Question Number 32361 by prof Abdo imad last updated on 23/Mar/18
$${let}\:{give}\:{a}>\mathrm{0}\:{find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{x}} }{\:\sqrt{{x}+{a}}}\:{dx}. \\ $$
Commented by abdo imad last updated on 25/Mar/18
$${the}\:{ch}.\sqrt{{x}+{a}}\:={t}\:\:{give}\:{x}+{a}\:={t}^{\mathrm{2}} \:\Rightarrow\:{x}={t}^{\mathrm{2}} −{a} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{x}} }{\:\sqrt{{x}+{a}}}\:{dx}\:=\:\int_{\sqrt{{a}}} ^{+\infty} \:\:\:\frac{{e}^{−\left({t}^{\mathrm{2}} −{a}\right)} }{{t}}\:\mathrm{2}{t}\:{dt} \\ $$$$=\mathrm{2}\:{e}^{{a}} \:\:\int_{\sqrt{{a}}} ^{+\infty} \:{e}^{−{t}^{\mathrm{2}} } \:{dt}\:=\mathrm{2}\:{e}^{{a}} \left(\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}^{\mathrm{2}} } \:−\int_{\mathrm{0}} ^{\sqrt{{a}}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\right) \\ $$$$=\mathrm{2}\:{e}^{{a}} \:\frac{\sqrt{\pi}}{\mathrm{2}}\:\:−\mathrm{2}\:{e}^{{a}} \:\:\int_{\mathrm{0}} ^{\sqrt{{a}}} \:\:{e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$={e}^{{a}} \sqrt{\pi}\:\:−\mathrm{2}\:{e}^{{a}} \:\int_{\mathrm{0}} ^{\sqrt{{a}}} \:\:{e}^{−{t}^{\mathrm{2}} } \:{dt}\:\:. \\ $$