What is the double integral of an exponential difference of sines?

Question

I am keen to know if there's a way of analytically evaluating the following integral: $$ \int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha|\sin(\theta)-\sin(\varphi)|}d\theta d\varphi $$ where $\alpha$ is a real constant. It arises when considering how arrays of charged wires interact with each other.

There are Bessel function representations for similar integrals when one does not have to take the modulus. Here, I simply don't know how to deal with the modulus.

I've tried rewriting it using $$ \sin(\theta)-\sin(\varphi)=2\sin\left(\frac{\theta-\varphi}{2}\right)\cos\left(\frac{\theta+\varphi}{2}\right) $$ and then transforming the coordinates to this rotated version. I think it's then fine to rotate the region of integration accordingly, owing to the double periodicity of the integrand, but then I get stuck.

Thanks in advance for any help.

Update: I've used Teresa Lisbon's suggestion of expanding out the exponential to find that $$ \int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha|\sin(\theta)-\sin(\varphi)|}d\theta d\varphi=4\pi^{2}J_{0}(\alpha)^{2}+32i\alpha_{p}F_{q}(1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};-\alpha^{2}) $$ Does anyone know if the hypergeometric function part can be simplified (ideally in terms of Bessel functions)? Thanks!

Answer 1

I'm not too sure if this is right, would be great if someone could confirm... $$\int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha|\sin(\theta)-\sin(\varphi)|}d\theta d\varphi$$ $$\int_{0}^{2\pi}e^{-i\alpha \sin(\varphi)}d\varphi\int_{0}^{2\pi}e^{i\alpha \sin(\theta)}d\theta $$ $$\int_{0}^{2\pi}e^{-i\alpha \sin(\varphi)}d\varphi \times \int_{0}^{2\pi}e^{i\alpha \sin(\theta)}d\theta $$ $$\int_{0}^{2\pi}\Bigl(\cos({-\alpha \sin(\varphi)})+i\sin({-\alpha \sin(\varphi)})\Bigr)d\varphi \times \int_{0}^{2\pi}\Bigl(\cos({\alpha \sin(\theta)})+i\sin({\alpha \sin(\theta)})\Bigr)d\theta $$

$$=\int_{0}^{2\pi}\Bigl(\cos({\alpha \sin\varphi})-i\sin({\alpha \sin\varphi})\Bigr)d\varphi \times \int_{0}^{2\pi}\Bigl(\cos({\alpha \sin\theta})+i\sin({\alpha \sin\theta})\Bigr)d\theta $$ P.S. This is just one case, considering the stuff under the modulo to be positive, there will obviously be another integral considering the stuff under modulo to be negative Hereafter, I believe you can make use of the Jacobi series expansions...

EDIT: Another interesting idea, might be wrong, kindy assist... $$\int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha|\sin(\theta)-\sin(\varphi)|}d\theta d\varphi$$

So, we get two cases, they are:

$$CASE.1.:\int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha\sin(\theta)-i\alpha \sin(\varphi)}d\theta d\varphi$$ AND $$CASE.2.:\int_{0}^{2\pi}\int_{0}^{2\pi}e^{-i\alpha\sin(\theta)+i\alpha \sin(\varphi)}d\theta d\varphi$$

Let's take case 1 for now: $$\int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha\sin(\theta)-i\alpha \sin(\varphi)}d\theta d\varphi=\int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha\sin(\theta)}\cdot e^{-i\alpha \sin(\varphi)}d\theta d\varphi$$

Consider only $$e^{i\alpha \sin \theta}$$ $$\sin \theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}$$ $$i\alpha \sin \theta=i\alpha \frac{e^{i\theta}-e^{-i\theta}}{2i}=\frac{\alpha}{2}\left(e^{i\theta}-e^{-i\theta}\right)$$ $$e^{i\alpha \sin \theta}=e^{\frac{\alpha}{2}\left(e^{i\theta}-e^{-i\theta}\right)}$$ $$=e^{\frac{\alpha e^{i\theta}}{2}}\cdot e^{\frac{-\alpha e^{-i\theta}}{2}}$$

Therefore, $$\int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha\sin(\theta)}\cdot e^{-i\alpha \sin(\varphi)}d\theta d\varphi=\int_{0}^{2\pi}\int_{0}^{2\pi}e^{\frac{\alpha e^{i\theta}}{2}{\frac{-\alpha e^{-i\theta}}{2}}}\cdot e^{\frac{\alpha e^{i\phi}}{2}{\frac{-\alpha e^{-i\phi}}{2}}}d\theta d\varphi$$

After this, if you take the expression in the exponent of $e$ to be $t$, you will get an integral of the form $$\int_{0}^{2\pi}\int_{0}^{0} F(t)F(\phi)dtd\phi$$

And then, according to the second point mentioned here, the entire thing turns out to be zero.

I have a gut feeling this is wrong, but I've been thinking of this question for quite some time and would appreciate opinions on the same. I was also advised to use Cauchy's Theorem, but I have no clue about it, so it would be great if someone could explain that as well.

Answer 2

You can express it this way \begin{align} &\int_0^{2\pi}\int_0^{2\pi}e^{i\alpha|\sin \theta -\sin \varphi|}d\theta d\varphi = \int_0^{2\pi}e^{i\alpha\sin \varphi} \left( \int_0^{2\pi} e^{i\alpha(\sin \theta -2\min(\sin\theta,\sin \varphi)}d\theta \right) d\varphi =\\ &=\int_0^{\frac{\pi}{2}}e^{i\alpha\sin \varphi} \left( \int_0^{\varphi} e^{-i\alpha\sin \theta }d\theta + e^{-2i\alpha\sin\varphi}\int_\varphi^{\pi-\varphi} e^{i\alpha\sin \theta}d\theta+\int_{\pi-\varphi}^{2\pi} e^{-i\alpha\sin \theta }d\theta \right) d\varphi +\dots \end{align}

the rest of the expression is analogous. I don't know anything about Bessel functions or anything of the sort, so I can't tell whether this will help you much.

Answer 3

Let's abbreviate by $I(\alpha)$ the $\alpha$-dependent integral in question.

Following the OP, the sine difference can be replaced according to $\sin\theta-\sin\varphi= 2\sin\left(\frac{\theta-\varphi}{2}\right) \cos\left(\frac{\theta+\varphi}{2}\right)$ giving $$ I(\alpha)=\int_{0}^{2\pi}\int_{0}^{2\pi}\exp\left(2i\alpha\left|\sin\left(\frac{\theta-\varphi}{2}\right) \cos\left(\frac{\theta+\varphi}{2}\right)\right|\right)\; d\theta\, d\varphi. $$ Setting $\theta=\varphi+2u$ yields $$ I(\alpha)=\int_{0}^{2\pi}\int_{-\frac{\varphi}{2}}^{\pi-\frac{\varphi}{2}}\exp\left(2i\alpha\left|\sin\left(u\right) \cos\left(\varphi+u\right)\right|\right)\; 2\,du\, d\varphi. $$ Using the periodicity of the integrand in $u$, the inner integration interval can be shifted to be independent of $\varphi$, namely $[0,\pi]$. After exchanging the integrations, the substitution $\varphi=v-u$ gives: $$ I(\alpha)=2\int_{0}^{\pi}\int_{u}^{2\pi+u}\exp\left(2i\alpha\left|\sin\left(u\right) \cos\left(v\right)\right|\right)\; dv\, du. $$ Using the periodicity of the integrand in $v$, the inner integration interval can be shifted to be independent of $u$: $$ I(\alpha)=2\int_{0}^{\pi}\int_{0}^{2\pi}\exp\left(2i\alpha\left|\sin\left(u\right) \cos\left(v\right)\right|\right)\; dv\, du. $$ Now expanding the exponential as suggested by Teresa Lisbon decouples the integrals $$ I(\alpha)=2\sum_{n=0}^\infty\frac{1}{n!}\int_{0}^{\pi}\int_{0}^{2\pi}\left(2i\alpha\left|\sin\left(u\right) \cos\left(v\right)\right|\right)^n\; dv\, du. $$ The integrals are known to be \begin{eqnarray*} \int_{0}^{\pi}\left|\sin\left(u\right)\right|^ndu&=&2\int_{0}^{\frac{\pi}{2}}\left(\sin\left(u\right)\right)^ndu =\frac{\Gamma\left(\frac{n+1}{2}\right)\sqrt{\pi}}{\Gamma\left(\frac{n+2}{2}\right)},\\ \int_{0}^{2\pi}\left|\cos\left(v\right)\right|^ndv&=&4\int_{0}^{\frac{\pi}{2}}\left(\sin\left(v\right)\right)^ndv =2\;\frac{\Gamma\left(\frac{n+1}{2}\right)\sqrt{\pi}}{\Gamma\left(\frac{n+2}{2}\right)}, \end{eqnarray*} giving $$ I(\alpha)=4\pi\sum_{n=0}^\infty\frac{\left(2i\alpha\right)^{n}}{n!}\left( \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}\right)^2. $$ Separating the summation by even and odd indices, $\sum_{n=0}^\infty f(n)=\sum_{n=0}^\infty f(2n)+\sum_{n=0}^\infty f(2n+1)$, the $\Gamma$-functions and factorials can be expressed by (rising) Pochhammer symbols, $$ \Gamma\left(n+a\right) = \left(a\right)^{(n)}\Gamma(a), \quad\left(a\in\left\{\frac{1}{2},1, \frac{3}{2}\right\}\right), $$ $$ (2n)!=2^{2n}\,n!\,\left(\frac{1}{2}\right)^{(n)},\qquad (2n+1)!=2^{2n}\,n!\,\left(\frac{3}{2}\right)^{(n)}, $$ and the result can be expressed by generalized hypergeometric functions: $$ I(\alpha)=4\pi^2{}_1F_2\left(\frac{1}{2};1,1;-\alpha^2\right) +32\,i\,\alpha\,{}_2F_3\left(1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};-\alpha^2\right). $$ The real part can be simplified to $4\pi^2\left(J_0(\alpha)\right)^2$, either by observing that for the even powers in the Taylor series of the exponential taking the modulus is irrelevant, and can be evaluated as real part of $\int_{0}^{2\pi}\int_{0}^{2\pi}\exp\left(i\alpha\left(\sin\left(\theta\right)- \sin\left(\varphi\right)\right)\right)\, d\theta\, d\varphi=\left(\int_{0}^{2\pi}\exp\left(i\alpha\sin\left(\theta\right)\right)\, d\theta\right)^2=\left(2\pi J_0(\alpha)\right)^2,$ or in the defining series of the hypergeometric function, one can use the identity $\sum_{n=0}^k{k\choose m}^2={2k\choose k}$ to get after a short calculation ${}_1F_2\left(\frac{1}{2};1,1;-\alpha^2\right)=\left(J_0(\alpha)\right)^2$. The additional question which was also raised in the OP, namely whether a similar simplification is also possible for the imaginary part, is open.