How to evaluate $\int_0^\infty \frac{\cos(ax)}{1+x^N}dx$ for $a,N\in\mathbb{R}$ and $N> 1$ Interesting challenge: Does anyone know how to evaluate the integral $$\int_0^\infty\frac{\cos(ax)}{1+x^N}\text{d}x$$ for $a,N\in\mathbb{R}$ and $N>1$?
The $N=2$ case is quite easily done using the Residue Theorem as well as for general $N>1$ for $a=0$. Both exploit symmetries when path integrating, which do not work in the general case.
Let me know if you have any ideas! Cheers!
 A: For even $N$ you can use the residue theorem as well. It suffices to consider $a>0$ and calculate the integral
$$\oint_{-\infty}^{\infty} \frac{e^{iax}}{1+x^N} \, {\rm d}x = \int_{0}^{\infty} \frac{2\cos(ax)}{1+x^N} \, {\rm d}x = 2\pi i \sum_{0 \leq k \leq \frac{N-1}{2}} {\rm Res}\left( \frac{e^{iax}}{1+x^N} \right)\Bigg|_{x=e^{\frac{i\pi+i2\pi k}{N}}} \\
=\frac{-2\pi i}{N} \sum_{k=0}^{N/2-1} e^{ia e^{\frac{i\pi+i2\pi k}{N}} + \frac{i\pi+i2\pi k}{N}} = \frac{-2\pi i}{N} \sum_{k=0}^{N/2-1} e^{ia \cos\left(\frac{\pi+2\pi k}{N}\right) + \frac{i\pi+i2\pi k}{N}}e^{-a\sin\left( \frac{\pi+2\pi k}{N} \right)} \\
=\frac{-2\pi i}{N} \sum_{k=0}^{N/2-1} \Bigg[\cos\left(a \cos\left(\frac{\pi+2\pi k}{N}\right) + \frac{\pi+2\pi k}{N}\right) \\ + i \sin\left(a \cos\left(\frac{\pi+2\pi k}{N}\right) + \frac{\pi+2\pi k}{N}\right)\Bigg]e^{-a\sin\left( \frac{\pi+2\pi k}{N} \right)} \\
=\frac{2\pi}{N} \sum_{k=0}^{N/2-1} \sin\left(a \cos\left(\frac{\pi+2\pi k}{N}\right) + \frac{\pi+2\pi k}{N}\right)e^{-a\sin\left( \frac{\pi+2\pi k}{N} \right)}$$
where the loop-integral on the LHS is closed in the upper half plane. The imaginary part is zero, since the LHS is real.
A: 
Just some working out in progress.

Well, we are trying to find the following integral:
$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right):=\int_0^\infty\frac{\cos\left(\text{k}x\right)}{\alpha^2+x^\text{n}}\space\text{d}x\tag1$$
To calculate the integral, we first note we may write that:
$$\frac{1}{\alpha^2+x^\text{n}}=\int_0^\infty\exp\left(-\left(\alpha^2+x^\text{n}\right)\text{y}\right)\space\text{dy}\tag2$$
And then we have:
$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right)=\int_0^\infty\left(\int_0^\infty\cos\left(\text{k}x\right)\exp\left(-\left(\alpha^2+x^\text{n}\right)\text{y}\right)\space\text{dy}\right)\space\text{d}x\tag3$$
Change the integration order assuming the value of the integral preserves:
$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right)=\int_0^\infty\exp\left(-\text{y}\alpha^2\right)\left(\int_0^\infty\cos\left(\text{k}x\right)\exp\left(-\text{y}x^\text{n}\right)\space\text{d}x\right)\space\text{dy}\tag4$$
Make use of the power series of $\cos x$:
$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right)=\int_0^\infty\exp\left(-\text{y}\alpha^2\right)\left(\int_0^\infty\sum_{\text{m}\space\ge\space0}\frac{\left(-1\right)^\text{m}\left(\text{k}x\right)^{2\text{m}}}{\left(2\text{m}\right)!}\cdot\exp\left(-\text{y}x^\text{n}\right)\space\text{d}x\right)\space\text{dy}\tag5$$
Which gives:
$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right)=\int_0^\infty\exp\left(-\text{y}\alpha^2\right)\left(\sum_{\text{m}\space\ge\space0}\frac{\left(-1\right)^\text{m}\text{k}^{2\text{m}}}{\left(2\text{m}\right)!}\int_0^\infty x^{2\text{m}}\exp\left(-\text{y}x^\text{n}\right)\space\text{d}x\right)\space\text{dy}\tag6$$
If you substitute:
$$\text{s}=\text{y}^\frac{2\text{m}+1}{\text{n}}\cdot x^{2\text{m}+1}\tag7$$
We will get:
$$\int_0^\infty x^{2\text{m}}\exp\left(-\text{y}x^\text{n}\right)\space\text{d}x=\frac{\text{y}^{-\left(\frac{2\text{m}}{\text{n}}+\frac{1}{\text{n}}\right)}}{2\text{m}+1}\int_0^\infty\exp\left(-\text{s}^\frac{\text{n}}{2\text{m}+1}\right)\space\text{ds}\tag8$$
Which is defined in terms of the incomplete gamma function:
$$\int_0^\infty\exp\left(-\text{s}^\frac{\text{n}}{2\text{m}+1}\right)\space\text{ds}=-\frac{2\text{m}+1}{\text{n}}\cdot\left[\Gamma\left(\frac{2\text{m}+1}{\text{n}},\text{s}^\frac{\text{n}}{2\text{m}+1}\right)\right]_0^\infty\tag9$$
So:
$$\int_0^\infty x^{2\text{m}}\exp\left(-\text{y}x^\text{n}\right)\space\text{d}x=\frac{\text{y}^{-\left(\frac{2\text{m}}{\text{n}}+\frac{1}{\text{n}}\right)}}{\text{n}}\cdot\left[\Gamma\left(\frac{2\text{m}+1}{\text{n}},\text{s}^\frac{\text{n}}{2\text{m}+1}\right)\right]_\infty^0\tag{10}$$
Which gives:
$$\int_0^\infty x^{2\text{m}}\exp\left(-\text{y}x^\text{n}\right)\space\text{d}x=\frac{\text{y}^{-\left(\frac{2\text{m}}{\text{n}}+\frac{1}{\text{n}}\right)}}{\text{n}}\cdot\Gamma\left(\frac{2\text{m}+1}{\text{n}}\right)\tag{11}$$
So:
$$\mathcal{I}_\text{n}\left(\text{k},\alpha\right)=\frac{1}{\text{n}}\sum_{\text{m}\space\ge\space0}\frac{\left(-1\right)^\text{m}\text{k}^{2\text{m}}}{\left(2\text{m}\right)!}\cdot\Gamma\left(\frac{2\text{m}+1}{\text{n}}\right)\int_0^\infty\exp\left(-\text{y}\alpha^2\right)\text{y}^{-\left(\frac{2\text{m}}{\text{n}}+\frac{1}{\text{n}}\right)}\space\text{dy}\tag{12}$$
A: The integral
$$f(a) = \int_0^\infty\frac{\cos(a x)}{1+x^N}\;dx$$
with $N= 1, 2, ...$ and $a\ge 0$ can be calculated explicitly in terms of known functions as follows.
Partial fraction decomposition leads to the representation
$$\frac{1}{1+ x^N} = \sum_{k=1}^{N} \frac{c_k}{z_k - x}\tag{1}$$
where
$$z_k=\exp\left( \frac {i \pi (2k+1)}{N}\right)\tag{2}$$
are the $N$ different roots of the equation $1+ x^N=0 $ and the coefficients are complex numbers given by
$$c_k = \lim_{x\to z_k} \frac{z_k - x}{1+x^N}=(-1)^N \frac{1}{\prod_{i=1, i\ne k}^{N} (z_i - z_k)}\tag{3}$$
Hence we are left with the integral
$$i(z,a)=\int_{0}^{\infty} \frac{\cos(a x)}{z-x}\;dx\tag{4}$$
Which can be solved giving
$$i(z,a) =  \frac{\pi}{2}\sin(a z) + \sin(a z)\text{ Si}(a z) + \cos(a z)\text{ Ci}(-a z)\tag{5}$$
Here $\text{ Si}$ and $\text{ Ci}$ are the integral sine and integral cosine, respectively (cf. https://en.wikipedia.org/wiki/Trigonometric_integral).
Putting things together we have found that the integral in question has a closed form in terms of known functions.
Remark: Just for information (not for practical use): Mathematica tells me that the integral can be expressed in compact form by the special function MeijerG.
