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Question Number 71663 by mathmax by abdo last updated on 18/Oct/19

$$calculate{A}_n={\int }_0^{\mathrm{\infty }}\frac{dx}{(x^2+1)(x^2+2)....(x^2+n)}$$$$withnintegrandn⩾1$$

Commented by mathmax by abdo last updated on 18/Oct/19

$$wehave2A_n={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{(x^2+1)(x^2+2)..(x^2+n)}let$$$$\phi \left(z\right)=\frac{1}{(z^2+1)(z^2+2)...(z^2+n)}=\frac{1}{\prod _{k=1}^n(z^2+k)}$$$$=\frac{1}{\prod _{k=1}^n(z-i\sqrt{k})(z+i\sqrt{k})}sothepolesof\phi are\stackrel{-}{+}i\sqrt{k}$$$$and{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \sum _{k=1}^{n}Res(\phi ,i\sqrt{k})$$$$Res(\phi ,i\sqrt{k})={lim}_{z\to i\sqrt{k}}(z-i\sqrt{k})\phi \left(z\right)$$$$=\frac{1}{\prod _{k=1}^n\left(2i\sqrt{k}\right)\prod _{j=1andj\ne k}^n(i\sqrt{k}-i\sqrt{j})}=\frac{1}{{\left(2i\right)}^n\sqrt{n!}\prod _{j=1andj\ne k}^n(i\sqrt{k}-i\sqrt{j})}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=\frac{2i\pi }{{\left(2i\right)}^n\sqrt{n!}\prod _{j=1}^n(i\sqrt{k}-i\sqrt{j})}\Rightarrow$$$$A_n=\sum _{k=1}^n\frac{i\pi }{{\left(2i\right)}^n\sqrt{n!}\prod _{j=1andj\ne k}^n(i\sqrt{k}-i\sqrt{j})}$$

Answered by mind is power last updated on 18/Oct/19

$$\frac{1}{({\mathrm{x}}^2+1).....({\mathrm{x}}^2+\mathrm{n})}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}(\frac{{\mathrm{a}}_{\mathrm{k}}}{(\mathrm{x}-\mathrm{i}\sqrt{\mathrm{k}})}+\frac{{\mathrm{b}}_{\mathrm{k}}}{(\mathrm{x}+\mathrm{i}\sqrt{\mathrm{k}}})$$$${\mathrm{a}}_{\mathrm{k}}={\lim }_{}(\mathrm{x}-\mathrm{i}\sqrt{\mathrm{k}}).\frac{1}{({\mathrm{x}}^2+1)......({\mathrm{x}}^2+\mathrm{n})}$$$$=\underset{\mathrm{l}=1,\mathrm{l}\ne \mathrm{k}}{\prod ^{\mathrm{n}}}\frac{1}{2\mathrm{i}\sqrt{\mathrm{k}}.(\mathrm{l}-\mathrm{k})}$$$${\mathrm{b}}_{\mathrm{k}}=\underset{\mathrm{l}=1,\mathrm{l}\ne \mathrm{k}}{\prod ^{\mathrm{n}}}\frac{1}{-2\mathrm{i}\sqrt{\mathrm{k}}(\mathrm{l}-\mathrm{k})}$$$$\frac{1}{({\mathrm{x}}^2+1).....({\mathrm{x}}^2+\mathrm{n})}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}(\underset{\mathrm{l}=1,\mathrm{l}\ne \mathrm{k}}{\prod ^{\mathrm{n}}}\frac{1}{2\mathrm{i}\sqrt{\mathrm{k}}(\mathrm{l}-\mathrm{k})}(.\frac{1}{\mathrm{x}-\mathrm{i}\sqrt{\mathrm{k}}}-\frac{1}{\mathrm{x}+\mathrm{i}\sqrt{\mathrm{k}}})$$$$=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\left(\underset{\mathrm{l}=1,\mathrm{k}\ne \mathrm{l}}{\prod ^{\mathrm{n}}}\frac{1}{(\mathrm{l}-\mathrm{k})({\mathrm{x}}^2+\mathrm{k})}\right)$$$${\mathrm{A}}_{\mathrm{n}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\underset{\mathrm{l}=1,\mathrm{l}\ne \mathrm{k}}{\prod ^{\mathrm{n}}}\frac{1}{(\mathrm{l}-\mathrm{k})}\underset{0}{\stackrel{+\mathrm{\infty }}{\int }}\frac{\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{k}}$$$$\int \frac{\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{k}}=\frac{1}{\sqrt{\mathrm{k}}}.{\tan }^{-1}\left(\frac{\mathrm{x}}{\sqrt{\mathrm{k}}}\right)+\mathrm{c}$$$${\int }_0^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{k}}=\frac{\pi }{2\sqrt{\mathrm{k}}}$$$${\mathrm{A}}_{\mathrm{n}}=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\underset{\mathrm{l}=1,\mathrm{l}\ne \mathrm{k}}{\prod ^{\mathrm{n}}}\frac{\pi }{2\sqrt{\mathrm{k}}.(\mathrm{l}-\mathrm{k})}$$$$$$$$$$$$$$$$$$$$$$

Commented by mathmax by abdo last updated on 18/Oct/19

$$thankyousir$$

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