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Question Number 6775 by Tawakalitu. last updated on 24/Jul/16

\int_{0}^{\infty} e^{t^{2}} e^{- s t} d t

p l e a s e h e l p

Answered by Yozzii last updated on 24/Jul/16

b = \int_{0}^{\infty} e^{t^{2} - s t} d t = \int_{0}^{\infty} e^{{(t - 0.5 s)}^{2} - 0.25 s^{2}} d t

b = e^{- 0.25 s^{2}} \int_{0}^{\infty} e^{{(t - 0.5 s)}^{2}} d t .

u = t - 0.5 s \Rightarrow d u = d t

∴ b = e^{- 0.25 s^{2}} (\int_{- 0.5 s}^{0} e^{u^{2}} d u + \int_{0}^{\infty} e^{u^{2}} d u)

\int_{0}^{\infty} e^{u^{2}} d u i s d i v e r g e n t w h i l e \int_{- 0.5 s}^{0} e^{u^{2}} d u

i s f i n i t e .

∴ F o r n o n z e r o s \in R, b = e^{- 0.25 s^{2}} {\int_{- 0.5 s}^{0} e^{u^{2}} d u + \int_{0}^{\infty} e^{u^{2}} d u}

d o e s n o t e x i s t .

I f s = 0, b = \int_{0}^{\infty} e^{t^{2}} d t w h i c h i s d i v e r g e n t a l s o .

Commented by Tawakalitu. last updated on 24/Jul/16

T h a n k s s o m u c h, T h a t m e a n s t h e q u e a s t i o n i s w r o n g ? ?

Commented by Yozzii last updated on 24/Jul/16

N o t n e c e s s a r i l y . T h e q u e s t i o n m a y h a v e

a s k e d t o s e e w h e t h e r t h e i m p r o p e r

i n t e g r a l i s c o n v e r g e n t o r d i v e r g e n t .

Commented by Tawakalitu. last updated on 24/Jul/16

T h a n k s

Commented by Tawakalitu. last updated on 25/Jul/16

T h e q u e s t i o n i s e v a l u a t e