Question Number 141929 by mathmax by abdo last updated on 24/May/21
$$\mathrm{find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\left(\mathrm{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)} \mathrm{dt} \\ $$
Answered by Dwaipayan Shikari last updated on 24/May/21
$$\xi\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {e}^{−\left({t}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{{t}^{\mathrm{2}} }\right)} {dt} \\ $$$$\xi'\left({a}\right)=−\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{a}}{{t}^{\mathrm{2}} }{e}^{−\left({t}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{{t}^{\mathrm{2}} }\right)} \:\:\:\:{dt}\:\:\:\:\:\:\:\:\:\:\frac{{a}}{{t}}={u}\Rightarrow−\frac{{a}}{{t}^{\mathrm{2}} }\Rightarrow\frac{{du}}{{dt}}\Rightarrow{dt}=−\frac{{t}^{\mathrm{2}} }{{a}}.{du} \\ $$$$\xi'\left({a}\right)=−\mathrm{2}\int_{\mathrm{0}} ^{\infty} {e}^{−\left(\frac{{a}^{\mathrm{2}} }{{u}^{\mathrm{2}} }+{u}^{\mathrm{2}} \right)} {du} \\ $$$$\xi'\left({a}\right)=−\mathrm{2}\xi\left({a}\right)\Rightarrow\xi\left({a}\right)={Ce}^{−\mathrm{2}{a}} \Rightarrow\xi\left(\mathrm{0}\right)={C}=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\xi\left({a}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}{e}^{−\mathrm{2}{a}} \\ $$$$\xi\left(\mathrm{1}\right)=\frac{\sqrt{\pi}}{\mathrm{2}{e}^{\mathrm{2}} } \\ $$