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Question Number 27342 by abdo imad last updated on 05/Jan/18

$$findf\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-x(1+t^2)}}{1+t^2}dtintermsofx$$$$withx⩾0andcalculate{\int }_0^{\mathrm{\infty }}{e}^{-t^2}dt.$$

Commented by abdo imad last updated on 08/Jan/18

$$afterverifyingthatfisderivablein[0,\propto [$$$$f^{,}\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{\partial }{\partial x}\left(\frac{e^{-x(1+t^2)}}{1+t^2}\right)dt={\int }_0^{\mathrm{\infty }}-e^{-x(1+t^2)}dt$$$$=-e^{-x}{\int }_0^{\mathrm{\infty }}{e}^{-{xt}^2}dtthech.\sqrt{x}t=ugive$$$$f^{,}\left(x\right)=-e^{-x}{\int }_0^{\mathrm{\infty }}{e}^{-u^2}\frac{du}{\sqrt{x}}=-\frac{e^{-x}}{\sqrt{x}}{\int }_0^{\mathrm{\infty }}{e}^{-u^2}du$$$$=-\frac{\sqrt{\pi }}{2}\frac{e^{-x}}{\sqrt{x}}(x>0)$$$$\Rightarrow f\left(x\right)=-\frac{\sqrt{\pi }}{2}{\int }_0^x\frac{e^{-t}}{\sqrt{t}}dt+\lambda afterthatweusethech.\sqrt{t}=u$$$$f\left(x\right)=-\frac{\sqrt{\pi }}{2}{\int }_0^{\sqrt{x}}\frac{e^{-u^2}}{u}2udu+\lambda$$$$=\lambda -\sqrt{\pi }{\int }_0^{\sqrt{x}}{e}^{-u^2}du$$$${lim}_{x->\propto }f\left(x\right)=0=\lambda -\sqrt{\pi }{\int }_0^{\propto }{e}^{-u^2}du=\lambda -\frac{\pi }{2}\Rightarrow \lambda =\frac{\pi }{2}$$$$finallyf\left(x\right)=\frac{\pi }{2}-\sqrt{\pi }{\int }_0^{\sqrt{x}}{e}^{-u^2}du$$$$wehavealotsofmethodtoprovethat{\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx=\frac{\sqrt{\pi }}{2}.$$

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