Define I(x)=int_x^infty e^(-y^2)dy. Evaluate int_0^infty e^(x^2)[I(x)]^2dx

Jensen Mclean 2022-09-07 Answered
Define
I ( x ) = x e y 2   d y
Evaluate
0 e x 2 [ I ( x ) ] 2   d x
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Answers (1)

Nadia Berry
Answered 2022-09-08 Author has 7 answers
To wit
I ( x ) = x 1 d u e x 2 u 2
which means that
I ( x ) 2 = x 2 1 d u 1 d v e x 2 ( u 2 + v 2 )
Now we can put this into the desired integral and reverse the order of integration to get
0 d x e x 2 I ( x ) 2 = 1 d u 1 d v 0 d x x 2 e x 2 ( u 2 + v 2 1 )
The inner integral converges because u 2 + v 2 1 0. We evaluate the inner integral and reduce the integral to a double integral:
0 d x e x 2 I ( x ) 2 = π 4 1 d u 1 d v ( u 2 + v 2 1 ) 3 / 2
The integral over v may be attacked by a trig substitution: v = u 2 1 tan t , d v = u 2 1 sec 2 t d t; the result is
1 d v ( u 2 + v 2 1 ) 3 / 2 = ( 1 1 u ) 1 u 2 1 = 1 u ( u + 1 )
The problem now reduces to the evaluation of
1 d u u ( u + 1 ) = lim r log ( r r + 1 ) + log 2 = log 2
Therefore, the desired integral has the value
0 d x e x 2 I ( x ) 2 = π 4 log 2
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