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Integral could lead to Hypergeometric function 
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#1
Nov813, 11:52 AM

P: 27

How can I perform this integral
\begin{equation} \int^∞_a dq \frac{1}{(q+b)} (q^2a^2)^n (qc)^n ? \end{equation} all parameters are positive (a, b, and c) and n>0. I tried using Mathemtica..but it doesn't work! if we set b to zero, above integral leads to the hypergeometric function! 


#2
Nov813, 12:16 PM

P: 761

Hi !
What do you think about the convergence, or not ? 


#3
Nov813, 12:41 PM

P: 27

The integral is convergent for some values of (n) which can be either positive or negative. As I mentioned, if I set b =0, the result have the form of hypergeometric function! But if b is not zero, Mathematica can't solve it. 


#4
Nov813, 02:09 PM

P: 615

Integral could lead to Hypergeometric function
To start, what on Earth is this for? We do we come up with such silly things to integrate? Second, let's see if we can simplify things considerably: $$\int\limits_{[a,+\infty)}\left(\frac{1}{q+b}(q^2a^2)^n(qc)^n\right)\mathrm{d}q=\int\limits_{[a,+\infty)}\left(\frac{1}{q+b}(qa)^n(q+a)^n(qc)^n\right)\mathrm{d}q.$$ I'm thinking we might just approach this by means of partial fractions. On cursory examination, I don't see a contour that would simplify things, so brute force might be necessary. 


#5
Nov913, 02:10 AM

P: 761

Would you mind give a non contradictory wording of the question about the sign of n. 


#6
Nov1013, 02:17 AM

P: 761

For which value of (n) the integral is convergent ? Clue : The integral is NOT convergent for any integer (n), either positive or negative or n=0. 


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