10
$\begingroup$

I found five other related integrals whose proofs I am studying now A, B, C, D, and E

$$\int^{2\pi}_0e^{\cos \theta}\cos(a\theta -\sin \theta)\,d \theta = \frac{2\pi}{a!}$$ $$\int_0^{2\pi} \exp(\cos(\theta)) \cos(\theta + \sin(\theta)) = 0$$ $$ \int_0^{2\pi} \exp(\alpha \cos(\theta))\cos(\sin(\theta)) = 2\pi I_0(\sqrt{1 - \alpha^2})$$ $$\int_0^{2\pi} \exp(x\cos(\theta) + y\sin(\theta))) = 2\pi I_0(\sqrt{x^2 + y^2})$$ $$ \int_0^\dfrac{\pi}{2}\beta^\alpha\exp\left(-\beta\cos(\theta)\right)d\theta = \dfrac{1}{2}\beta^\alpha\pi\left(J_0(\beta)-L_0(\beta)\right)$$

I was also able to find a very general statement in Gradshteyn as entry number 3.338. $$\int_{-\pi}^{\pi} \frac{\exp{\frac{a + b\sin x + c \cos x}{1 + p \sin x + q \cos x}}}{1 + p \sin x + q \cos x} dx = \frac{2\pi e^{-\alpha}I_0(\beta)}{\sqrt{1 - p^2 - q^2}}$$ $$\textrm{where } \alpha = \frac{bp + cq -a}{1 - p^2 - q^2},\; \beta = \sqrt{\alpha^2 - \frac{a^2 - b^2 - c^2}{1 - p^2 - q^2}}$$

But the simplest approach of using integration by parts to reduce my problem to one of these does not work.

Background Here's some background into why I am interested in this integral, let $v = [x, y] \in \mathbb{R}^2$ and $r = [\cos(\theta), \sin(\theta)] \in \mathbb{R}^2$, Consider the value of $$\underset{\theta \tilde{} \textrm{Hill}}{E}[\exp(v^Tr)]$$ This is the expected value of exponential of the projection of a random vector chosen using the Hill distribution, where the "Hill" is an unnormalized distribution that linearly increases from $0$ at $-\pi$ to $1$ at $0$ and then decreases linearly from $0 \textrm{ to } \pi$. Discarding normalizing factor of Hill, This expectation will become:

$$ \int_{-\pi}^{0} (\theta + \pi)\exp(x\cos\theta + y\sin\theta) d\theta + \int_{0}^{\pi} (\pi - \theta) \exp(x\cos\theta + y\sin\theta) d\theta $$

Now, there are simplifying unnormalized distributions I could assume in my model, instead of Hill, such as Uniform from 0 to $2\pi$, or $\exp(\cos(\theta))$ both of these distribution allow analytical calculation of the above expectation just based on the identities written below, but I want to know which distributions I can compute this expectation for (Can I do this for Hill?) I will guess that I can only do it for distributions that have some finite decomposition in terms of spherical harmonics. Unfortunately, my knowledge is lacking in complex analysis and spherical harmonics so I can't quickly assess my options.

$\endgroup$
  • $\begingroup$ $\theta$ showing up outside the exponentials and trigs don't give me much hope that there's a tidy closed form. $\endgroup$ – J. M. is a poor mathematician Jan 21 '17 at 5:23
1
$\begingroup$

Not hard to see that \begin{align*} \int_0^{2 \pi } \theta \cdot e^{x \cos \theta +y \sin \theta } \, \mathrm{d}\theta &=\int_{0}^{2\pi }\theta \sum_{k=0}^{\infty }\frac{\left ( x \cos \theta +y \sin \theta \right )^{k}}{k!}\, \mathrm{d}\theta \\ &=\int_{0}^{2\pi }\theta \sum_{k=0}^{\infty }\frac{1}{k!}\sum_{j=0}^{k}\binom{k}{j}y^{j}\sin^{j}\theta x^{k-j}\cos^{k-j}\theta \, \mathrm{d}\theta \\ &=\sum_{k=0}^{\infty }\frac{1}{k!}\sum_{j=0}^{k}\binom{k}{j}y^{j}x^{k-j}\int_{0}^{2\pi }\theta \sin^{j}\theta \cos^{k-j}\theta \, \mathrm{d}\theta \end{align*} For the last integral, with the help of Mathematica we get the following complex result \begin{align*} \int_{0}^{2\pi }\theta \sin^{m}\theta \cos^{n}\theta \, \mathrm{d}\theta =&\frac{(-1)^{m+n} \pi ^2 \csc \left(\frac{n \pi }{2}\right) \csc \left(\frac{\pi m}{2}+\frac{n \pi }{2}\right) \Gamma \left(\frac{n}{2}+\frac{1}{2}\right)}{4 \Gamma \left(\frac{1}{2}-\frac{m}{2}\right) \Gamma \left(\frac{m}{2}+\frac{n}{2}+1\right)}+\frac{\pi ^2 \csc \left(\frac{n \pi }{2}\right) \csc \left(\frac{\pi m}{2}+\frac{n \pi }{2}\right) \Gamma \left(\frac{n}{2}+\frac{1}{2}\right)}{4 \Gamma \left(\frac{1}{2}-\frac{m}{2}\right) \Gamma \left(\frac{m}{2}+\frac{n}{2}+1\right)}\\ &+(-1)^{m+n} 2^{\frac{m}{2}-\frac{1}{2}} \cos \left(\frac{m \pi }{2}\right) \Gamma \left(\frac{m}{2}+\frac{1}{2}\right) \Gamma \left(-\frac{m}{2}-n-\frac{1}{2}\right) \Gamma (n+1) \, _2F_1\left(\frac{1-m}{2},\frac{m+1}{2};\frac{m}{2}+n+\frac{3}{2};\frac{1}{2}\right)\\ &+\frac{(-1)^{m+2 n} 2^{\frac{m}{2}-\frac{1}{2}} \pi \Gamma \left(\frac{m}{2}+\frac{1}{2}\right) \Gamma (n+1) \, _2F_1\left(\frac{1-m}{2},\frac{m+1}{2};\frac{m}{2}+n+\frac{3}{2};\frac{1}{2}\right)}{\Gamma \left(\frac{m}{2}+n+\frac{3}{2}\right)}\\ &+\frac{(-1)^{m+n} \Gamma \left(\frac{n}{2}\right) \Gamma \left(\frac{m}{2}+1\right) \, _3F_2\left(\frac{1}{2},1,\frac{m}{2}+1;\frac{3}{2},1-\frac{n}{2};1\right)}{2 \Gamma \left(\frac{m}{2}+\frac{n}{2}+1\right)}\\ &+\frac{\Gamma \left(\frac{n}{2}\right) \Gamma \left(\frac{m}{2}+1\right) \, _3F_2\left(\frac{1}{2},1,\frac{m}{2}+1;\frac{3}{2},1-\frac{n}{2};1\right)}{2 \Gamma \left(\frac{m}{2}+\frac{n}{2}+1\right)}\\ &+\frac{(-1)^n \Gamma \left(\frac{m}{2}\right) \Gamma \left(\frac{n}{2}+1\right) \, _3F_2\left(\frac{1}{2},1,\frac{n}{2}+1;\frac{3}{2},1-\frac{m}{2};1\right)}{2 \Gamma \left(\frac{m}{2}+\frac{n}{2}+1\right)}\\ &+\frac{(-1)^{m+2 n} \Gamma \left(\frac{m}{2}\right) \Gamma \left(\frac{n}{2}+1\right) \, _3F_2\left(\frac{1}{2},1,\frac{n}{2}+1;\frac{3}{2},1-\frac{m}{2};1\right)}{2 \Gamma \left(\frac{m}{2}+\frac{n}{2}+1\right)}\\ &+\frac{(-1)^n \pi \Gamma \left(\frac{m}{2}+\frac{1}{2}\right) \left(\pi \csc \left(\frac{m \pi }{2}\right) \csc \left(\frac{\pi m}{2}+\frac{n \pi }{2}\right)+\pi \sec \left(\frac{n \pi }{2}\right)\right)}{4 \Gamma \left(\frac{1}{2}-\frac{n}{2}\right) \Gamma \left(\frac{m}{2}+\frac{n}{2}+1\right)}\\ &+\frac{(-1)^{m+2 n} \pi \Gamma \left(\frac{m}{2}+\frac{1}{2}\right) \left(\pi \csc \left(\frac{m \pi }{2}\right) \csc \left(\frac{\pi m}{2}+\frac{n \pi }{2}\right)+\pi \sec \left(\frac{n \pi }{2}\right)\right)}{4 \Gamma \left(\frac{1}{2}-\frac{n}{2}\right) \Gamma \left(\frac{m}{2}+\frac{n}{2}+1\right)}\\ &+\frac{(-2)^{m+n} \pi ^{5/2} \Gamma \left(\frac{m}{2}+\frac{1}{2}\right) \sec \left(\frac{\pi m}{2}+n \pi \right)}{\Gamma \left(\frac{1}{2}-\frac{n}{2}\right) \Gamma \left(-\frac{m}{2}-\frac{n}{2}+\frac{1}{2}\right) \Gamma (m+n+1)} \end{align*} Looking at this horrible result, If I'm not doing the wrong way, I don't there will be a closed form for the integral, as least it won't be too short.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.