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Question Number 122323 by mathmax by abdo last updated on 15/Nov/20

$$\mathrm{calculate}{\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+2)....({\mathrm{x}}^2+\mathrm{n})}$$$$\mathrm{wth}\mathrm{n}\mathrm{integr}\mathrm{natural}\mathrm{and}\mathrm{n}⩾1$$

Commented by Dwaipayan Shikari last updated on 16/Nov/20

$${\int }_0^{\mathrm{\infty }}(\frac{1}{x^2+1}-\frac{1}{x^2+2})dx$$$$=\frac{\pi }{2}(1-\frac{1}{\sqrt{2}})$$$${\int }_0^{\mathrm{\infty }}\frac{1}{(x^2+1)(x^2+2)(x^2+3)}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{1}{(x^2+1)}-\frac{1}{(x^2+3)}-{\int }_0^{\mathrm{\infty }}\frac{1}{(x^2+2)}-\frac{1}{(x^2+3)}$$$$=\frac{1}{2}(\frac{\pi }{2}-\frac{\pi }{2\sqrt{3}})-(\frac{\pi }{2\sqrt{2}}-\frac{\pi }{2\sqrt{3}})$$$$=\frac{\pi }{4}-\frac{\pi }{2\sqrt{2}}+\frac{\pi }{4\sqrt{3}}$$$${\int }_0^{\mathrm{\infty }}\frac{1}{(x^2+1)(x^2+2)(x^2+3)(x^2+4)}dx$$$$=\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{1}{(x^2+1)(x^2+2)(x^2+3)}-\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{1}{(x^2+2)(x^2+3)(x^2+4)}$$$$=\frac{\pi }{12}-\frac{\pi }{6\sqrt{2}}+\frac{\pi }{6\sqrt{3}}-\frac{1}{6}{\int }_0^{\mathrm{\infty }}\frac{1}{x^2+2}-\frac{1}{x^2+4}+\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{1}{x^2+3}-\frac{1}{x^2+4}$$$$=\frac{\pi }{12}-\frac{\pi }{6\sqrt{2}}+\frac{\pi }{6\sqrt{3}}-\frac{1}{6}(\frac{\pi }{2\sqrt{2}}-\frac{\pi }{2\sqrt{4}})+\frac{1}{3}(\frac{\pi }{2\sqrt{3}}-\frac{\pi }{2\sqrt{4}})$$$$=\frac{\pi }{12}-\frac{\pi }{4\sqrt{2}}+\frac{\pi }{3\sqrt{3}}$$$$....$$$$$$

Answered by mathmax by abdo last updated on 16/Nov/20

$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{u}\right)=\frac{1}{(\mathrm{u}+1)(\mathrm{u}+2)....(\mathrm{u}+\mathrm{n})}=\frac{1}{\prod _{\mathrm{k}=1}^{\mathrm{n}}(\mathrm{u}+\mathrm{k})}$$$$\mathrm{F}\left(\mathrm{u}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{\mathrm{a}}_{\mathrm{k}}}{\mathrm{u}+\mathrm{k}}\mathrm{with}{\mathrm{a}}_{\mathrm{k}}={\lim }_{\mathrm{u}\to -\mathrm{k}}(\mathrm{u}+\mathrm{k})\mathrm{F}\left(\mathrm{u}\right)$$$$\mathrm{we}\mathrm{have}\mathrm{F}\left(\mathrm{u}\right)=\frac{1}{(\mathrm{u}+1)....(\mathrm{u}+\mathrm{k}-1)(\mathrm{u}+\mathrm{k})(\mathrm{u}+\mathrm{k}+1)...(\mathrm{u}+\mathrm{n})}$$$$(\mathrm{u}+\mathrm{k})\mathrm{F}\left(\mathrm{u}\right)=\frac{1}{(\mathrm{u}+1)(\mathrm{u}+2)...(\mathrm{u}+\mathrm{k}-1)(\mathrm{u}+\mathrm{k}+1)....(\mathrm{u}+\mathrm{n})}\Rightarrow$$$${\mathrm{a}}_{\mathrm{k}}=\frac{1}{(-\mathrm{k}+1)(-\mathrm{k}+2)....(-1)(-\mathrm{k}+\mathrm{k}+1)(-\mathrm{k}+\mathrm{k}+2)...(-\mathrm{k}+\mathrm{n})}$$$$=\frac{1}{{(-1)}^{\mathrm{k}-1}(\mathrm{k}-1)!}\times \frac{1}{(\mathrm{n}-\mathrm{k})!}=\frac{{(-1)}^{\mathrm{k}-1}}{(\mathrm{k}-1)!(\mathrm{n}-\mathrm{k})!}\Rightarrow$$$$\mathrm{F}\left(\mathrm{u}\right)=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}-1}}{(\mathrm{k}-1)!(\mathrm{n}-\mathrm{k})!(\mathrm{u}+\mathrm{k})}$$$${2\mathrm{A}}_{\mathrm{n}}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\mathrm{F}\left({\mathrm{x}}^2\right)\mathrm{dx}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}-1}}{(\mathrm{k}-1)!(\mathrm{n}-\mathrm{k})!}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{k}}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{k}}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{(\mathrm{x}-\mathrm{i}\sqrt{\mathrm{k}})(\mathrm{x}+\mathrm{i}\sqrt{\mathrm{k}})}=2\mathrm{i}\pi \mathrm{Res}(\mathrm{f},\mathrm{i}\sqrt{\mathrm{k}})$$$$=2\mathrm{i}\pi \times \frac{1}{2\mathrm{i}\sqrt{\mathrm{k}}}=\frac{\pi }{\sqrt{\mathrm{k}}}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\frac{\pi }{2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}-1}}{(\mathrm{k}-1)!(\mathrm{n}-\mathrm{k})!\sqrt{\mathrm{k}}}$$

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