A complicated integral to be evaluated by complex contour method

  • #1
Bengy
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TL;DR Summary
Evaluating a complicated integral
I encounter a very complicated integral in my research project, and it can only be done by complex contour integral. I have tried many other methods like integration by parts and substitution, which all fail. In the integral, all a, alpha and beta are positive real numbers. I have found a similar youtube video which has the same denominator as mine, but the numerator with cosine and exponential is still an issue. I have enclosed the integral as an attachment. Can anyone offer me some hint ? :)

Integral.png
 
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  • #2
For complicated integrals, look for some integration tables like in a CRC math tables book or Schaums outlines math tables book.

They organize integrals by their components like the x2a2 and from there you might a matching integral and solution.
 
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  • #3
The problem can not be solved analytically.
 
  • #4
for a>0 β0 if not mistaken
I=eα22βa Re0π2dθ  e2βa2(cosec θiα2βa)2
For α=β=0,I=π2a which is an upper limit of the integral.

[EDIT]
I=18aiCdzz{ exp[B(zz1)2+Azz1]+exp[B(zz1)2Azz1] }
where closed contour C is the unit circle,
A=4aα, B=8a2β
but I cannot go further.
With the result of @renormalize post #6 for A=0
Cdzz exp[B(zz1)2]=2πi erfc(B2)
It seems as if erfc(B2) is the residue at pole z=0. An interesting relation.
 
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  • #5
anuttarasammyak said:
For α=β=0,I=π2a which is an upper limit of the integral.
This is correct.

The definite integral can be solved analytically by using the substitution method when α=β=0 and a>0.

I=acos(2αx)xx2a2e2βx2dx=a1xx2a2dx

t2=x2a21tdt=xa2dx|a|0

I=1a01t2+1dt=1aarctant|0=π2a

By the way, I still believe that the problem can not be solved analytically when a, α and β are positive real numbers.
 
  • #6
Gavran said:
By the way, I still believe that the problem can not be solved analytically when a, α and β are positive real numbers.
As long as we're posting simpler integrals that can be done, I'd like to nominate:ae2βx2xx2a2dx=π2aerfc(2βa)(a>0,β>0)(courtesy of Mathematica)
But I see no way to leverage this result to analytically evaluate the OP's original integral.
 
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  • #7
Thank you very much for all the replies! Indeed it is a very difficult problem. I find that Mathematica can also give an exact result if I omit the exponential term.
 
  • #8
renormalize said:
As long as we're posting simpler integrals that can be done, I'd like to nominate:(courtesy of Mathematica)
But I see no way to leverage this result to analytically evaluate the OP's original integral.
Thanks, but the cosine term is important for my case, as my problem has oscillatory behaviour
 
  • #9
I'm not sure if this will be useful for your purposes, but here’s an expansion in :



For those interested, here’s a method to prove the integral: :

Use , then



Then use :



Use , then and:

 
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  • #10
Thank you very much for your effort! It provides more insight to the question.
 
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