# Definite integral involving trigonometric functions and algebraic function

Consider integral with real $a, y$ and where $y> 1$

$$I(a, y) = \int_0^{y} \mathrm{d}x \frac{1}{1-x^2}~ \frac{\sin(ax) - (ax)\cos(ax)}{(ax)^3}$$

where singularity at $x=1$ is dealt with by taking principal value.

Mathematica spits out long and ugly result involving sine integral and cosine integral.

What methods to try to get solution in more beautiful form?

• I think that you are lucky to get an answer. If I may ask, in which context did you find this integral ? – Claude Leibovici Oct 6 '16 at 10:51
• The solution isn't so ugly. It is a linear combinaiton of the form $\sum c_k\,Ci(x-r_k)+s_k\,Si(x-r_k)$ where $r_k=-1,0,1$ plus a few fractions with $\sin$ and $\cos$. – Yves Daoust Oct 11 '16 at 12:34
• @Claude Leibovici In the quantum mechanics! – Nigel1 Oct 12 '16 at 12:15

$$\frac1{x^3(1-x^2)}=\frac1{x^3}+\frac1x-\frac1{2(x+1)}-\frac1{2(x-1)}$$ and $$\frac1{x^2(1-x^2)}=\frac1{x^2}+\frac1{2(x+1)}-\frac1{2(x-1)}.$$
The powers in the denominators can be decreased by parts. In the end, you get a sum of integrands of the form $\sin x/(x-a)$ or $\cos x/(x-a)$, which are indeed sine and cosine integrals.