Question Number 90292 by mathmax by abdo last updated on 22/Apr/20
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{e}^{{zx}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx}\:{with}\:{z}\:{from}\:{C}\:{and}\:{Re}\left({z}\right)<\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 23/Apr/20
$${let}\:{f}\left({z}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}−{e}^{{zx}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx}\:\Rightarrow\:{f}^{'} \left({z}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:{e}^{{zx}^{\mathrm{2}} } {dx} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\sqrt{−{z}}{x}\right)^{\mathrm{2}} } {dx}\:=_{\sqrt{−{z}}{x}\:={u}} \:\:−\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{u}^{\mathrm{2}} } \frac{{du}}{\:\sqrt{−{z}}} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{−{z}}}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}^{\mathrm{2}} } {du}\:=−\frac{\pi}{\mathrm{2}\sqrt{−{z}}}\:\Rightarrow{f}\left({z}\right)=\pi\sqrt{−{z}}\:+{c} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}\:={c}\:\Rightarrow{f}\left({z}\right)\:=\pi\sqrt{−{z}}\:\:\:{if}\:\:{z}\:=−\alpha\:+{i}\beta\:\:\:\:\:\:\left(\alpha>\mathrm{0}\right) \\ $$$$−{z}\:=\alpha−{i}\beta\:\:=\sqrt{\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\:{e}^{−{iarctan}\left(\frac{\beta}{\alpha}\right)} \:\Rightarrow\sqrt{−{z}}=\left(\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\beta}{\alpha}\right)} \\ $$$$\Rightarrow{f}\left({z}\right)\:=^{\mathrm{4}} \sqrt{\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\:{e}^{−\frac{{i}}{\mathrm{2}}{arctan}\left(\frac{\beta}{\alpha}\right)} \\ $$$$ \\ $$