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I am working on a detection problem which finally, I have to solve the following 2-D integral:

$$\int\limits_{a}^{b} \int\limits_{c}^{d}e^{A\sin(x)\cos(y-B)}\, \mathrm{d}x \, \mathrm{d}y \ ,$$

where $a,b,c,d$ ,$A$ and $B$ are constants. Does anyone can help me with the analytical solution for this integral?

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    $\begingroup$ Your question might be more appropriate for the math stackexchange. $\endgroup$
    – MBaz
    Oct 22 '20 at 17:07
  • $\begingroup$ I have got no answer there. $\endgroup$
    – Mahdi
    Oct 22 '20 at 17:19
  • $\begingroup$ You should look up integral identities for Bessel Functions. $\endgroup$
    – Andy Walls
    Oct 22 '20 at 17:19
  • $\begingroup$ It would be turned out to a Bessel function if the integral be in a period (which is not in this case). $\endgroup$
    – Mahdi
    Oct 22 '20 at 17:55
  • $\begingroup$ Essentially the same question has been cross-posted at Math.SE math.stackexchange.com/q/3871844/441161 $\endgroup$
    – Andy Walls
    Oct 24 '20 at 14:01
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Since, the Taylor series expansion for $e^z$ about $z=0$ is

$$e^z = \sum_{n=0}^{\infty}\dfrac{z^n}{n!}$$

then, ignoring the question of convergence, you can say

$$\begin{align*} \int_a^b \int_c^d e^{A\sin(x)\cos(y-B)}dx dy &= \int_a^b \int_c^d \sum_{n=0}^{\infty}\dfrac{(A\cos(y-B))^n}{n!}\sin^n x \; dx \, dy\\ \\ &= \int_a^b \sum_{n=0}^{\infty}\dfrac{(A\cos(y-B))^n}{n!}\int_c^d\sin^n x \; dx \, dy\\ \\ &= \sum_{n=0}^{\infty}\dfrac{A^n}{n!}\int_a^b \cos^n(y-B)\int_c^d\sin^n x \; dx \, dy\\ \\ &= \sum_{n=0}^{\infty}\dfrac{A^n}{n!}\left(\int_{a-B}^{b-B} \cos^n u\;du\right)\left(\int_c^d\sin^n x \; dx \right)\\ \\ &= \left(u\big|_{a-B}^{b-B}\right)\left(x\big|_c^d\right) + A\left(\sin u\big|_{a-B}^{b-B}\right)\left(-\cos x\big|_c^d\right) \\ &\quad + \sum_{n=2}^{\infty}\dfrac{A^n}{n!}\left(\int_{a-B}^{b-B} \cos^n u\;du\right)\left(\int_c^d\sin^n x \; dx \right)\\ \end{align*}$$

Using integration by parts, the reduction formulas for the integrals of $\sin^nx$ and $\cos^nu$ work out to be

$$\int \sin^nx \; dx = -\dfrac{1}{n}\sin^{n-1} x \cos x +\dfrac{n-1}{n}\int \sin^{n-2} x \;dx$$ and $$\int \cos^nu \;du = \dfrac{1}{n}\cos^{n-1} u \sin u +\dfrac{n-1}{n}\int \cos^{n-2} u\; du$$

For $|A|\gg 0$, you could use the Taylor series expansion of $e^z$ about $z=A$ for a series that converges more rapidly for that value of $A$.

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  • $\begingroup$ Greatttttt @Andy. Thank you so much. As you mentioned, there is convergence issue here which prevents us from getting a closed form solution (for large n). Can we interpret this as a closed form solution? Do you have other ideas to get a closed form solution? It is important because I want to compute that for definite bands. $\endgroup$
    – Mahdi
    Oct 24 '20 at 8:28
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    $\begingroup$ An infinite series is not a closed form solution. The best one can likely hope for here is an expression in terms of a Hypergeometric series/function, which isn't really closed either. What I have given you is a series whose terms are computable, with some recursive computation of the factors in the terms, that avoids numerical integration. You can either compute the number of terms in this series you need to the needed precision to do the task, or you can do the numerical integration of the original integral. $\endgroup$
    – Andy Walls
    Oct 24 '20 at 13:50

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