two dimensional integration of a trigonometric function

Question

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?

Answer 1

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 $A$.