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answer-to-25824-we-have-a-x-2-e-x-2-ln-a-so-for-a-gt-1-ln-a-ln-a-1-2-2-gt-gt-gt-gt-R-a-x-2-R-e-x-ln-a-1-2-2-dx-and-with-the-changement-t-x-ln-a-




Question Number 25932 by abdo imad last updated on 16/Dec/17
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...
answerto25824wehaveax2=ex2ln(a)sofora>1ln(a)=((ln(a))1/2)2>>>>Rax2=Re(x(ln(a)1/2)2dxandwiththechangementt=x(ln(a)1/2>>>>x=t(ln(a))1/2wehaveRax2dx=π1/2(ln(a))1/2if0<a<1ln(a)<0andtheintegraleisdivergente

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