{\frac{\left( \begin{matrix} 2j \\ j \\ \end{matrix} \right)}{\left( 2j+1 \right)2^{3j}}\cdot \frac{\sqrt{\pi }\Gamma \left( j+1 \right)}{2\Gamma \left( j+\frac{3}{2} \right)}}}$

$\displaystyle{=\frac{\sqrt{2}}{2}\sum\limits_{j=0}^{+\infty }{\frac{\left( \begin{matrix} 2j \\ j \\ \end{matrix} \right)}{\left( 2j+1 \right)2^{3j}}\cdot \frac{\sqrt{\pi }j!}{2\frac{\left( 2\left( j+1 \right) \right)!}{2^{2\left( j+1 \right)}\left( j+1 \right)!}\sqrt{\pi }}}=\sqrt{2}\sum\limits_{j=0}^{+\infty }{\frac{\left( 2j \right)!}{\left( j! \right)^{2}\left( 2j+1 \right)2^{j}}\cdot \frac{j!\left( j+1 \right)!}{\left( 2j+1+1 \right)!}}}$

$\displaystyle{=\sqrt{2}\sum\limits_{j=0}^{+\infty }{\frac{\left( 2j \right)!}{\left( j! \right)^{2}\left( 2j+1 \right)2^{j}}\cdot \frac{j!\left( j+1 \right)!}{\left( 2j+2 \right)\left( 2j+1 \right)\left( 2j \right)!}}=\frac{\sqrt{2}}{2}\sum\limits_{j=0}^{+\infty }{\frac{1}{\left( 2j+1 \right)^{2}2^{j}}}}$

$\displaystyle{=\frac{\sqrt{2}}{2}\sum\limits_{j=0}^{+\infty }{\frac{\left( \frac{1}{2} \right)^{j}}{\left( 2j+1 \right)^{2}}}=\frac{\sqrt{2}}{2}\left( 1+\frac{\frac{1}{2}}{3^{2}}+\frac{\left( \frac{1}{2} \right)^{2}}{5^{2}}+\frac{\left( \frac{1}{2} \right)^{3}}{7^{2}}+\frac{\left( \frac{1}{2} \right)^{4}}{9^{2}}+\frac{\left( \frac{1}{2} \right)^{5}}{11^{2}}+... \right)}$

$\displaystyle{=\frac{1}{\sqrt{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{3}}{3^{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{5}}{5^{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{7}}{7^{2}}+\frac{\left( \frac{1}{2} \right)^{9}}{9^{2}}+\frac{\left( \frac{1}{2} \right)^{11}}{11^{2}}+...}$

$\displaystyle{=\left( \frac{1}{\sqrt{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{2}}{2^{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{3}}{3^{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{4}}{4^{2}}+\frac{\left( \frac{1}{\sqrt{2}} \right)^{5}}{5^{2}}+... \right)-\frac{1}{2^{2}}\left( \frac{\frac{1}{2}}{1}+\frac{\left( \frac{1}{2} \right)^{2}}{2^{2}}+\frac{\left( \frac{1}{2} \right)^{3}}{3^{2}}+\frac{\left( \frac{1}{2} \right)^{4}}{4^{2}}+\frac{\left( \frac{1}{2} \right)^{5}}{5^{2}}+... \right)}$

$\displaystyle{=Li_{2}\left( \frac{1}{\sqrt{2}} \right)-\frac{1}{2^{2}}\left( \frac{1}{2}+\frac{\left( \frac{1}{2} \right)^{2}}{2^{2}}+\frac{\left( \frac{1}{2} \right)^{3}}{3^{2}}+\frac{\left( \frac{1}{2} \right)^{4}}{4^{2}}+\frac{\left( \frac{1}{2} \right)^{5}}{5^{2}}+... \right)}$

$\displaystyle{=Li_{2}\left( \frac{1}{\sqrt{2}} \right)-\frac{1}{2^{2}}Li_{2}\left( \frac{1}{2} \right)=Li_{2}\left( \frac{1}{\sqrt{2}} \right)-\frac{1}{4}\cdot \frac{\pi ^{2}-6\log ^{2}2}{12}}$

$\displaystyle{I_{2}=\int\limits_{0}^{\frac{\pi }{2}}{\arcsin ^{2}\left( \frac{\sin x}{\sqrt{2}} \right)dx}=\sum\limits_{j=1}^{+\infty }{\frac{2^{j-1}}{j^{2}\left( \begin{matrix} 2j \\ j \\ \end{matrix} \right)}\int\limits_{0}^{\frac{\pi }{2}}{sin^{2j}xdx}}=\sum\limits_{j=1}^{+\infty }{\frac{2^{j-1}}{j^{2}\left( \begin{matrix} 2j \\ j \\ \end{matrix} \right)}\cdot \frac{\sqrt{\pi }\Gamma \left( j+\frac{1}{2} \right)}{2\Gamma \left( j+1 \right)}}}$

$\displaystyle{=\frac{\pi }{2}\sum\limits_{j=1}^{+\infty }{\frac{2^{j-1}}{j^{2}\left( \begin{matrix} 2j \\ j \\ \end{matrix} \right)}\cdot \frac{\left( 2j \right)!}{2^{2j}\left( j! \right)^{2}}}=\frac{\pi }{2}\sum\limits_{j=1}^{+\infty }{\frac{2^{j-1}}{j^{2}}\cdot \frac{1}{2^{2j}}}=\frac{\pi }{4}\sum\limits_{j=1}^{+\infty }{\frac{1}{j^{2}2^{j}}}=\frac{\pi }{4}\left( -\frac{1}{2}\log ^{2}2+\frac{\zeta \left( 2 \right)}{2} \right)}$

$\displaystyle{=\frac{\pi }{8}\left( -\log ^{2}2+\frac{\pi ^{2}}{6} \right)}$

$\displaystyle{\int\limits_{0}^{1}{\frac{\left( \arcsin x^{2} \right)^{2}}{\sqrt{1-x^{2}}}dx}=\frac{\pi ^{3}}{8}-2\pi \left( Li_{2}\left( \frac{1}{\sqrt{2}} \right)-\frac{1}{4}\cdot \frac{\pi ^{2}-6\log ^{2}2}{12} \right)+4\cdot \frac{\pi }{8}\left( -\log ^{2}2+\frac{\pi ^{2}}{6} \right)}$

$\displaystyle{=\frac{\pi ^{3}}{4}-\frac{3\pi }{4}\cdot \log ^{2}2-2\pi Li_{2}\left( \frac{1}{\sqrt{2}} \right)}$

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