Integration Questions

Question Number 106285 by bemath last updated on 04/Aug/20

$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{dx}\:? \\$$

Answered by bobhans last updated on 04/Aug/20

$$\mathrm{recall}\:\underset{−\infty} {\overset{\infty} {\int}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{dx}\:=\:\sqrt{\pi} \\$$$$\Leftrightarrow\:\int_{−\infty} ^{\mathrm{0}} \mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx}\:+\:\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{dx}\:=\:\sqrt{\pi}\: \\$$$$\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{is}\:\mathrm{even}\:\mathrm{function} \\$$$$\mathrm{so}\:\underset{−\infty} {\overset{\mathrm{0}} {\int}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{dx}\:=\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{dx} \\$$$$\mathrm{finally}\:\mathrm{we}\:\mathrm{get}\:\Rightarrow\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{dx}\:=\:\sqrt{\pi} \\$$$$\therefore\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{dx}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}}.\:\bigstar\: \\$$

Answered by MAB last updated on 04/Aug/20

$${I}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{dx}\: \\$$$${I}^{\mathrm{2}} =\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}\right)\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{y}^{\mathrm{2}} } {dy}\right) \\$$$${I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dxdy} \\$$$${let}\:{x}={rcos}\left(\theta\right)\:{and}\:{y}={rsin}\left(\theta\right) \\$$$${where}\:{r}\geqslant\mathrm{0}\:{and}\:\mathrm{0}\leqslant\theta\leqslant\pi/\mathrm{2} \\$$$${dxdy}={rdrd}\theta \\$$$${I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \int_{\mathrm{0}} ^{\infty} {re}^{−{r}^{\mathrm{2}} } {drd}\theta \\$$$${I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}\left[{e}^{−{r}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\infty} {d}\theta \\$$$${I}^{\mathrm{2}} =\frac{\pi}{\mathrm{4}} \\$$$${hence}\:{I}=\frac{\sqrt{\pi}}{\mathrm{2}} \\$$