find-the-value-of-0-e-tx-2-cosx-dx-with-t-gt-0-

Question Number 28694 by abdo imad last updated on 29/Jan/18

f i n d t h e v a l u e o f \int_{0}^{\infty} e^{- {t x}^{2}} c o s x d x w i t h t > 0 .

Commented by abdo imad last updated on 29/Jan/18

l e t p u t I = \int_{0}^{\infty} e^{- {t x}^{2}} c o s x d x

I = \frac{1}{2} \int_{- \infty}^{+ \infty} e^{- {t x}^{2} + i x} d x b e c a u s e \int_{- \infty}^{+ \infty} e^{- {t x}^{2}} s i n x d x = 0 b u t

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

= \int_{- \infty}^{+ \infty} e^{- {(\sqrt{t} x - \frac{i}{2 \sqrt{t}})}^{2} - \frac{1}{4 t}} d x (c h . \sqrt{t} x - \frac{i}{2 \sqrt{t}} = u)

= e^{- \frac{1}{4 t}} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{t}} = \frac{\sqrt{π}}{\sqrt{t}} e^{- \frac{1}{4 t}} (\int_{- \infty}^{+ \infty} e^{- u^{2}} d u = \sqrt{π}) s o

I = \frac{\sqrt{π}}{2 \sqrt{t}} e^{- \frac{1}{4 t}} .