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I'm looking to find a closed-form (or series) solution for the definite integral

$$\int^{\pi/2}_0 \sin(x) \cos(x) \sin(a*\cos(x)) \exp(i*b*\sin(x)) dx$$

where $a$ and $b$ are real constants and $i$ is the imaginary unit.

I got a solution for the case $b=0$, involving spherical Bessel functions. But the general case $b \not = 0$ complicates matters and eludes me. Substituting tan(x/2)=u does not seem to produce a useful lead. Any suggestion(s) as to the solution (e.g. via symbolic algebra tool) or what method/substitution could be used?

Lucozade
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1 Answers1

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If you write $$e^{i b \sin (x)}=\sum_{n=0}^\infty\frac {i^n}{n!} \sin^n(x)\,b^n$$ you face integrals $$I_n=\int_0^{\frac \pi 2}\cos (x)\, \sin ^{n+1}(x)\, \sin (a \cos (x))\,dx$$ which have general solution in terms of generalized hypergeometric function but which simplify in terms of Bessel J functions.

$$I_n=\frac{\sqrt{\pi }}{a}\,\left(\frac{2}{a}\right)^{\frac{n-1}{2}}\Gamma \left(\frac{n+2}{2}\right)\, J_{\frac{n+3}{2}}(a)$$

Then the solution is $$\large\color{blue}{I=\frac{\pi}{\sqrt{2a} }\sum_{n=0}^\infty \,\frac{J_{\frac{n+3}{2}}(a)}{\Gamma \left(\frac{n+1}{2}\right)}\,\,t^n} \qquad \text{where}\quad \color{blue}{t=i\frac{b}{\sqrt{2a}}}$$

For an illustration purposes (to show the convergence), with $a=\pi$ and $b=e$

$$\left( \begin{array}{cc} p & \sum_{n=0}^p \\ 0 & +0.31830989+0.00000000 \, i \\ 1 & +0.31830989+0.65977312 \, i \\ 2 & -0.39661531+0.65977312 \, i \\ 3 & -0.39661531+0.12678791 \, i \\ 4 & -0.09150410+0.12678791 \, i \\ 5 & -0.09150410+0.26910237 \, i \\ 6 & -0.14763987+0.26910237 \, i \\ 7 & -0.14763987+0.24989264 \, i \\ 8 & -0.14182965+0.24989264 \, i \\ 9 & -0.14182965+0.25146819 \, i \\ 10 & -0.14221701+0.25146819 \, i \\ 11 & -0.14221701+0.25138105 \, i \\ 12 & -0.14219894+0.25138105 \, i \\ 13 & -0.14219894+0.25138453 \, i \\ 14 & -0.14219957+0.25138453 \, i \\ 15 & -0.14219957+0.25138442 \, i \\ \end{array} \right)$$

Claude Leibovici
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