Integrals Involving Contour Integration and Trigonometric Numerators I've been examining integrals of the form
$$I=\int_0^\infty\frac{x^\mu\cos(ax)}{1+x^{2p}}dx$$
and
$$J=\int_0^\infty\frac{x^\mu\sin(ax)}{1+x^{2p}}dx$$
for $-1<\mu<2p-1$, $a\in\mathbb{R}$, and $p\in\mathbb{N}$. Without knowing any answers beforehand, I went about trying to evaluate these integrals using contour integration, where I instead seek the evaluation of
$$K=\int_0^\infty\frac{x^\mu e^{iax}}{1+x^{2p}}dx$$
and $K=I+iJ$. The contour that I selected is a semicircle in the upper half plane, where we avoid the possible singularity at $0$ by doing a small semicircle clockwise about $0$ (I can upload a picture if necessary). Keeping branch cuts in mind, the integral along the negative real axis is equivalent to $\bar{K}e^{i\mu\pi}$, since the contour runs along an angle of $\pi$. The upper arc of the contour vanishes as the radius gets large (as does the smaller semicircle when it gets small) and we have
$$K+\bar{K}e^{i\mu\pi}=\oint_C\frac{z^\mu e^{iaz}}{1+z^{2p}}dz$$
If I multiply this by $e^{-i\mu\pi/2}/2$, this then reads
$$\int_0^\infty\frac{x^\mu\cos(ax-\mu\pi/2)}{1+x^{2p}}dx=\frac{e^{-i\mu\pi/2}}{2}\oint_C\frac{z^\mu e^{iaz}}{1+z^{2p}}dz$$
I can compute the contour integral directly, although it is quite messy, so I will simply ignore writing out the icky formula and simply state that I can evaluate
$$\int_0^\infty\frac{x^\mu\cos(ax-\mu\pi/2)}{1+x^{2p}}dx=f(\mu,a,p)$$
The annoying part of this is that I am only able to evaluate $I$ and $J$ for particular values of $\mu$; namely, we know $I$ when $\mu$ is even and $J$ when $\mu$ is odd, but nothing inbetween. I was wondering if there was a way to recover $I$ or $J$ from this method. I have tried differentiating with respect to $a$ to see if there is a relation that I can use, but that didn't seem to work. If anyone has an idea, or an alternate method, it would be much appreciated. I also don't even know if these integrals have a closed form that can be extracted symbolically.
Thank you for reading!
 A: Not an answer
Applying $x\mapsto -x$ on $I$, we have 
$$I=(-1)^\mu\int^0_{-\infty}\frac{x^\mu\cos ax}{1+x^{2p}}dx$$
Thus, $$I\left(1+\frac1{(-1)^\mu}\right)=\int^\infty_{-\infty}\frac{x^\mu\cos ax}{1+x^{2p}}dx$$
Similarly, $$J\left(1-\frac1{(-1)^\mu}\right)=\int^\infty_{-\infty}\frac{x^\mu\sin ax}{1+x^{2p}}dx$$
Hence, 
$$I\left(1+\frac1{(-1)^\mu}\right)+iJ\left(1-\frac1{(-1)^\mu}\right)=\underbrace{\int^\infty_{-\infty}\frac{x^\mu e^{iax}}{1+x^{2p}}dx}_{F}\qquad(*)$$

$F$ can be evaluated by residue theorem easily, giving $$F=-
\frac{\pi i}{p}\sum^p_{n=1}\omega_n^{\mu+1} \exp(ia\omega_n)
\qquad\text{where}\qquad
\omega_n=\exp\left(\frac{2n-1}{2p}\pi i\right)
$$

Anyway, let $F=M+iN$ where $M,N\in\mathbb R$. 
Expanding $(*)$ and comparing real and imaginary parts give
$$(1+\cos\pi\mu)I+(-\sin\pi\mu)J=M$$
$$(-\sin\pi\mu)I+(1-\cos\pi\mu)J=N$$
Unfortunately this system does not give the solution to $I,J$. However, it does give some relations between $I,J$ and $M,N$, namely
$$I=\frac{M+\sin\pi\mu\cdot J}{1+\cos\pi\mu}$$
$$\frac{M}{1+\cos\pi\mu}+\frac {N}{\sin\pi\mu}=0$$
Hope that this is a little helpful.
