# Proving an equality of the series

I need to prove the next equality,

$$\int \limits_{0}^{\frac{\pi}{2}}\int \limits_{0}^{\frac{\pi}{2}} \cos(2a \sin(x) \sin(y))dxdy = \left(\int \limits_{0}^{\frac{\pi}{2}}\cos(a \sin(z))dz \right)^2$$ for each $a \in \mathbb{R}$.

I am proving it as follows.

1. I used a series expansion for the left integral:

$$\cos(2a \sin(x)\sin(y)) = 1 + \sum_{i = 1}^{\infty}\left( (-1)^{i}\frac{(2a)^{2i}}{(2i)!}\sin^{2i}(x)\sin^{2i}(y)\right) \qquad (1).$$

After integrating $(1)$ will be as $$\frac{\pi^{2}}{4}\left(1 + \sum_{i = 1}^{\infty}\left( \frac{(-1)^{i}a^{2i}(2i)!}{2^{2i}(i!)^{4}}\right)\right).$$

1. I used a series expansion for the right integral:

$$\int \limits_{0}^{\frac{\pi}{2}}\cos(a \sin(z))dz = \frac{\pi}{2}\left(1 + \sum_{i = 1}^{\infty}\left( \frac{(-1)^{i}a^{2i}}{2^{2i}(i!)^{2}}\right)\right).$$

How to prove an equality of the next expression,

$$\left(\frac{\pi}{2}\left( 1 + \sum_{i = 1}^{\infty}\left( \frac{(-1)^{i}a^{2i}}{2^{2i}(i!)^{2}}\right)\right) \right)^{2} \qquad (2),$$

$$\frac{\pi^{2}}{4}\left(1 + \sum_{i = 1}^{\infty}\left( \frac{(-1)^{i}a^{2i}(2i)!}{2^{2i}(i!)^{4}}\right)\right),$$

by gift to the square of $(2)$?

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Both series are hypergeometric, and are easy to evaluate by writing them in the canonical forms: $$\frac{\pi^2}{4} \sum_{i=0}^\infty \frac{(-1)^i \cdot a^{2i} \cdot (2i)!}{4^{i} (i!)^4} = \frac{\pi^2}{4} \sum_{i=0}^\infty \frac{(1/2)_i}{(1)_i \cdot (1)_i} \frac{(-a^2)^i}{i!} = \frac{\pi^2}{4} \cdot {}_1 F_{2}\left( \frac{1}{2} ; 1,1 ; -a^2\right) = \left(\frac{\pi}{2} \cdot J_0(a)\right)^2$$ Similarly: $$\frac{\pi}{2}\sum_{i=0}^\infty \frac{(-1)^i a^{2i}}{2^{2i} (i!)^2} = \frac{\pi}{2}\sum_{i=0}^\infty \frac{1}{(1)_i} \frac{(-a^2/4)^i}{i!} = \frac{\pi}{2} \cdot {}_0F_1\left(; 1; -\frac{a^2}{4}\right) = \frac{\pi}{2} \cdot J_0(a)$$ Hence the identity is established.
The following hypergeometric identity were used: for ${}_1F_2$ this and for ${}_0F_1$ this.