# how to calculate the integral $\int_{0}^{+\infty}\displaystyle\frac{x^\beta\cos(ax)}{x^2-b^2}dx$

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 !.

• I suppose that it could lead to some hypergometric functions. – Claude Leibovici Sep 15 '20 at 10:03
• It must be a $Principal\ Value$, I guess. – Felix Marin Sep 15 '20 at 14:55
• Probably irrelevant if you're only interested in obtaining the asymptotics, but the linked question gives a way to rigorously derive the closed form. $\int_0^\infty x^\beta/(x^2 + 1) e^{-a b x} dx$ is an integral of a product of two Meijer G-functions. There is a general result (the Mellin convolution method) that says the integral is again a G-function and another general result (Slater's theorem) that says this G-function can be expanded into a sum of generalized hypergeometric functions if $\beta \neq 0$ and $\beta \neq 1$. – Maxim Oct 6 '20 at 12:39
• @Maxim Thank you for letting me know, I am very much interested in the result you mentioned, will definitely check it out! BTW, can you please recommend some learning materials about the asymptotic approach, I am very new in this area and want to systematically learn it. Thank you! – Yolbarsop Oct 6 '20 at 13:30
• I'd recommend Bender, Orszag, Advanced Mathematical Methods for Scientists and Engineers. – Maxim Oct 6 '20 at 15:25

## 1 Answer

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

$$\int_{0}^{\infty}{x^{\large\beta}\sin\left(ax\right) \over x^{2} - b^{2}}\,dx=-a^{1-\beta } \cos \left(\frac{\pi \beta }{2}\right) \Gamma (\beta -1) \, _1F_2\left(1;\frac{2-\beta }{2},\frac{3-\beta}{2};-\frac{a^2 b^2}{4} \right)-$$ $$\frac{\pi}{2} b^{\beta -1} \cot \left(\frac{\pi \beta }{2}\right) \sin (a b)$$

• The integrals diverge because of the pole at $x = b$. The principal value integrals exist: $$\operatorname {v. \! p.} \int_0^\infty \frac {x^\beta \cos a x} {x^2 - b^2} dx = \\ a^{1 - \beta} \hspace {1.5 px} \Gamma(\beta - 1) \hspace {1.5 px} {_1 \hspace {-1.5 px} F_2} {\left( 1; \frac {2 - \beta} 2, \frac {3 - \beta} 2; -\frac {a^2 b^2} 4 \right)} \sin \frac {\pi \beta} 2 + \frac \pi 2 b^{\beta - 1} \cos(a b) \tan \frac {\pi \beta} 2, \\ a > 0, \quad b > 0, \quad -1 < \beta < 2.$$ The first term is the termwise integral of your divergent sum (with the lower limit $n = 0$). – Maxim Sep 15 '20 at 22:09
• @Maxim. Thank you ! I was hoping to get around the problem using series but I realize that, once more, this was stupid. – Claude Leibovici Sep 16 '20 at 3:04
• @Claude Leibovici Thank you so much for your detailed answer, this is awesome. I have very limited knowledge of the hypergeometric function. If I want to get the asymptotic behavior of 1F2(a,b,c,z) for large |z|. Is there any reference I can find the answer? Thank you! – Yolbarsop Sep 17 '20 at 8:44
• @Maxim. Thank you very much for your important note on this problem. – Yolbarsop Sep 17 '20 at 8:45
• @Yolbarsop See DLMF 16.11. Since $z < 0$, you'll get $A ((\sqrt z)^\alpha e^{B \sqrt z} + (-\sqrt z)^\alpha e^{-B \sqrt z}) + o(|z|^{\alpha/2})$. If you're interested in the derivation, that should probably be a separate question. – Maxim Sep 17 '20 at 12:17