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

∞∫0x−iax2+bx+1dx,

where a∈R, b∈[0,2), and 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?

**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*.

**Attribution***Source : Link , Question Author : Community , Answer Author : Jack D’Aurizio*