We were told today by our teacher (I suppose to scare us) that in certain schools for physics in Soviet Russia there was as an entry examination the following integral given
$$\int\limits_{0}^{\infty} \frac{x^{-\mathfrak{i}a}}{x^2+bx+1} \,\mathrm{d}x\,,$$
where $a \in \mathbb{R}$, $b \in [0,2)$, and $\mathfrak{i}$ is the imaginary unit. And since we are doing complex analysis at the moment, it can, according to my teacher, be calculated using complex methods.
I was wondering how this could work? It seems hard to me to find a good curve to apply the residue theorem for this object, I suppose. Is there a trick to compute this integral?
Best Answer
$\phantom{}$ Dear MSE users, this is the new episode of Mister Feynman and Monsieur Laplace versus contour integration.
Tonight we have a scary integral, but we may immediately notice that $$ \mathcal{L}(x^{-ia})(s) = s^{ia-1}\Gamma(1-ia),\qquad \mathcal{L}^{-1}\left(\frac{1}{x^2+bx+1}\right)(s) =\frac{e^{Bs}-e^{\overline{B}s}}{\sqrt{b^2-4}}$$ where $B$ is the root of $x^2+bx+1$ with a positive imaginary part. By the properties of the Laplace transform, the original integral is converted into $$\frac{\Gamma(1-ia)}{\sqrt{b^2-4}}\int_{0}^{+\infty}s^{ia-1}\left(e^{Bs}-e^{\overline{B}s}\right)\,ds$$ which can be evaluated in terms of the $\Gamma$ function.
Due to the reflection formula, the final outcome simplifies into $$ \frac{\left(B^{-i a}-\overline{B}^{-i a}\right) \pi }{\sinh(\pi a)\sqrt{4-b^2}} $$ and we may notice that $B=\exp\left[i\arccos\frac{b}{2}\right]$ allows a further simplification.
Outro: poor children.