Integrating Double Integral using Substitution and Integration by Parts

  • Thread starter Kuma
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In summary, the conversation is about a difficult double integral problem involving the use of integration by parts and substitution. The person has set up the integration by parts and substitution but is stuck on finding the correct answer. They are also unsure if the reverse chain rule applies in this case. The suggestion is made to post the entire problem for better understanding.
  • #1
Kuma
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Homework Statement


Okay this is part of a much bigger double integral, but I have hit a roadblock here.

[tex]\int[/tex]x1/u (top), 0 (bottom). (1-(x1/u))^a2 -1 ((x1/u) - x1)^a3 - 1


Homework Equations



integration by parts/substitution?


The Attempt at a Solution



so first i set up an integration by parts here with
du = (1-(x1/u))^a2 -1

and

v = ((x1/u) - x1)^a3 - 1

i found out dv and to find out u i had to integrate that first part. So now I am stuck because this is a reverse chain rule i believe, so i set up a substitution with t = 1-(x1/u) and found out dt = x1/u^2 du. However i keep getting the wrong answer, any help?
 
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  • #2
Kuma said:

Homework Statement


Okay this is part of a much bigger double integral, but I have hit a roadblock here.

[tex]\int[/tex]x1/u (top), 0 (bottom). (1-(x1/u))^a2 -1 ((x1/u) - x1)^a3 - 1


Homework Equations



integration by parts/substitution?


The Attempt at a Solution



so first i set up an integration by parts here with
du = (1-(x1/u))^a2 -1

and

v = ((x1/u) - x1)^a3 - 1

i found out dv and to find out u i had to integrate that first part. So now I am stuck because this is a reverse chain rule i believe, so i set up a substitution with t = 1-(x1/u) and found out dt = x1/u^2 du. However i keep getting the wrong answer, any help?
Maybe post the whole integration problem so we could see how you got to this point?
:smile:
 
  • #3
Yes, I think it would help to see the rest of the problem. If you were integrating over x1 you could express the result in terms of an incomplete beta function. However from reading your comments, the integral seems to be over u, which makes it more complicated.
 

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