Question Number 196145 by Erico last updated on 18/Aug/23
$$\mathrm{Calculer}\:\underset{\:\frac{\mathrm{1}}{\alpha}} {\int}^{\:+\infty} {e}^{−\alpha{t}^{\mathrm{2}} +\mathrm{2}{t}} {dt} \\ $$
Answered by Mathspace last updated on 20/Aug/23
$${t}=\frac{\mathrm{1}}{\alpha}+{x}\:\Rightarrow \\ $$$$\int_{\frac{\mathrm{1}}{\alpha}} ^{\infty} \:{e}^{−\alpha{t}^{\mathrm{2}} +\mathrm{2}{t}} {dt}=\int_{\mathrm{0}} ^{\infty} {e}^{−\alpha\left(\frac{\mathrm{1}}{\alpha}+{x}\right)^{\mathrm{2}} +\mathrm{2}\left(\frac{\mathrm{1}}{\alpha}+{x}\right)} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {e}^{−\alpha\left\{\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }+\frac{\mathrm{2}{x}}{\alpha}+{x}^{\mathrm{2}} \right\}+\frac{\mathrm{2}}{\alpha}+\mathrm{2}{x}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\frac{\mathrm{1}}{\alpha}−\mathrm{2}{x}−\alpha{x}^{\mathrm{2}} +\frac{\mathrm{2}}{\alpha}+\mathrm{2}{x}} {dx} \\ $$$$={e}^{\frac{\mathrm{1}}{\alpha}} \:\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\alpha{x}^{\mathrm{2}} } {dx}\:\:\left(\sqrt{\alpha}{x}={u}\right) \\ $$$$={e}^{\frac{\mathrm{1}}{\alpha}} \int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}^{\mathrm{2}} } \frac{{du}}{\:\sqrt{\alpha}} \\ $$$$=\frac{{e}^{\frac{\mathrm{1}}{\alpha}} }{\:\sqrt{\alpha}}.\frac{\sqrt{\pi}}{\mathrm{2}}\:\Rightarrow \\ $$$${I}=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\alpha}}\:{e}^{\frac{\mathrm{1}}{\alpha}} \:\:\:\left(\alpha>\mathrm{0}\right) \\ $$
Commented by Mathspace last updated on 20/Aug/23
$${sorry}\:{I}=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\alpha}}\:{e}^{−\frac{\mathrm{1}}{\alpha}} \\ $$