let-give-a-gt-0-find-0-e-x-x-a-dx-

Question Number 32361 by prof Abdo imad last updated on 23/Mar/18

l e t g i v e a > 0 f i n d \int_{0}^{\infty} \frac{e^{- x}}{\sqrt{x + a}} d x .

Commented by abdo imad last updated on 25/Mar/18

t h e c h . \sqrt{x + a} = t g i v e x + a = t^{2} \Rightarrow x = t^{2} - a

\int_{0}^{\infty} \frac{e^{- x}}{\sqrt{x + a}} d x = \int_{\sqrt{a}}^{+ \infty} \frac{e^{- (t^{2} - a)}}{t} 2 t d t

= 2 e^{a} \int_{\sqrt{a}}^{+ \infty} e^{- t^{2}} d t = 2 e^{a} (\int_{0}^{\infty} e^{- t^{2}} - \int_{0}^{\sqrt{a}} e^{- t^{2}} d t)

= 2 e^{a} \frac{\sqrt{π}}{2} - 2 e^{a} \int_{0}^{\sqrt{a}} e^{- t^{2}} d t

= e^{a} \sqrt{π} - 2 e^{a} \int_{0}^{\sqrt{a}} e^{- t^{2}} d t .