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Question Number 148249 by mnjuly1970 last updated on 26/Jul/21

Commented by Kamel last updated on 26/Jul/21

$$$$$${\mathrm{\Omega }}_n={\int }_0^{+\mathrm{\infty }}\frac{cos\left(\pi x\right)}{(x^2+1^2)(x^2+2^2)....(x^2+n^2)}dx$$$$\frac{1}{(x^2+1^2)(x^2+2^2)....(x^2+n^2)}=\underset{k=1}{\sum ^n}\frac{a_k}{(x^2+k^2)}...\left(E\right)$$$$X=x^2,\left(E\right)\times (x^2+k^2),X\to -k^2$$$$a_k=\frac{1}{(1^2-k^2)(2^2-k^2)...({(k-1)}^2-k^2)({(k+1)}^2-k^2)({(k+2)}^2-k^2)...(n^2-k^2)}$$$$=\frac{{(-1)}^{k-1}k!\left(2k\right)}{(k-1)!(2k-1)!(n-k)!(2k+1)(2k+2)(2k+3)...(n+k)}$$$$=\frac{2{(-1)}^{k-1}{k}^2}{(n+k)!(n-k)!}$$$${\mathrm{\Omega }}_n=2\underset{k=1}{\sum ^n}\frac{k^2{(-1)}^{k-1}}{(n+k)!(n-k)!}{\int }_0^{+\mathrm{\infty }}\frac{cos\left(\pi x\right)}{x^2+k^2}dx$$$${\int }_0^{+\mathrm{\infty }}\frac{cos\left(\pi x\right)}{x^2+k^2}dx=\frac{\pi }{2k}{e}^{-k\pi }$$$$\therefore {\mathrm{\Omega }}_n=\pi \underset{k=1}{\sum ^n}\frac{{(-1)}^{k-1}{ke}^{-k\pi }}{(n+k)!(n-k)!}$$$$\therefore {\int }_0^{+\mathrm{\infty }}\frac{cos\left(\pi x\right)}{(x^2+1^2)(x^2+2^2)....(x^2+n^2)}dx=\pi \underset{k=1}{\sum ^n}\frac{{(-1)}^{k-1}{ke}^{-k\pi }}{(n+k)!(n-k)!}$$

Answered by Olaf_Thorendsen last updated on 26/Jul/21

$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\pi x\right)}{\underset{k=1}{\prod ^n}(x^2+k^2)}dx$$$$\mathrm{\Omega }=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\cos \left(\pi x\right)}{\underset{k=1}{\prod ^n}(x^2+k^2)}dx$$$$\mathrm{Let}f\left(x\right)=\frac{\cos \left(\pi x\right)}{\underset{k=1}{\prod ^n}(x^2+k^2)}dx$$$$\mathrm{The}\mathrm{function}f\mathrm{is}\mathrm{even}:$$$$\mathrm{\Omega }=\frac{1}{2}\mathrm{Re}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x}}{\underset{k=1}{\prod ^n}(x^2+k^2)}dx=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\pi x}}{\underset{k=1}{\prod ^n}(x^2+k^2)}dx$$$$\mathrm{\Omega }=2i\pi \underset{k=1}{\sum ^n}\mathrm{Res}(f\left(x\right),ik)$$$$\mathrm{Res}(f\left(x\right),ik)=\underset{x\to \mathrm{i}k}{\lim }\left(\frac{e^{i\pi x}}{(x+ik)\underset{\underset{p\ne k}{p=1}}{\prod ^n}(x^2+p^2)}\right)$$$$\mathrm{Res}(f\left(x\right),ik)=\frac{e^{-k\pi }}{\left(2ik\right)\underset{\underset{p\ne k}{p=1}}{\prod ^n}(p^2-k^2)}$$$$\mathrm{\Omega }=\pi \underset{k=1}{\sum ^n}\frac{e^{-k\pi }}{k\underset{\underset{p\ne k}{p=1}}{\prod ^n}(p^2-k^2)}$$$$...tobecontinued...$$

Answered by mathmax by abdo last updated on 26/Jul/21

$${\mathrm{\Psi }}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{\cos \left(\pi \mathrm{x}\right)}{({\mathrm{x}}^2+1^2)({\mathrm{x}}^2+2^2).....({\mathrm{x}}^2+{\mathrm{n}}^2)}\mathrm{dx}\Rightarrow$$$$2{\mathrm{\Psi }}_{\mathrm{n}}=\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{i}\pi \mathrm{x}}}{({\mathrm{x}}^2+1^2)({\mathrm{x}}^2+2^2)....({\mathrm{x}}^2+{\mathrm{n}}^2)}\mathrm{dx}\right)$$$$\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{i}\pi \mathrm{z}}}{({\mathrm{z}}^2+1^2)({\mathrm{z}}^2+2^2)....({\mathrm{z}}^2+{\mathrm{n}}^2)}\Rightarrow$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{i}\pi \mathrm{z}}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})(\mathrm{z}+2\mathrm{i})(\mathrm{z}-2\mathrm{i})....(\mathrm{z}-\mathrm{ni})(\mathrm{z}+\mathrm{ni})}$$$$\mathrm{the}\mathrm{poles}\mathrm{of}\phi \mathrm{are}\stackrel{-}{+}\mathrm{ki}\mathrm{rdsidus}\mathrm{theorem}\mathrm{give}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{Res}(\phi ,\mathrm{ki})$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{i}\pi \mathrm{z}}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}-2\mathrm{i})....(\mathrm{z}-\mathrm{ni})(\mathrm{z}+\mathrm{i})(\mathrm{z}+2\mathrm{i})...(\mathrm{z}+\mathrm{ni})}$$$$=\frac{{\mathrm{e}}^{\mathrm{i}\pi \mathrm{z}}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}-2\mathrm{i})....(\mathrm{z}-(\mathrm{k}-1)\mathrm{i})(\mathrm{z}-\mathrm{ki})(\mathrm{z}-(\mathrm{k}+1)\mathrm{i})...(\mathrm{z}-\mathrm{ni})\prod _{\mathrm{k}=1}^{\mathrm{n}}(\mathrm{z}+\mathrm{ki})}$$$$\Rightarrow \mathrm{Res}(\phi ,\mathrm{ki})=\frac{{\mathrm{e}}^{\mathrm{i}\pi \left(\mathrm{ki}\right)}}{(\mathrm{ki}-\mathrm{i})(\mathrm{ki}-2\mathrm{i})....\left(\mathrm{i}\right)(-\mathrm{i})...(\mathrm{ki}-\mathrm{ni})\prod _{\mathrm{k}=1}^{\mathrm{n}}\left(2\mathrm{ki}\right)}$$$$=\frac{{\mathrm{e}}^{-\pi \mathrm{k}}}{{\mathrm{i}}^{\mathrm{k}-1}(\mathrm{k}-1)!{(-\mathrm{i})}^{\mathrm{n}-\mathrm{k}}(\mathrm{n}-\mathrm{k})!{\left(2\mathrm{i}\right)}^{\mathrm{n}}\mathrm{n}!}$$$$=\frac{{\mathrm{e}}^{-\mathrm{k}\pi }}{{(-1)}^{\mathrm{n}-\mathrm{k}}{\mathrm{i}}^{\mathrm{n}-1}{\left(\mathrm{i}\right)}^{\mathrm{n}}{2}^{\mathrm{n}}(\mathrm{k}-1)!(\mathrm{n}-\mathrm{k})!\mathrm{n}!}$$$$=\mathrm{i}\frac{{\mathrm{e}}^{-\mathrm{k}\pi }}{{(-1)}^{\mathrm{k}}{2}^{\mathrm{n}}(\mathrm{k}-1)!(\mathrm{n}-\mathrm{k})!\mathrm{n}!}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=\frac{2\mathrm{i}\pi }{\mathrm{n}!\times 2^{\mathrm{n}}}\mathrm{i}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}{\mathrm{e}}^{-\mathrm{k}\pi }}{(\mathrm{k}-1)!(\mathrm{n}-\mathrm{k})!}$$$$=-\frac{\pi }{\mathrm{n}!2^{\mathrm{n}-1}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}{\mathrm{e}}^{-\mathrm{k}\pi }}{(\mathrm{k}-1)!(\mathrm{n}-\mathrm{k})!}$$$$=\frac{\pi }{\mathrm{n}!2^{\mathrm{n}-1}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}-1}{\mathrm{e}}^{-\mathrm{k}\pi }}{(\mathrm{k}-1)!(\mathrm{n}-\mathrm{k})!}=2{\mathrm{\Psi }}_{\mathrm{n}}\Rightarrow$$$${\mathrm{\Psi }}_{\mathrm{n}}=\frac{\pi }{2^{\mathrm{n}}\mathrm{n}!}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}-1}{\mathrm{e}}^{-\mathrm{k}\pi }}{(\mathrm{k}-1)!(\mathrm{n}-\mathrm{k})!}$$

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