Question Number 53601 by maxmathsup by imad last updated on 23/Jan/19
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{x}^{\mathrm{2}} } −{e}^{−{x}} }{{x}}\:{dx}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jan/19
$$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {x}^{−\mathrm{1}} {dx}−\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{−\mathrm{1}} {dx} \\ $$$${x}=\sqrt{{t}}\:\:{dx}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{t}}\:}{dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{t}} ×}{\sqrt{{t}}}×\frac{{dt}}{\mathrm{2}\sqrt{{t}}}−\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} ×\frac{{dx}}{{x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} ×\frac{{dt}}{{t}}−\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} ×\frac{{dx}}{{x}} \\ $$$${gamma}\:{functiin}=\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{{n}−\mathrm{1}} {dx}=\lceil\left({n}\right) \\ $$$$\lceil\left({n}+\mathrm{1}\right)={n}\lceil\left({n}\right)={n}!\:{when}\:{n}>\mathrm{0} \\ $$$${so}\:{we}\:{can}\:{not}\:{use}\:{gamma}\:{function}.. \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{ax}} }{{x}}{dx} \\ $$$$\frac{{dI}_{\mathrm{1}} }{{da}}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{ax}} ×−{x}}{{x}}{dx}=−\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} {dx}=−\mathrm{1}×\mid\frac{{e}^{−{ax}} }{−{a}}\mid_{\mathrm{0}} ^{\infty} \\ $$$$\frac{{dI}_{\mathrm{1}} }{{da}}=\frac{\mathrm{1}}{{a}}\left(\frac{\mathrm{1}}{{e}^{\infty} }−\frac{\mathrm{1}}{{e}^{\mathrm{0}} }=−\frac{\mathrm{1}}{{a}}\right. \\ $$$${I}_{\mathrm{1}} =−{lna}+{c} \\ $$$${we}\:{have}\:{to}\:{find}\:{value}\:{of}\:{c}\:\:\:{wait}.... \\ $$