a-Let-I-0-e-x-x-2-dx-Show-that-it-is-legitimate-to-take-the-derivative-of-I-and-also-I-0-Then-show-that-

Question Number 136476 by Ar Brandon last updated on 22/Mar/21

(a) Let

I (α) = \int_{0}^{\infty} e^{- {(x - \frac{α}{x})}^{2}} dx

Show that it is legitimate to take the derivative of I (α) and also I^{'} (α) =

0 . Then show that

I (α) = \frac{\sqrt{π}}{2} .

(b) Use (a) to prove

\int_{0}^{\infty} e^{- (x^{2} + α^{2} x^{- 2})} dx = \frac{\sqrt{π}}{2} e^{- 2 α} .

Answered by Dwaipayan Shikari last updated on 22/Mar/21

I (α) = \int_{0}^{\infty} e^{- x^{2} - \frac{α^{2}}{x^{2}}} d x

I^{'} (α) = - 2 \int_{0}^{\infty} \frac{α}{x^{2}} e^{- x^{2} - \frac{α^{2}}{x^{2}}} d x \frac{α}{x} = u \Rightarrow \frac{- α}{x^{2}} = \frac{d u}{d x}

I^{'} (α) = - 2 \int_{0}^{\infty} e^{- \frac{α^{2}}{u^{2}} - u^{2}} d u \Rightarrow I^{'} (α) = - 2 I (α)

I (α) = {C e}^{- 2 α}

I (0) = C = \int_{0}^{\infty} e^{- x^{2}} d x = \frac{\sqrt{π}}{2}

I (α) = \frac{\sqrt{π}}{2} e^{- 2 α}