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Question Number 53601 by maxmathsup by imad last updated on 23/Jan/19

$$calculate{\int }_0^{\mathrm{\infty }}\frac{e^{-x^2}-e^{-x}}{x}dx.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jan/19

$${\int }_0^{\mathrm{\infty }}{e}^{-x^2}{x}^{-1}dx-{\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{-1}dx$$$$x=\sqrt{t}dx=\frac{1}{2\sqrt{t}}dt$$$${\int }_0^{\mathrm{\infty }}\frac{e^{-t}\times }{\sqrt{t}}\times \frac{dt}{2\sqrt{t}}-{\int }_0^{\mathrm{\infty }}{e}^{-x}\times \frac{dx}{x}$$$$\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{-t}\times \frac{dt}{t}-{\int }_0^{\mathrm{\infty }}{e}^{-x}\times \frac{dx}{x}$$$$gammafunctiin={\int }_0^{\mathrm{\infty }}{e}^{-x}{x}^{n-1}dx=\lceil \left(n\right)$$$$\lceil (n+1)=n\lceil \left(n\right)=n!whenn>0$$$$sowecannotusegammafunction..$$$$I_1={\int }_0^{\mathrm{\infty }}\frac{e^{-ax}}{x}dx$$$$\frac{{dI}_1}{da}={\int }_0^{\mathrm{\infty }}\frac{e^{-ax}\times -x}{x}dx=-{\int }_0^{\mathrm{\infty }}{e}^{-ax}dx=-1\times \mid \frac{e^{-ax}}{-a}{\mid }_0^{\mathrm{\infty }}$$$$\frac{{dI}_1}{da}=\frac{1}{a}(\frac{1}{e^{\mathrm{\infty }}}-\frac{1}{e^0}=-\frac{1}{a}$$$$I_1=-lna+c$$$$wehavetofindvalueofcwait....$$

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