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Question Number 34320 by abdo mathsup 649 cc last updated on 04/May/18

$$calculate{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2+1-i}$$

Commented by math khazana by abdo last updated on 05/May/18

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2+1-i}=Iweconsiderthecomplex$$$$function\phi \left(z\right)=\frac{1}{z^2+1-i}polesof\phi ?$$$$\phi \left(z\right)=\frac{1}{z^2-(-1+i)}=\frac{1}{z^2-\sqrt{2}{e}^{i\frac{3\pi }{4}}}$$$$=\frac{1}{(z-2^{\frac{1}{4}}{e}^{i\frac{3\pi }{8}})(z+2^{\frac{1}{4}}{e}^{i\frac{3\pi }{8}})}sothepolesof\phi are$$$$z_0=2^{\frac{1}{4}}{e}^{i\frac{3\pi }{8}}and{z}_1=-2^{\frac{1}{4}}{e}^{i\frac{3\pi }{8}}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,z_0)$$$$Res(\phi ,z_0)=\frac{1}{2z_0}=\frac{1}{2^{\frac{5}{4}}{e}^{i\frac{3\pi }{8}}}=2^{-\frac{5}{4}}{e}^{-i\frac{3\pi }{8}}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi .2^{-\frac{5}{4}}{e}^{-i\frac{3\pi }{8}}$$$$=2^{-\frac{1}{4}}i\pi (cos\left(\frac{3\pi }{8}\right)-isin\left(\frac{3\pi }{8}\right))=I.$$$$$$

Commented by math khazana by abdo last updated on 05/May/18

$$remarkwehave$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2+1-i}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2+1+i}{{(x^2+1)}^2+1}dx$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2+1}{{(x^2+1)}^2+1}dx+i{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{{(x^2+1)}^2+1}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2+1}{{(x^2+1)}^2+1}dx=\pi 2^{-\frac{1}{4}}sin\left(\frac{3\pi }{8}\right)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{{(x^2+1)}^2+1}=\pi 2^{-\frac{1}{4}}cos\left(\frac{3\pi }{8}\right).$$

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