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Question Number 139708 by qaz last updated on 30/Apr/21

$$Prove:1+2\underset{n=1}{\sum ^{\mathrm{\infty }}}{a}^n\cos \left(nx\right)=\frac{1-a^2}{1-2a\cos x+a^2},(\mid a\mid <1)$$ $$Andcalculate::{\int }_0^{\pi }\frac{x^2}{1+8\sin {}^{2}x}dx=\frac{5{\pi }^3}{36}-\frac{\pi }{6}{ln}^{2}2$$

Answered by Dwaipayan Shikari last updated on 30/Apr/21

$$1+\underset{n=1}{\sum ^{\mathrm{\infty }}}{a}^n{e}^{inx}+a^n{e}^{-inx}=1+\underset{n=1}{\sum ^{\mathrm{\infty }}}{\left({ae}^{ix}\right)}^n+{\left({ae}^{-ix}\right)}^n$$ $$=1+\frac{{ae}^{ix}}{1-{ae}^{ix}}+\frac{{ae}^{-ix}}{1-{ae}^{-i}}=1+\frac{{ae}^{ix}-a^2+{ae}^{-ix}-a^2}{1-2acos\left(x\right)+a^2}$$ $$=1+\frac{2acos\left(x\right)-2a^2}{1-2acosx+a^2}=\frac{1-a^2}{1-2acos\left(x\right)+a^2}$$

Commented byqaz last updated on 30/Apr/21

$$thankyou.sir$$

Answered by mindispower last updated on 30/Apr/21

$$::$$ $$={\int }_0^{\pi }\frac{x^2}{1+8(1-{cos}^2\left(x\right))}dx$$ $$={\int }_0^{\pi }\frac{x^2}{(3-2\sqrt{2}cos\left(x\right))(3+2\sqrt{2}cos\left(x\right))}$$ $$=\frac{1}{6}({\int }_0^{\pi }\frac{x^2}{3-2\sqrt{2}cos\left(x\right)}dx+{\int }_0^{\pi }\frac{x^2}{3+2\sqrt{2}cos\left(x\right)}dx)$$ $$=\frac{1}{6}(A+B)$$ $$6A={\int }_0^{\pi }\frac{1-{\left(\frac{1}{\sqrt{2}}\right)}^2}{1+\frac{1}{{\left(\sqrt{2}\right)}^2}-2.\frac{1}{\sqrt{2}}cos\left(x\right)}{x}^{2}dx$$ $$6B={\int }_0^{\pi }\frac{1-{(-\frac{1}{\sqrt{2}})}^2}{1+{(-\frac{1}{\sqrt{2}})}^2+2.-\frac{1}{\sqrt{2}}cos\left(x\right)}{x}^{2}dx$$ $$A=\frac{1}{6}{\int }_0^{\pi }(x^2+2\sum _{n⩾1}{\left(\frac{1}{\sqrt{2}}\right)}^n{x}^{2}cos\left(nx\right))$$ $$B=\frac{1}{6}{\int }_0^{\pi }(x^2+2\sum _{n⩾1}{(-\sqrt{2})}^n{x}^{2}cos\left(nx\right))$$ $${\int }_0^{\pi }{x}^{2}cos\left(nx\right)dx=-\frac{2}{n}{\int }_0^{\pi }xsin\left(nx\right)dx$$ $$=\frac{2}{n^2}\left[\pi {(-1)}^n\right]$$ $$A=\frac{{\pi }^3}{18}+\frac{1}{3}\sum _{n⩾1}{\left(\frac{1}{\sqrt{2}}\right)}^n.\frac{2\pi }{n^2}\left({(-1)}^n\right)$$ $$B=\frac{{\pi }^3}{18}+\frac{1}{3}\sum _{n⩾1}{(-\frac{1}{\sqrt{2}})}^n.\frac{2}{n^2}\left({(-1)}^n\right)$$ $$A+B=\frac{{\pi }^3}{9}+\frac{1}{3}\sum _{n⩾1}{\left(\frac{1}{2}\right)}^n.\frac{\pi }{n^2}$$ $${li}_2\left(z\right)=\sum _{n⩾1}\frac{z^n}{n^2};\mathrm{\forall }\mid z\mid ⩽1$$ $$=\frac{{\pi }^3}{9}+\frac{\pi }{3}{Li}_2\left(\frac{1}{2}\right),{li}_2\left(\frac{1}{2}\right)=\frac{{\pi }^2}{12}-\frac{{ln}^2\left(2\right)}{2}$$ $$=\frac{{\pi }^3}{9}+\frac{\pi }{3}(\frac{{\pi }^2}{12}-\frac{{ln}^2\left(2\right)}{2})=\frac{5{\pi }^3}{36}-\frac{\pi {ln}^2\left(2\right)}{6}$$ $$$$

Commented bymnjuly1970 last updated on 30/Apr/21

$$bravobravosirpower...$$

Commented bymindispower last updated on 01/May/21

$$withepleasurSir$$

Answered by mathmax by abdo last updated on 01/May/21

$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{a}}^{\mathrm{n}}\cos \left(\mathrm{nx}\right)=\mathrm{Re}(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{a}}^{\mathrm{n}}{\mathrm{e}}^{\mathrm{inx}}-1)$$ $$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\left({\mathrm{ae}}^{\mathrm{ix}}\right)}^{\mathrm{n}}=\frac{1}{1-{\mathrm{ae}}^{\mathrm{ix}}}=\frac{1}{1-\mathrm{acosx}-\mathrm{iasinx}}=\frac{1-\mathrm{acosx}+\mathrm{iasinx}}{{(1-\mathrm{acosx})}^2+{\mathrm{a}}^2{\sin }^2\mathrm{x}}$$ $$\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{a}}^{\mathrm{n}}\cos \left(\mathrm{nx}\right)=\frac{1-\mathrm{acosx}}{1-2\mathrm{acosx}+{\mathrm{a}}^2}-1\Rightarrow$$ $$1+2\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{a}}^{\mathrm{n}}\cos \left(\mathrm{nx}\right)=1+\frac{2-2\mathrm{acosx}}{1-2\mathrm{acosx}+{\mathrm{a}}^2}-2$$ $$=\frac{2-2\mathrm{acosx}}{1-2\mathrm{acosx}+{\mathrm{a}}^2}-1=\frac{2-2\mathrm{acosx}-1+2\mathrm{acosx}-{\mathrm{a}}^2}{{\mathrm{a}}^2-2\mathrm{acosx}+1}=\frac{1-{\mathrm{a}}^2}{{\mathrm{a}}^2-2\mathrm{acosx}+1}$$

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