let-f-t-0-e-ax-e-bx-x-2-e-tx-2-dx-with-t-gt-0-1-calculate-f-t-2-find-a-simple-form-of-f-t-3-find-the-value-of-0-e-2x-e-x-x-2-e-3x-2-dx-

Question Number 35821 by prof Abdo imad last updated on 24/May/18

l e t f (t) = \int_{0}^{\infty} \frac{e^{- a x} - e^{- b x}}{x^{2}} e^{- {t x}^{2}} d x w i t h t > 0

1) c a l c u l a t e f^{^{'}} (t)

2) f i n d a s i m p l e f o r m o f f (t)

3) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{e^{- 2 x} - e^{- x}}{x^{2}} e^{- 3 x^{2}} d x

Commented by prof Abdo imad last updated on 31/May/18

f (t) = \int_{- \infty}^{+ \infty} \frac{e^{- a x} - e^{- b x}}{x^{2}} e^{- {t x}^{2}} w i t h t > 0

Commented by prof Abdo imad last updated on 31/May/18

3) f i n d t h e v a l u e o f \int_{- \infty}^{+ \infty} \frac{e^{- 2 x} - e^{- x}}{x^{2}} e^{- 3 x^{2}} d x

Commented by abdo mathsup 649 cc last updated on 01/Jun/18

w e h a v e f^{^{'}} (t) = - \int_{- \infty}^{+ \infty} (e^{- a x} - e^{- b x}) e^{- {t x}^{2}} d x

= \int_{- \infty}^{+ \infty} e^{- b x - {t x}^{2}} d x - \int_{- \infty}^{+ \infty} e^{- a x - {t x}^{2}} d x b u t

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

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

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

= \frac{e^{\frac{a^{2}}{4 t}}}{\sqrt{t}} \sqrt{π} \Rightarrow f (t) = \sqrt{π} \int_{.}^{t} \frac{e^{\frac{a^{2}}{4 u}}}{\sqrt{u}} d u + λ

c h a n g e m e n t \sqrt{u} = x g i v e

f (t) = \sqrt{π} \int_{.}^{\sqrt{t}} \frac{e^{\frac{a^{2}}{4 x^{2}}}}{x} 2 x d x + λ

= 2 \sqrt{π} \int_{.}^{\sqrt{t}} e^{\frac{a^{2}}{4 x^{2}}} d x + λ

Commented by abdo mathsup 649 cc last updated on 01/Jun/18

e r r o r f r o m l i n e 6

\int_{- \infty}^{+ \infty} e^{- {t x}^{2} - a x} d x = \frac{\sqrt{π}}{\sqrt{t}} e^{\frac{a^{2}}{4 t}} a l s o

\int_{- \infty}^{+ \infty} e^{- {t x}^{2} - b x} d x = \frac{\sqrt{π}}{\sqrt{t}} e^{\frac{b^{2}}{4 t}} \Rightarrow

f^{^{'}} (t) = \sqrt{π} {\frac{e^{\frac{b^{2}}{4 t}}}{\sqrt{t}} - \frac{e^{\frac{a^{2}}{4 t}}}{\sqrt{t}}} \Rightarrow

f (t) = \sqrt{π} \int_{1}^{t} \frac{e^{\frac{b^{2}}{4 u}} - e^{\frac{a^{2}}{4 u}}}{\sqrt{u}} d u + c

=_{\sqrt{u} = x} \sqrt{π} \int_{1}^{\sqrt{t}} \frac{e^{\frac{b^{2}}{4 x^{2}}} - e^{\frac{a^{2}}{4 x^{2}}}}{x} 2 x d x + c

= 2 \sqrt{π} \int_{1}^{\sqrt{t}} {e^{\frac{b^{2}}{4 x^{2}}} - e^{\frac{a^{2}}{4 x^{2}}}} d x + c

c = f (1)