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?

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    $\begingroup$ I think that you are lucky to get an answer. If I may ask, in which context did you find this integral ? $\endgroup$ – Claude Leibovici Oct 6 '16 at 10:51
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    $\begingroup$ 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 $\cos$. $\endgroup$ – Yves Daoust Oct 11 '16 at 12:34
  • $\begingroup$ @Claude Leibovici In the quantum mechanics! $\endgroup$ – Nigel1 Oct 12 '16 at 12:15

You start with the decompositions in partial fractions

$$\frac1{x^3(1-x^2)}=\frac1{x^3}+\frac1x-\frac1{2(x+1)}-\frac1{2(x-1)}$$ 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.


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