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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!

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  • $\begingroup$ The Bessel integrals only work for integer $\alpha$. An analytic form is very unlikely. $\endgroup$
    – user65203
    Apr 19, 2021 at 12:28
  • $\begingroup$ I don't think that's true. The integrals in dlmf.nist.gov/10.9 are valid even when z is not an integer. $\endgroup$
    – Chris
    Apr 19, 2021 at 15:58
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    $\begingroup$ Just a guess and nothing more : but thinking about the exponential series and $\int_0^{2 \pi} |\sin \theta - \sin (\varphi)|^{2n} d \theta d \phi$ , in terms of $n$ seems to have a nice formula involving $\pi^2$. I don't quite know where this comes from. $\endgroup$ May 13, 2021 at 4:34
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    $\begingroup$ When starting writing down my answer, I saw the most recent update. My answer is almost the same, with an additional factor $\alpha$ in the imaginary part and $p=2, q=3$, of course, thus giving for it $$32\,i\,\alpha\;{}_2F_3\left(1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};-\alpha^2\right).$$ The real part is the same. But I did not try to simplify the remaining generalized hypergeometric function. Should I still write down my derivation as an answer? $\endgroup$
    – Uwe
    May 13, 2021 at 13:02

3 Answers 3

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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.

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    $\begingroup$ Can't believe my eyes $\endgroup$ May 13, 2021 at 19:42
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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.

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    $\begingroup$ My concern is that the regions of integration will have to be split into triangles, so the limits of the 'inner' integral will have to contain the other variable. $\endgroup$
    – Chris
    Apr 19, 2021 at 12:00
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    $\begingroup$ Due to the modulus sign your answer is wrong. It would have been correct with square brackets instead of that absolute value in the exponent. I've misread the question in the same way as you did (?) and gave you an upvote. I've deleted my own answer, which was built on top of yours. $\endgroup$ May 10, 2021 at 18:38
  • $\begingroup$ I've mentioned that i've considered only one case in my answer, and that there'll be a counterpart...will it still be wrong? $\endgroup$ May 11, 2021 at 2:34
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    $\begingroup$ The stuff inside the modulus sign will always have regions where it is positive and where it is negative depending on the relative values of theta and phi. It will never be the case that the the stuff inside the modulus is always positive or always negative. $\endgroup$
    – Chris
    May 11, 2021 at 6:36
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    $\begingroup$ It is already known that the integral can be done when there is no modulus sign (one obtains Bessel functions). I specifically need to know how to deal with the modulus sign. I don't think Cauchy's theorem will help here because the modulus function is nowhere analytic. Also your substitution isn't valid because the function describing your substitution isn't monotonic. $\endgroup$
    – Chris
    May 11, 2021 at 19:51
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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.

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    $\begingroup$ I wouldn't then know how to evaluate the 'indefinite' inner integrals. $\endgroup$
    – Chris
    May 12, 2021 at 8:29
  • $\begingroup$ @Chris I thought that at least you got rid of the inconvenient absolute value, but of course, if there is no way to compute those, it's of little use. $\endgroup$
    – miguelsxvi
    May 12, 2021 at 18:11

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