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Question Number 59893 by aliesam last updated on 15/May/19

Commented by maxmathsup by imad last updated on 15/May/19

$$methodusingtheformuae{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{sin\left(\pi a\right)}if0<a<1$$$$letI={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2+4}{x^4+16}\Rightarrow I{=}_{x=2t}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{4t^2+4}{16t^4+16}\left(2\right)dt=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{t^2+1}{t^4+1}dt$$$$={\int }_0^{\mathrm{\infty }}\frac{t^2+1}{t^4+1}dt={\int }_0^{\mathrm{\infty }}\frac{t^2}{t^4+1}dt+{\int }_0^{\mathrm{\infty }}\frac{dt}{t^4+1}$$$$changement{t}^4=ugivet=u^{\frac{1}{4}}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{t^2}{t^4+1}dt={\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{2}}}{1+u}\frac{1}{4}{u}^{\frac{1}{4}-1}du$$$$=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{3}{4}-1}}{1+u}du=\frac{1}{4}\frac{\pi }{sin\left(\frac{3\pi }{4}\right)}=\frac{\pi }{4.\frac{1}{\sqrt{2}}}=\frac{\pi \sqrt{2}}{4}$$$$samechangementgive{\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^4}={\int }_0^{\mathrm{\infty }}\frac{1}{4}\frac{u^{\frac{1}{4}-1}}{1+u}du=\frac{1}{4}\frac{\pi }{sin\left(\frac{\pi }{4}\right)}=\frac{\pi }{4\frac{1}{\sqrt{2}}}=\frac{\pi \sqrt{2}}{4}$$$$\Rightarrow I=\frac{\pi \sqrt{2}}{4}+\frac{\pi \sqrt{2}}{4}=\frac{2\pi \sqrt{2}}{4}=\frac{\pi \sqrt{2}}{2}\Rightarrow ★I=\frac{\pi }{\sqrt{2}}★$$

Commented by malwaan last updated on 15/May/19

$$greatsir$$$$but;cananyonesolveit$$$$withoutusingthisformulae$$

Commented by maxmathsup by imad last updated on 16/May/19

$$yesthisintegralisalsosolvedbyresidustheoremiwillpostthemethod$$$$soon...$$

Commented by malwaan last updated on 16/May/19

$$thankssir$$$$Iamwaiting...$$

Commented by maxmathsup by imad last updated on 17/May/19

$$residusmethodletA={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2+4}{x^4+16}dx\Rightarrow A{=}_{x=2t}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{4t^2+4}{16(t^4+1)}2dt$$$$\Rightarrow 2A={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{t^2+1}{t^4+1}dtletW\left(z\right)=\frac{z^2+1}{z^4+1}polesofW!$$$$W\left(z\right)=\frac{z^2+1}{(z^2-i)(z^2+i)}=\frac{z^2+1}{(z-e^{\frac{i\pi }{4}})(z+e^{\frac{i\pi }{4}})(z-e^{-\frac{i\pi }{4}})(z+e^{-\frac{i\pi }{4}})}sothepolesofW$$$$are\stackrel{-}{+}{e}^{\frac{i\pi }{4}}and\stackrel{-}{+}{e}^{-\frac{i\pi }{4}}residustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi \{Res(W,e^{\frac{i\pi }{4}})+Res(W,-e^{-\frac{i\pi }{4}})\}$$$$Res(W,e^{\frac{i\pi }{4}})={lim}_{z\to e^{\frac{i\pi }{4}}}(z-e^{\frac{i\pi }{4}})W\left(z\right)=\frac{i+1}{2e^{\frac{i\pi }{4}}\left(2i\right)}=\frac{1+i}{4i}{e}^{-\frac{i\pi }{4}}$$$$Res(W,-e^{-\frac{i\pi }{4}})={lim}_{z\to -e^{-\frac{i\pi }{4}}}(z+e^{-\frac{i\pi }{4}})W\left(z\right)=\frac{-i+1}{-2e^{-\frac{i\pi }{4}}(-2i)}=\frac{(1-i)e^{\frac{i\pi }{4}}}{4i}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi (\frac{1+i}{4i}{e}^{-\frac{i\pi }{4}}+\frac{1-i}{4i}{e}^{\frac{i\pi }{4}}\}=\frac{\pi }{2}\left(2Re(1+i)e^{-\frac{i\pi }{4}}\right)$$$$=\pi Re\left\{(1+i)e^{-\frac{i\pi }{4}}\right\}but(1+i)e^{-i\frac{\pi }{4}}=\sqrt{2}{e}^{\frac{i\pi }{4}}{e}^{-\frac{i\pi }{4}}=\sqrt{2}\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=\pi \sqrt{2}\Rightarrow$$$$2A=\pi \sqrt{2}\Rightarrow A=\frac{\pi \sqrt{2}}{2}\Rightarrow A=\frac{\pi }{\sqrt{2}}.$$

Commented by malwaan last updated on 18/May/19

$$Thankyousomuchsir$$

Commented by maxmathsup by imad last updated on 18/May/19

$$youarewelcome.$$

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