Question Number 34297 by math khazana by abdo last updated on 03/May/18
$${find}\:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{z}\:{t}^{\mathrm{2}} } {dt}\:\:\:{with}\:{z}={r}\:{e}^{{i}\theta} \:\:\in\:{C}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 04/May/18
$${let}\:{put}\:{I}\:=\int_{−\infty} ^{+\infty} \:{e}^{−{zt}^{\mathrm{2}} } {dt}\:\Rightarrow\:{I}\:=\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{re}^{{i}\theta} {t}^{\mathrm{2}} } {dt} \\ $$$${I}\:=\:\int_{−\infty} ^{+\infty} \:\:\:{e}^{−\left(\sqrt{{r}}\:{e}^{{i}\frac{\theta}{\mathrm{2}}} \:{t}\right)^{\mathrm{2}} } {dt}\:\:\:{changement}\:\sqrt{{r}}\:\:{e}^{{i}\frac{\theta}{\mathrm{2}}} \:\:{t}\:={u} \\ $$$${give}\:{I}\:\:=\:\:\frac{\mathrm{1}}{\sqrt{{r}}}\:{e}^{−{i}\frac{\theta}{\mathrm{2}}} \:\:\int_{−\infty} ^{++\infty} \:\:{e}^{−{u}^{\mathrm{2}} } {du} \\ $$$${I}\:=\:\frac{\sqrt{\pi}}{\sqrt{{r}}}\:\:{e}^{−{i}\frac{\theta}{\mathrm{2}}} =\:\frac{\sqrt{\pi}}{\sqrt{{r}}}\:\:\left(\:{cos}\left(\frac{\theta}{\mathrm{2}}\right)\:−{i}\:{sin}\left(\frac{\theta}{\mathrm{2}}\right)\right)\:\:\left({r}>\mathrm{0}\right) \\ $$
Answered by candre last updated on 03/May/18
$${I}=\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{zt}^{\mathrm{2}} } {dt} \\ $$$${I}^{\mathrm{2}} =\left(\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{zt}^{\mathrm{2}} } {dt}\right)^{\mathrm{2}} =\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{zx}^{\mathrm{2}} } {dx}\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{zy}^{\mathrm{2}} } {dy} \\ $$$$=\underset{−\infty} {\overset{+\infty} {\int}}\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{z}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dxdy} \\ $$$$\left({x},{y}\right)=\rho\left(\mathrm{cos}\:\varphi,\mathrm{sin}\:\varphi\right) \\ $$$$\frac{\partial\left({x},{y}\right)}{\partial\left(\rho,\varphi\right)}{d}\rho{d}\varphi=\begin{vmatrix}{\frac{\partial{x}}{\partial\rho}}&{\frac{\partial{y}}{\partial\rho}}\\{\frac{\partial{x}}{\partial\varphi}}&{\frac{\partial{y}}{\partial\varphi}}\end{vmatrix}{d}\rho{d}\varphi=\begin{vmatrix}{\mathrm{cos}\:\varphi}&{\mathrm{sin}\:\varphi}\\{−\rho\mathrm{sin}\:\varphi}&{\rho\mathrm{cos}\:\varphi}\end{vmatrix}{d}\rho{d}\varphi \\ $$$$=\rho{d}\rho{d}\varphi \\ $$$$−\infty<{x}<+\infty,−\infty<{y}<+\infty\equiv\mathrm{0}\leqslant\rho<\infty,\mathrm{0}\leqslant\varphi<\mathrm{2}\pi \\ $$$${I}^{\mathrm{2}} =\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\underset{\mathrm{0}} {\overset{+\infty} {\int}}{e}^{−{z}\rho^{\mathrm{2}} } \rho{d}\rho{d}\varphi \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}{d}\varphi\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{z}\rho^{\mathrm{2}} } \rho{d}\rho \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}{d}\varphi=\mathrm{2}\pi \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{z}\rho^{\mathrm{2}} } \rho{d}\rho=\frac{\mathrm{1}}{\mathrm{2}{z}}\underset{−\infty} {\overset{\mathrm{0}} {\int}}{e}^{{u}} {du}=\frac{\mathrm{1}}{\mathrm{2}{z}} \\ $$$${u}=−{z}\rho^{\mathrm{2}} \\ $$$${du}=−\mathrm{2}{z}\rho{d}\rho \\ $$$${I}^{\mathrm{2}} =\frac{\pi}{{z}} \\ $$$${I}=\sqrt{\frac{\pi}{{z}}}\:\:\:\:\:\left(\Re\left({z}\right)>\mathrm{0}\right) \\ $$
Commented by math khazana by abdo last updated on 04/May/18
$${your}\:{method}\:{is}\:{correct}\:{sir}\:{candart}...{thanks}... \\ $$