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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}

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  • $\begingroup$ it may be beneficial to note that $$\Gamma(s,x)+\gamma(s,x)=\Gamma(s)$$ where $\gamma(s,x)$ is the lower incomplete gamma function. Fortunately we have a representation for $\gamma(s,x)$ with the hypergeometric function $_1F_1$: $$\gamma(s,x)=s^{-1}x^{s}e^{-x}\,_1F_1(1;1+s;x)$$ You can learn more here: mathworld.wolfram.com/IncompleteGammaFunction.html $\endgroup$ – clathratus Dec 19 '18 at 4:43
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    $\begingroup$ Thanks @clathratus! Yes, I hoping I can employ those. I have been able to find a single 2nd Order ODE and a paired 1st Order ODE System that solves this, so am going to try all fronts and hope for the best haha :-) $\endgroup$ – user150203 Dec 19 '18 at 4:55

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