Solution of Integral

Question

I want to solve the following integral

$$ \frac{\alpha \beta}{2} \int_0^\pi \cos\theta \sec^{2}\theta(\tan(\theta/2))^{-\beta-1} (1+\gamma(\tan(\theta/2))^{-\beta})^{-\frac{\alpha}{\gamma}-1}d\theta$$ after subtituting $(\tan(\theta/2))^{-\beta}=z$

I got this $$\int_0^\infty \frac{{(1+\gamma z)}{}^{-(\frac{\alpha}{\gamma}+1)}}{1+z^{-\frac{2}{\beta}}}dz$$

where $\alpha, \beta and \gamma>0.$ How to solve above integral? Kindly help me in this regards.

@Did I simply need someone help me to solve the stated integral. — SAAN, Mar 12 '15 at 10:50
What have you already tried? What have you already thought of? — Pedro, Mar 12 '15 at 10:51
@Did since it was pretty obvious what the question was, namely: how can i solve this integral? However, I find that the person could give more information about what he/she has already tried and the methods the persons already thought of, in order to help him or her optimally. — Pedro, Mar 12 '15 at 10:53
@Pedro thanks for comments, I initially start with larger problem, but stuck on that stage. No hint is working. — SAAN, Mar 12 '15 at 10:53
@Pedro There is, in effect, no question here. If there was, perhaps the OP would have added anything personal to their post and this would have produced an acceptable one. (But you know all this, don't you?) — Did, Mar 12 '15 at 10:56
Have you thought already of some methods? What have you tried to do? People expect that you have shown some effort or that you can present a list of things you might think can help to solve the problem, but you don't know how to implement it. Where did you found the problem? Is it from a book? Is it something you made up yourself? So that people know whether there actually should be a solution and that it is actually solvable. If you don't put this information in your post, your question might be downvoted by people. — Pedro, Mar 12 '15 at 11:00

Lucian · Answer 1 · 2019-08-21T10:05:17.227

0

For $B\in\mathbb N$ we have $~\displaystyle\int_0^\infty\frac{(1+Ax)^B}{1+x^C}~dx~=~\frac\pi C\cdot\sum_{k=0}^B~{B\choose k}~\frac{A^k}{\sin\bigg((k+1)~\dfrac\pi C\bigg)}~,~$ which

can be easily shown by expanding the numerator using the binomial theorem and letting

$t=\dfrac1{1+x^C}~,~$ then recognizing the expression of the beta function in the new integral,

and employing Euler's reflection formula for the $\Gamma$ function to simplify the result.

edited Aug 21 '19 at 10:05

answered Mar 12 '15 at 11:47

Lucian

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But the RHS diverges for every integer C? – Did Mar 12 '15 at 12:25
@Did: For convergence we must have $C>B+1$, obviously, so $\dfrac{k+1}C$ will always be strictly in between $0$ and $1$, thus avoiding the problematic points $\sin0=\sin\pi=0.$ – Lucian Mar 12 '15 at 20:41
This should be mentioned clearly in any answer. – Did Mar 12 '15 at 22:35
1

@Did: And why is that ? To prevent the reader from God-forbid ever developing any critical thought and analytic skills ? – Lucian Mar 12 '15 at 23:00
Pfff... $ $ $ $ – Did Mar 12 '15 at 23:12
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@Lucian For $B=-(\frac{\alpha}{\gamma}+1)$ this solution does not work. – SAAN Mar 13 '15 at 10:12
@Azeem: Your initial post did not mention anything about the parameters in question being positive. – Lucian Mar 13 '15 at 11:08

Solution of Integral

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