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Question Number 37601 by prof Abdo imad last updated on 15/Jun/18

$$letgiveninyehrnatural⩾1findtbevalueof$$$$A_n={\int }_0^{\mathrm{\infty }}\frac{dx}{(x^2+1)(x^2+2)....(x^2+n)}$$

Commented by math khazana by abdo last updated on 17/Jun/18

$$2A_n={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{(x^2+1)(x^2+2)....(x^2+n)}$$$$let\phi \left(z\right)=\frac{1}{(z^2+1)(z^2+2).....(z^2+n)}$$$$\phi \left(z\right)=\frac{1}{\prod _{k=1}^n(z^2+k)}=\frac{1}{\prod _{k=1}^n(z-i\sqrt{k})(z+i\sqrt{k})}$$$$=\frac{1}{\prod _{k=1}^n(z-i\sqrt{k})\prod _{k=1}^n(z+i\sqrt{k})}sothepolesof$$$$\phi arei\sqrt{k}and-i\sqrt{k}withk\in \{1,2,...,n\}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \sum _{z_k\in w^{+}}Res(\phi ,z_k)with$$$$w^{+}=\{z\in C/Im\left(z\right)>0\}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \sum _{k=1}^{n}Res(\phi ,i\sqrt{k})$$$$Res(\phi ,i\sqrt{k})=\frac{1}{p^{{}^{\prime }}\left(i\sqrt{k}\right)}with$$$$p\left(x\right)=(x^2+1)(x^2+2)...(x^2+n)$$$$letq\left(t\right)=(t+1)(t+2)....(t+n)\Rightarrow$$$$q^{{}^{\prime }}\left(t\right)=\sum _{k=1}^n\prod _{p=1_{p\ne -k}}^n(x+p)$$$$p\left(x\right)=q\left(x^2\right)\Rightarrow p^{{}^{\prime }}\left(x\right)=2x{q}^{{}^{\prime }}\left(x^2\right)\Rightarrow$$$$p^{{}^{\prime }}\left(x\right)=2x\sum _{k=1}^n\prod _{p=1_{p\ne -k}}^n(x^2+p)\Rightarrow$$$$p^{{}^{\prime }}\left(i\sqrt{m}\right)=2i\sqrt{m}\sum _{k=1}^n\prod _{p=1_{p\ne -k}}^n(p-m)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \sum _{m=1}^n\frac{1}{2i\sqrt{m}}\left\{\sum _{k=1}^n\prod _{p=1_{p\ne -k}}^n(p-m)\right\}$$$$=\frac{\pi }{\sqrt{m}}\sum _{m=1}^n\left\{\sum _{k=1}^n\prod _{p=1_{p\ne -k}}^n(p-m)\right\}.$$

Commented by math khazana by abdo last updated on 17/Jun/18

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=\frac{\pi }{\sqrt{m}}\sum _{m=1}^n{\left\{\sum _{m=1}^n\prod _{p=1}^n(p-m)\right\}}^{-1}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \sum _{m=1}^n\frac{1}{2i\sqrt{m}\sum _{k=1}^n\prod _{p=1_{p\ne -k}}^n(p-m)}$$$$=\sum _{m=1}^n\frac{\pi }{\sqrt{m}}{\left\{\sum _{k=1}^n\prod _{p=1_{p\ne -k}}^n(p-m)\right\}}^{-1}$$

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