answer to 25824 we have a^(−x^2 ) = e^(−x^2_ ln(a)) so for a>1 ln(a)=( (ln(a))^(1/2) )^2 >>>>∫_R a^(−^ x^2 ) = ∫_R e^(−(x (ln(a)^(1/2) )^2 ) dx and with the changement t=x (ln(a)^(1/2) >>>>x=t ( ln(a))^(−1/2) we have ∫_R a^(−x^2 ) dx = π^(1/2) (ln(a))^(−1/2) ...if 0<a<1 ln(a)<0 and the integrale is divergente...


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Question Number 25932 by abdo imad last updated on 16/Dec/17

$a n s w e r t o 25824 w e h a v e a^{- x^{2}} = e^{- x^{2_{}} l n (a)} s o f o r a > 1$ $l n (a) = {({(l n (a))}^{1 / 2})}^{2} >>>> \int_{R} a^{-^{} x^{2}} = \int_{R} e^{- (x {(l n {(a)}^{1 / 2})}^{2}} d x$ $a n d w i t h t h e c h a n g e m e n t t = x (l n {(a)}^{1 / 2} >>>> x = t {(l n (a))}^{- 1 / 2}$ $w e h a v e \int_{R} a^{- x^{2}} d x = π^{1 / 2} {(l n (a))}^{- 1 / 2} . . . i f 0 < a < 1 l n (a) < 0$ $a n d t h e i n t e g r a l e i s d i v e r g e n t e . . .$
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