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I met this integral while I was trying to model element-specific transmission electron microscopy images of core-shell nanocrystals where the shell component diffuses into the core:

$$\int_{1}^{b} \frac{\sin(a x)}{\sqrt{x^2-1}} dx \space\space\space [a > 0, b > 1]$$

Being just a chemist, I tried naive approaches such as (in order) Wolfram Alpha, Mathematica, integration by parts, Gradshteyn&Ryzhik. All I got is that

$$\int_{1}^{\infty} \frac{\sin(a x)}{\sqrt{x^2-1}} dx = \frac{\pi}{2} J_0(a) \space\space\space[a > 0]$$

from the last approach, of course :-) (GR, 3.753.3).

My question is: is there a solution to the former integral or should I confide in numerical integration?

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    $\begingroup$ You have the closed form solution as $\frac{\pi}{2}J_0(a)$. That is as good as it gets. $\endgroup$ – Mark Viola Dec 19 '16 at 17:26
  • $\begingroup$ Bessel function $\endgroup$ – Bumblebee Dec 19 '16 at 17:53
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    $\begingroup$ Thanks for your comment. Yes, a Bessel function is welcome! Unfortunately, the integral from GR has [1,+Infinity] as integration limits. The integral I am interested in has [1,b] as integration limits, where b is a real number larger than 1. In this case, I could not find an analytical solution. Such a solution would save some programming and testing effort. $\endgroup$ – Alessandro Ponti Dec 20 '16 at 8:24
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The answer you have (the zeroth order Bessel function of the first kind) is no less valid a solution than sin(x), ln(x), or exp(x), which I would presume are acceptable answers.

Then it is a question of how you generate the values you need in a reliable way. For example, look at Numerical Recipes in Fortran 77 for a discussion of this.

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    $\begingroup$ Thanks for your comment. Yes, a Bessel function is welcome!The integral from GR has [1,+Infinity] as integration limits. The integral I am interested in has [1,b] as integration limits, where b is a real number larger than 1. $\endgroup$ – Alessandro Ponti Dec 20 '16 at 8:18
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For $b=\infty$, your integral matches the "Mehler–Sonine" form (see http://dlmf.nist.gov/10.9)

$$\int_0^\infty\sin(x\cosh(t))dt=\frac\pi2J_0(x).$$

For a finite upper bound, the integral is the indefinite version of the above and will be more complicated than a Bessel function, as it has an extra degree of freedom.

Don't expect an easy closed-from.

Be careful when performing numerical integration, as the function is alternating. It might be useful to decompose the integrand in a cardinal sine and a remainder.

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  • $\begingroup$ The DLMF result is the same as the second integral in the OP, after a substitution with $\cosh$. The "alternating" part can be easily dealt with by modern quadrature methods (e.g. Clenshaw-Curtis, Levin); otherwise, one can always split the integral among the zeroes of $\sin ax$ and sum afterwards. $\endgroup$ – J. M. is a poor mathematician Dec 20 '16 at 8:52
  • $\begingroup$ Thank you for your comment. I had a look to DLMF and Erdelyi's book and I got the impression that no closed/series form of the indefinite integral is known. JM: I did not know Clenshaw-Curtis method: thanks for the suggestion. $\endgroup$ – Alessandro Ponti Dec 20 '16 at 13:30

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