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Question Number 67850 by mathmax by abdo last updated on 01/Sep/19

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

Commented by mathmax by abdo last updated on 01/Sep/19

$$letI={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2-z}letz={re}^{i\theta }\Rightarrow I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x^2-{\left(\sqrt{r}{e}^{\frac{i\theta }{2}}\right)}^2}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{(x-\sqrt{r}{e}^{\frac{i\theta }{2}})(x+\sqrt{r}{e}^{\frac{i\theta }{2}})}=\frac{1}{2\sqrt{r}{e}^{\frac{i\theta }{2}}}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\{\frac{1}{x-\sqrt{r}{e}^{\frac{i\theta }{2}}}-\frac{1}{x+\sqrt{r}{e}^{\frac{i\theta }{2}}}\}dx$$$$=\frac{1}{2\sqrt{r}{e}^{\frac{i\theta }{2}}}\{{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x-\sqrt{r}{e}^{\frac{i\theta }{2}}}-{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x-(-\sqrt{r}{e}^{\frac{i\theta }{2}})}\}$$$$wehaveprovedthatfora\in C{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x-a}=i\pi ifIm\left(a\right)>0and$$$$-i\pi ifIm\left(a\right)<0soif\theta >0\Rightarrow$$$$I=\frac{1}{2\sqrt{r}{e}^{\frac{i\theta }{2}}}\{i\pi -(-i\pi )\}=\frac{2i\pi }{2\sqrt{r}{e}^{\frac{i\theta }{2}}}=\frac{i\pi }{\sqrt{r}{e}^{\frac{i\theta }{2}}}$$$$if\theta <0wegetI=\frac{1}{2\sqrt{r}{e}^{\frac{i\theta }{2}}}\{-i\pi -\left(i\pi \right)\}=-\frac{i\pi }{\sqrt{r}{e}^{\frac{i\theta }{2}}}$$

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