How to calculate the following definite integrals $?$: $$ \int_{0}^{\infty}{x^{\large\beta}\cos\left(ax\right) \over x^{2} - b^{2}}\,\mathrm{d}x\,\,\,\mbox{and}\,\,\, \int_{0}^{\infty}{x^{\large\beta}\sin\left(ax\right) \over x^{2} - b^{2}}\,\mathrm{d}x\,,\,\,\,\,\, \mbox{assuming}\ a,b > 0\ \mbox{and} \left\vert\,{\beta}\,\right\vert < 1. $$ I found the following integral from the table of integrals: $$ \int_{0}^{\infty}{x^{\large\beta}\cos\left(ax - \beta\pi/2\right) \over x^{2} - b^{2}}\,\mathrm{d}x = -\,{\pi \over 2}\,b^{\beta - 1} \sin\left(ab - {\pi\beta \over 2}\right) $$ but couldn't find those two. Do they exist $?$. if yes, how can I find the answer $?$.
Thank you all !.
As you probably noticed, there are a lot of problems with numerical integration.
As I wrote in comments, there is more than likely some hypergeometric function appearing for the first integral.
Hoping no miastake, I wrote $$\int_{0}^{+\infty}\displaystyle\frac{x^\beta\cos(ax)}{x^2-b^2}dx=\int_{0}^{+\infty} \sum_{n=0}^\infty b^{2n} \cos (a x) x^{\beta -2 n-2}\,dx$$ $$I_n=\int_{0}^{+\infty}\cos (a x) x^{\beta -2 n-2}\,dx=-a^{2 n+1-\beta} \sin \left(\pi n-\frac{\pi \beta }{2}\right) \Gamma (-2 n+\beta -1)$$ which, effectively, gives $$\int_{0}^{+\infty}\displaystyle\frac{x^\beta\cos(ax)}{x^2-b^2}dx=a^{1-\beta } \sin \left(\frac{\pi \beta }{2}\right) \Gamma (\beta -1) \, _1F_2\left(1;\frac{2-\beta }{2},\frac{3-\beta}{2};-\frac{a^2b^2}{4} \right)$$
Edit
@Maxim pointed out in comments a serious mistake in my approach. As he/she wrote, with the prinicipal value, to this result must be added the term $$\frac{\pi}{2} b^{\beta -1} \tan \left(\frac{\pi \beta }{2}\right) \cos (a b)$$
Similarly
$$\int_{0}^{\infty}{x^{\large\beta}\sin\left(ax\right) \over x^{2} - b^{2}}\,dx=-a^{1-\beta } \cos \left(\frac{\pi \beta }{2}\right) \Gamma (\beta -1) \, _1F_2\left(1;\frac{2-\beta }{2},\frac{3-\beta}{2};-\frac{a^2 b^2}{4} \right)-$$ $$\frac{\pi}{2} b^{\beta -1} \cot \left(\frac{\pi \beta }{2}\right) \sin (a b)$$