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Question Number 29975 by abdo imad last updated on 14/Feb/18

$$letgive0<\alpha <1$$ $$1)provethat\pi coth\left(\pi \alpha \right)-\frac{1}{\alpha }=\sum _{n=1}^{\mathrm{\infty }}\frac{2\alpha }{{\alpha }^2+n^2}.$$ $$2)byintegrationon[0,1]find\prod _{n=1}^{\mathrm{\infty }}(1+\frac{1}{n^2}).$$

Commented byabdo imad last updated on 16/Feb/18

$$1)letdeveloppthe2\pi periodicfunctionf\left(x\right)=ch\left(\alpha x\right)$$ $$f\left(x\right)=\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_{n}cos\left(nx\right)$$ $$a_n=\frac{2}{T}{\int }_{\left[T\right]}f\left(x\right)cos\left(nx\right)dx=\frac{2}{2\pi }{\int }_{-\pi }^{\pi }ch\left(\alpha x\right)cos\left(nx\right)dx$$ $$=\frac{2}{\pi }{\int }_0^{\pi }ch\left(\alpha x\right)cos\left(nx\right)dx\Rightarrow \frac{\pi }{2}{a}_n={\int }_0^{\pi }\frac{e^{\alpha x}+e^{-\alpha x}}{2}cos\left(nx\right)dx$$ $$\pi a_n={\int }_0^{\pi }{e}^{\alpha x}cos\left(nx\right)dx+{\int }_0^{\pi }{e}^{-\alpha x}cos\left(nx\right)dx=I\left(\alpha \right)+I(-\alpha )$$ $$I\left(\alpha \right)=Re\left({\int }_0^{\pi }{e}^{\alpha x+inx}dx\right)=Re\left({\int }_0^{\pi }{e}^{(\alpha +in)x}dx\right)but$$ $${\int }_0^{\pi }{e}^{(\alpha +in)x}dx={\left[\frac{1}{\alpha +in}{e}^{(\alpha +in)x}\right]}_0^{\pi }=\frac{1}{\alpha +in}(e^{\alpha \pi }{(-1)}^n-1)$$ $$=\frac{\alpha -in}{{\alpha }^2+n^2}({(-1)}^n{e}^{\alpha \pi }-1)\Rightarrow I\left(\alpha \right)=\frac{\alpha ({(-1)}^n{e}^{\alpha \pi }-1)}{{\alpha }^2+n^2}$$ $$I(-\alpha )=\frac{-\alpha ({(-1)}^n{e}^{-\alpha \pi }-1)}{{\alpha }^2+n^2}$$ $$\pi a_n=\frac{\alpha {(-1)}^n(e^{\alpha \pi }-e^{-\alpha \pi })}{{\alpha }^2+n^2}=\frac{2\alpha {(-1)}^{n}sh\left(\alpha \pi \right)}{{\alpha }^2+n^2}\Rightarrow$$ $$a_n=\frac{2}{\pi }\alpha sh\left(\alpha \pi \right)\frac{{(-1)}^n}{{\alpha }^2+n^2}and{a}_0=\frac{2}{\pi }\frac{sh\left(\alpha \pi \right)}{\alpha }\Rightarrow$$ $$ch\left(\alpha x\right)=\frac{sh\left(\alpha \pi \right)}{\pi \alpha }+\frac{2\alpha sh\left(\alpha \pi \right)}{\pi }\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{{\alpha }^2+n^2}cos\left(nx\right)$$ $$x=\pi \Rightarrow ch\left(\pi \alpha \right)=\frac{sh\left(\alpha \pi \right)}{\pi \alpha }+\frac{2\alpha sh\left(\alpha \pi \right)}{\pi }\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\alpha }^2+n^2}\Rightarrow$$ $$coth\left(\pi \alpha \right)=\frac{1}{\pi \alpha }+\frac{2\alpha }{\pi }\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{\alpha }^2+n^2}\Rightarrow$$ $$\pi coth\left(\pi \alpha \right)-\frac{1}{\alpha }=\sum _{n=1}^{\mathrm{\infty }}\frac{2\alpha }{{\alpha }^2+n^2}.$$

Commented byabdo imad last updated on 16/Feb/18

$$wehave{\int }_0^1(\pi coth\left(\pi \alpha \right)-\frac{1}{\alpha })d\alpha =\sum _{n=1}^{\mathrm{\infty }}{\int }_0^1\frac{2\alpha }{{\alpha }^2+n^2}d\alpha$$ $$=\sum _{n=1}^{\mathrm{\infty }}{\left[ln({\alpha }^2+n^2)\right]}_0^1=\sum _{n=1}^{\mathrm{\infty }}ln(1+n^2)-ln\left(n^2\right)=\sum _{n=1}^{\mathrm{\infty }}ln(1+\frac{1}{n^2})$$ $$=ln\left(\prod _{n=1}^{\mathrm{\infty }}(1+\frac{1}{n^2})\right)\Rightarrow$$ $$\prod _{n=1}^{\mathrm{\infty }}(1+\frac{1}{n^2})=e^{{\int }_0^1(\pi coth\left(\pi \alpha \right)-\frac{1}{\alpha })d\alpha }butch.\alpha \pi =tgive$$ $${\int }_0^1\frac{\alpha \pi coth\left(\alpha \pi \right)-1}{\alpha }d\alpha ={\int }_0^{\pi }\frac{tcotht-1}{\frac{t}{\pi }}\frac{dt}{\pi }$$ $$={\int }_0^{\pi }\frac{tcoth\left(t\right)-1}{t}dt={\int }_0^{\pi }(\frac{e^t-e^{-t}}{e^t+e^{-t}}-\frac{1}{t})dt....becontinued...$$ $$\pi$$

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