# two dimensional integration of a trigonometric function

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?

• Your question might be more appropriate for the math stackexchange.
– MBaz
Oct 22, 2020 at 17:07
• I have got no answer there. Oct 22, 2020 at 17:19
• You should look up integral identities for Bessel Functions. Oct 22, 2020 at 17:19
• It would be turned out to a Bessel function if the integral be in a period (which is not in this case). Oct 22, 2020 at 17:55
• Essentially the same question has been cross-posted at Math.SE math.stackexchange.com/q/3871844/441161 Oct 24, 2020 at 14:01

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 $$A$$.

• 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. Oct 24, 2020 at 8:28
• 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. Oct 24, 2020 at 13:50