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Question Number 93873 by mathmax by abdo last updated on 15/May/20

$$find{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{{(x^2+2x+4)}^3}$$

Commented by mathmax by abdo last updated on 16/May/20

$$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{{(x^2+2x+4)}^3}\Rightarrow I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{{({(x+1)}^2+3)}^3}$$$${=}_{x+1=\sqrt{3}t}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\sqrt{3}dt}{27{(t^2+1)}^3}=\frac{\sqrt{3}}{27}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{{(t^2+1)}^3}$$$$let\phi \left(z\right)=\frac{1}{{(z^2+1)}^3}\Rightarrow \phi \left(z\right)=\frac{1}{{(z-i)}^3{(z+i)}^3}sothepolesof\phi are$$$$iand-i\left(triples\right)and{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)$$$$Res(\phi ,i)={lim}_{z\to i}\frac{1}{(3-1)!}{\left\{{(z-i)}^3\phi \left(z\right)\right\}}^{\left(2\right)}$$$$=\frac{1}{2}{lim}_{z\to i}{\left\{{(z+i)}^{-3}\right\}}^{\left(2\right)}=\frac{1}{2}{lim}_{z\to i}{\{-3{(z+i)}^{-4}\}}^{\left(1\right)}$$$$=\frac{1}{2}{lim}_{z\to i}12{(z+i)}^{-5}=6{\left(2i\right)}^{-5}=\frac{6}{{\left(2i\right)}^5}=\frac{6}{32i}=\frac{3}{16i}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \times \frac{3}{16i}=\frac{3\pi }{8}\Rightarrow I=\frac{\sqrt{3}}{27}\times \frac{3\pi }{8}=\frac{\pi \sqrt{3}}{72}$$$$★I=\frac{\pi \sqrt{3}}{72}★$$

Answered by Ar Brandon last updated on 15/May/20

$${\mathrm{L}}_{\mathrm{n}}=\int \frac{\mathrm{Ax}+\mathrm{B}}{{({\mathrm{x}}^2+\mathrm{px}+\mathrm{q})}^{\mathrm{n}}}\mathrm{dx}\Rightarrow {\mathrm{L}}_{\mathrm{n}}=(\mathrm{B}-\frac{\mathrm{Ap}}{2}){\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{[{(\mathrm{x}+\frac{\mathrm{p}}{2})}^2+{\left(\sqrt{\mathrm{q}-\frac{{\mathrm{p}}^2}{4}}\right)}^2]}^{\mathrm{n}}}\mathrm{dx}$$$$\mathrm{A}=0,\mathrm{B}=1,\mathrm{p}=2,\mathrm{q}=4,\mathrm{n}=3$$$$\Rightarrow {\mathrm{L}}_3={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^2+2\mathrm{x}+4)}^3}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{[{(\mathrm{x}+1)}^2+{\left(\sqrt{3}\right)}^2]}^3}$$$${\mathrm{J}}_{\mathrm{n}}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{{[{\mathrm{x}}^2+{\mathrm{u}}^2]}^{\mathrm{n}}}\mathrm{dx}\Rightarrow {\mathrm{J}}_{\mathrm{n}}=\frac{2\mathrm{n}-3}{{2\mathrm{u}}^2(\mathrm{n}-1)}{\mathrm{J}}_{\mathrm{n}-1}$$$${\mathrm{J}}_1={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{(\mathrm{x}+1)}^2+{\left(\sqrt{3}\right)}^2}=\frac{\sqrt{3}}{3}\lfloor {\tan }^{-1}\left[\frac{\mathrm{x}+1}{\sqrt{3}}\right]{\rfloor }_{-\mathrm{\infty }}^{+\mathrm{\infty }}=\frac{\pi \sqrt{3}}{3}$$$${\mathrm{J}}_2=\frac{\pi \sqrt{3}}{18}\Rightarrow {\mathrm{J}}_3=\frac{3}{12}\cdot \frac{\pi \sqrt{3}}{18}=\frac{\pi \sqrt{3}}{72}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^2+2\mathrm{x}+4)}^3}=\frac{\pi \sqrt{3}}{72}$$

Commented by Ar Brandon last updated on 15/May/20

In case of necessity please check my solution to Q.93639 for Ln and Jn ��

Commented by mathmax by abdo last updated on 16/May/20

$$thankyousir.$$

Commented by Ar Brandon last updated on 16/May/20

cheers ��

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