let-a-gt-0-find-the-value-of-f-a-0-e-t-2-a-t-2-dt-

Question Number 37635 by math khazana by abdo last updated on 16/Jun/18

l e t a > 0 f i n d t h e v a l u e o f

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

Commented by prof Abdo imad last updated on 17/Jun/18

w e h a v e 2 f (a) = \int_{- \infty}^{+ \infty} e^{- {{(t - \frac{\sqrt{a}}{t})}^{2} + 2 \sqrt{a}}} d t

= e^{- 2 \sqrt{a}} \int_{- \infty}^{+ \infty} e^{- {(t - \frac{\sqrt{a}}{t})}^{2}} d t c h a n g e m e n t

t - \frac{\sqrt{a}}{t} = x g i v e t^{2} - \sqrt{a} = x t \Rightarrow

t^{2} - x t - \sqrt{a} = 0

Δ = x^{2} + 4 \sqrt{a} \Rightarrow t_{1} = \frac{x + \sqrt{x^{2} + 4 \sqrt{a}}}{2}

t_{2} = \frac{x - \sqrt{x^{2} + 4 \sqrt{a}}}{2} l e t t a k e t = \frac{x + \sqrt{x^{2} + 4 \sqrt{a}}}{2} \Rightarrow

d t = \frac{1}{2} {1 + \frac{x}{\sqrt{x^{2} + 4 \sqrt{a}}}} \Rightarrow

2 f (a) = \frac{e^{- 2 \sqrt{a}}}{2} \int_{- \infty}^{+ \infty} e^{- x^{2}} {1 + \frac{x}{\sqrt{x^{2} + 4 \sqrt{a}}}} d x

= \frac{e^{- 2 \sqrt{a}}}{2} \int_{- \infty}^{+ \infty} e^{- x^{2}} d x + \frac{e^{- 2 \sqrt{a}}}{2} \int_{- \infty}^{+ \infty} \frac{x e^{- x^{2}}}{\sqrt{x^{2} + 4 \sqrt{a}}} d x

= \frac{\sqrt{π} e^{- 2 \sqrt{a}}}{2} + 0 \Rightarrow

f (a) = \frac{e^{- 2 \sqrt{a}}}{4} \sqrt{π} .