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Question Number 67851 by mathmax by abdo last updated on 01/Sep/19

$$find\int \frac{dx}{x^2-z}withzfromC.$$

Commented by MJS last updated on 01/Sep/19

$${\mathrm{what}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{problem}/\mathrm{mistake}\mathrm{if}\mathrm{we}\mathrm{just}$$$$\mathrm{calculate}\mathrm{it}\mathrm{as}\mathrm{if}z\in \mathbb{R}?$$$$\int \frac{dx}{x^2-z}=\frac{1}{2\sqrt{z}}\ln \frac{x-\sqrt{z}}{x+\sqrt{z}}+C$$$$\mathrm{for}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{x^2-z}\mathrm{this}\mathrm{would}\mathrm{give}-\frac{\pi }{2\sqrt{z}}\mathrm{i}$$

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

$$letz=r{e}^{i\theta }\Rightarrow \int \frac{dx}{x^2-z}=\int \frac{dx}{x^2-{\left(\sqrt{r}{e}^{\frac{i\theta }{2}}\right)}^2}$$$$=\int \frac{dx}{(x-\sqrt{r}{e}^{\frac{i\theta }{2}})(x+\sqrt{r}{e}^{\frac{i\theta }{2}})}=\frac{1}{2\sqrt{r}{e}^{\frac{i\theta }{2}}}\int (\frac{1}{x-\sqrt{r}{e}^{\frac{i\theta }{2}}}-\frac{1}{x+\sqrt{r}{e}^{\frac{i\theta }{2}}})dx$$$$=\frac{e^{-\frac{i\theta }{2}}}{2\sqrt{r}}ln\left(\frac{x-\sqrt{r}{e}^{\frac{i\theta }{2}}}{x+\sqrt{r}{e}^{\frac{i\theta }{2}}}\right)+C.$$

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

$$sirmjsyouransweriscorrect.$$

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

$${\int }_0^{\mathrm{\infty }}\frac{dx}{x^2-z}=\frac{e^{-\frac{i\theta }{2}}}{2\sqrt{r}}{\left[ln\left(\frac{x-\sqrt{r}{e}^{\frac{i\theta }{2}}}{x+\sqrt{r}{e}^{\frac{i\theta }{2}}}\right)\right]}_0^{+\mathrm{\infty }}=\frac{e^{-\frac{i\theta }{2}}}{2\sqrt{r}}(-ln(-1))=\frac{i\pi }{2\left(2\sqrt{r}\right)e^{\frac{i\theta }{2}}}$$$$=\frac{i\pi }{4\sqrt{r}{e}^{\frac{i\theta }{2}}}withz=r{e}^{i\theta }.$$

Commented by MJS last updated on 01/Sep/19

$$\mathrm{thank}\mathrm{you}$$$$\mathrm{I}\mathrm{just}{\mathrm{wasn}}^{\prime }\mathrm{t}\mathrm{sure}\mathrm{if}\mathrm{this}\mathrm{was}\mathrm{allowed}\mathrm{in}\mathbb{C}$$

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