Solution of Integral

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.

• You want that, but what is your question? – Did Mar 12 '15 at 10:49
• @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
• @Pedro Nice to know. Why? – Did Mar 12 '15 at 10:51
• 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

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.

• 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
• @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
• @Lucian For $B=-(\frac{\alpha}{\gamma}+1)$ this solution does not work. – SAAN Mar 13 '15 at 10:12