Integral could lead to Hypergeometric function

In summary, the integral is convergent for some values of (n), which can be either positive or negative, but not for any integer value of (n), including 0. It can also be expressed in the form of a hypergeometric function when b is set to 0, but Mathematica is unable to solve it for non-zero values of b.
  • #1
DMESONS
27
0
How can I perform this integral



\begin{equation}
\int^∞_a dq \frac{1}{(q+b)} (q^2-a^2)^n (q-c)^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!
 
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  • #2
Hi !

What do you think about the convergence, or not ?
 
  • #3
JJacquelin said:
Hi !

What do you think about the convergence, or not ?

Thanks for your comment.

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
DMESONS said:
How can I perform this integral



\begin{equation}
\int^∞_a dq \frac{1}{(q+b)} (q^2-a^2)^n (q-c)^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!
I'll preface this by asking what it is for, but I'll try to provide a partial solution, too.

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:

$$\int\limits_{[a,+\infty)}\left(\frac{1}{q+b}(q^2-a^2)^n(q-c)^n\right)\mathrm{d}q=\int\limits_{[a,+\infty)}\left(\frac{1}{q+b}(q-a)^n(q+a)^n(q-c)^n\right)\mathrm{d}q.$$

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
DMESONS said:
all parameters are positive (a, b, and c) and n>0.
DMESONS said:
The integral is convergent for some values of (n) which can be either positive or negative.
Hi !
Would you mind give a non contradictory wording of the question about the sign of n.
 
  • #6
DMESONS said:
The integral is convergent for some values of (n) which can be either positive or negative.

Hi !
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.
 

FAQ: Integral could lead to Hypergeometric function

1. What is an integral?

An integral is a mathematical concept that represents the area under a curve in a given interval. It is a fundamental concept in calculus and is used to calculate quantities such as displacement, velocity, and acceleration.

2. What is a Hypergeometric function?

A Hypergeometric function is a special type of mathematical function that arises in problems involving counting and probability. It is a generalization of the binomial coefficient and is used to solve various problems in statistics, physics, and engineering.

3. How does an integral lead to a Hypergeometric function?

When evaluating an integral, there are often instances where the resulting function cannot be expressed in terms of elementary functions. In these cases, the integral is often written in terms of a Hypergeometric function, which provides a more general and concise representation.

4. What are the applications of Hypergeometric functions?

Hypergeometric functions have a wide range of applications in different fields of science and engineering. Some examples include solving problems in quantum mechanics, statistical analysis, and solving differential equations in physics and engineering.

5. Are there any limitations to using Hypergeometric functions?

While Hypergeometric functions are useful in many applications, they have certain limitations. For example, they may not be applicable to all types of integrals, and their use may require specific knowledge in advanced mathematics. Additionally, they may not always provide a closed-form solution, which can make their use challenging in certain scenarios.

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