Compute the integral: $$\int_0^{\infty} \frac{x^p}{x^2 + 2x\cos\lambda+1}dx, (-1 < p < 1, -\pi < \lambda < \pi)$$ with complex analysis techinques.
What I've tried in this problem (because the problem is located among other problems that I've solved using the $\textit{circle that avoids positive part of x-axes}$) is to find residues in the singularities and that would give me solution. But, in case of $0 < p < 1$ this integral does not have same properties as in case $- 1 < p < 0$ (we had classified them as separate types), so I am clueless here. Any help is appreciated. Should I use any particular curve that is non-standard, or I am missing something? Thank you.

$\textbf{Edit:}$ Also, it is clear to me that the $p$ is chosen in such way that the integral converges.

  • $\begingroup$ See this answer from this link: math.stackexchange.com/a/890358/349501 $\endgroup$ – Shashi Jan 24 '18 at 0:38
  • $\begingroup$ What you also can do is to consider the same integral but with $(-x) ^p$ and do the integration on a dog bone contour with the "bone" on the positive real axis. Use the Principal Log for that. $\endgroup$ – Shashi Jan 24 '18 at 0:45
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    $\begingroup$ It might be worth to have a look at the following answer for the latter suggestion. The exponent is imaginary but the technique remains the same. math.stackexchange.com/questions/2529614/… $\endgroup$ – Shashi Jan 24 '18 at 0:51
  • $\begingroup$ @Shashi that is idea and contour I was looking for. Thank you. $\endgroup$ – Nemanja Beric Jan 24 '18 at 5:33

By the Laplace transform

$$ \int_{0}^{+\infty}\frac{x^p}{x^2+2x\cos\lambda+1}=\frac{\Gamma(p+1)}{\sin\lambda}\int_{0}^{+\infty}s^{-p-1}e^{-s\cos\lambda}\sin(s\sin\lambda)\,ds $$ and by the integral representation for the $\Gamma$ function and the reflection formula the RHS equals $$ \frac{-\Gamma(p+1)\,\Gamma(-p)\sin(p\lambda)}{\sin\lambda}=\frac{\pi\sin(p\lambda)}{\sin(\lambda)\sin(\pi p)}.$$

  • $\begingroup$ Is there any other way to solve it? We haven't done Laplace transforms during any of analysis courses. $\endgroup$ – Nemanja Beric Jan 23 '18 at 22:57
  • $\begingroup$ The same can be achieved by contour integration, once a suitable contour has been found. The main reasons I prefer to use the Laplace transform when possible is due to its simplicity and the fact that, like Feyman's trick, it allows to skip the hunt for the right contour. So, even if you have not met it during your courses, it is definitely worth studying. $\endgroup$ – Jack D'Aurizio Jan 23 '18 at 23:07

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