# analytic result of an integral

I am working on a physics project and I encounter an integral that I need to get analytic results about. Otherwise I will have to numerically compute the second integral, which significantly increases the amount of computer work.

Here is the integral: $\int_0^\pi e^{ia\cos\phi}(\sin\phi)^2 d\phi$.

I am not sure if it can be analytically computed. It does look like definition of the spherical Bessel function of first kind if it's $\sin\phi\cos\phi$ rather than $(\sin\phi)^2$. Any help would be appreciated.

• Igor Rivin has pointed the right direction. Thank you for your attention. – HanaKaze Jan 27 '16 at 22:05

The above result is complete. I think,it can be done in the following way. I consider that $$a \in \mathbb{R}$$ We know, from the representation of Bessel function that $$\exp(ia\cos \phi)=\sum_{m=-\infty}^{m=\infty}i^m J_m(a) \exp(im\phi)$$ such that $$\cos(a \cos \phi)=J_0(a)+2\sum_{m=1}^{\infty}(-1)^m J_{2m}(a)\cos(2m\phi)$$ and $$\sin(a \cos \phi)=2\sum_{m=0}^{\infty}(-1)^m J_{2m+1}(a)\cos((2m+1)\phi)$$, where $$J_m$$ represents Bessel function of first kind with order $$m$$.
As the integrand is the even function, for simplicity,we can write it as $$\frac{1}{2}\int_0^{2\pi}\bigg(J_0(a)+2\sum_{m=1}^{\infty}(-1)^m J_{2m}(a)\cos 2m\phi+ 2 i\sum_{m=0}^{\infty}(-1)^m J_{2m+1}(a)\cos(2m+1)\phi\bigg)\frac{(1-\cos 2\phi)}{2}\,d\phi$$ where, the imaginary part will be $$+$$ve or $$-$$ve depending on $$a$$ is $$+$$ve or $$-$$ve. Using the fact that $$\int_0^{2\pi}\cos nx \,dx=\int_0^{2\pi}\sin nx \, dx=0$$ and $$\int_0^{2\pi}\cos nx \sin mx \, dx=\pi \delta_{nm}$$, it becomes, $$\frac{1}{4}\int_0^{2\pi}\bigg(J_0(a)+2 J_2(a)\cos^2 (2\phi)\bigg)\,d\phi$$ Now, using $$\cos^2 2\phi= \frac{1+\cos 4\phi}{2}$$, the final result will be, $$\frac{1}{2}(\pi J_0(a)+\pi J_2(a))=\frac{\pi J_1(a)}{a}$$ using the recurrence relation $$J_{n-1}(x)+J_{n+1}(x)=\frac{2nJ_n(x)}{x}$$Where, $$a$$ can be taken forboth the $$+ve$$ and $$-ve$$ values, and accordingly the answer will be finally $$\frac{\pi J_1(|a|)}{|a|}$$ as shown above.
Mathematica says: $$\frac{\pi J_1(\left| a\right| )}{\left| a\right| }.$$