Find an approximate equivalent. int_0^(+oo) x log(1+x^2)e^(−Bx) dx

Celeste Barajas 2022-11-21 Answered
This integral sounds quite complex and I could not find an approximate equivalent.
0 + x log ( 1 + x 2 ) e B x d x
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Answers (1)

jennasyliang4tr
Answered 2022-11-22 Author has 15 answers
For starters, substitute x 2 x ,   as follows :
I   =   0   x   ln ( 1 + x 2 )   e B x   d x   =   1 2 0 ln ( 1   +   x 2 )   e B x 2   d ( x 2 )   =     =   1 2   0 ln ( 1 + x )   e B x   d x   =   1 2   [ 0 ln ( 1 + A x )   e B x   d x ] A = 1   =   J ( 1 ) 2
Now, employ differentiation under the integral sign :
J ( A )   =   0 x 1 + A x   e B x   d x   =   d 2 d B 2   0 1 1 + A x   e B x   d x   =     =   d 2 d B 2   { 1 A   [ sin B A ( π 2  Si  B A ) 2   cos B A  Ci  B A ] } ,
which, when integrated with regard to A, yields the rather beautiful expression :
J ( A ) 2   =   d 2 d B 2 [  Ci 2   B A +  Si 2   B A π  Si  B A ] ,
which, when evaluated at A=1, gives   I   =   J ( 1 ) 2   =   d 2 d B 2   [  Ci 2   B +  Si 2   B π  Si  B ] ,
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