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.

  • $\begingroup$ You want that, but what is your question? $\endgroup$ – Did Mar 12 '15 at 10:49
  • $\begingroup$ @Did I simply need someone help me to solve the stated integral. $\endgroup$ – SAAN Mar 12 '15 at 10:50
  • $\begingroup$ What have you already tried? What have you already thought of? $\endgroup$ – Pedro Mar 12 '15 at 10:51
  • $\begingroup$ @Pedro Nice to know. Why? $\endgroup$ – Did Mar 12 '15 at 10:51
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    $\begingroup$ 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. $\endgroup$ – Pedro Mar 12 '15 at 11:00

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.

  • $\begingroup$ But the RHS diverges for every integer C? $\endgroup$ – Did Mar 12 '15 at 12:25
  • $\begingroup$ @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 $\sin0=\sin\pi=0.$ $\endgroup$ – Lucian Mar 12 '15 at 20:41
  • $\begingroup$ This should be mentioned clearly in any answer. $\endgroup$ – Did Mar 12 '15 at 22:35
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    $\begingroup$ @Did: And why is that ? To prevent the reader from God-forbid ever developing any critical thought and analytic skills ? $\endgroup$ – Lucian Mar 12 '15 at 23:00
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    $\begingroup$ @Lucian For $B=-(\frac{\alpha}{\gamma}+1)$ this solution does not work. $\endgroup$ – SAAN Mar 13 '15 at 10:12

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