I recently was able to show:
\begin{equation} \int_{0}^{\infty} \frac{e^{-kx^n}}{x^n + a}\:dx = e^{ak}a^{\frac{1}{n} - 1}\frac{\Gamma\left(\frac{1}{n} \right)\Gamma\left(1 - \frac{1}{n}, ak \right)}{n} \end{equation}
For $a,k > 0$ and $n > 1$. I wanted to change the function in the numerator to $\cos\left(kx^n\right)$. In doing so, I used the property that
\begin{equation} \cos\left(mx^n\right) = \Re\left[ e^{-mx^ni}\right] \end{equation}
And so,
\begin{equation} \int_{0}^{\infty} \frac{\cos\left(kx^n\right)}{x^n + a}\:dx = \Re\left[ \int_{0}^{\infty} \frac{e^{-mx^ni}}{x^n + a}\:dx \right] \end{equation}
If we let $m' = mi$ we have:
\begin{equation} \int_{0}^{\infty} \frac{\cos\left(mx^n\right)}{x^n + a}\:dx = \Re\left[ \int_{0}^{\infty} \frac{e^{-m'x^n}}{x^n + a}\:dx \right] \end{equation}
We use the solution as per the integral above:
\begin{equation} \int_{0}^{\infty} \frac{\cos\left(mx^n\right)}{x^n + a}\:dx = \Re\left[ \int_{0}^{\infty} \frac{e^{-m'x^n}}{x^n + a}\:dx \right] = \Re\left[e^{am'}a^{\frac{1}{n} - 1}\frac{\Gamma\left(\frac{1}{n} \right)\Gamma\left(1 - \frac{1}{n}, am' \right)}{n}\right] \\ = \frac{a^{\frac{1}{n} - 1}\Gamma\left(\frac{1}{n} \right)}{n}\Re\left[e^{ami}\Gamma\left(1 - \frac{1}{n}, ami \right)\right] \end{equation}
Obviously the $e^{ami}$ is simple to deal with. Where I'm having issues is finding the Real component of Upper Incomplete Gamma Function.
Does anyone have any idea on how to do this?
Edit, i.e. is there any way to solve for $\alpha, \beta$ for:
\begin{equation} \Gamma\left(1 - \frac{1}{n}, ami \right) = \alpha + \beta i \end{equation}