Evaluate the modified Bessel function in large limit $\int_0^{2\pi}\exp(-C)\cdot\exp\left(a\sin\phi+b\cos\phi\right)\,d\phi$

Question

I am trying to evaluate the following integral: \begin{align} \int_0^{2\pi}\exp(-C)\cdot\exp\left(a\sin\phi+b\cos\phi\right)\,d\phi\,, \end{align} where $C,a,b$ are large constants. I know that this eventually reduces to the modified Bessel function of the first kind. However, since $C, a, b$ are very large (the term "$-C+a\sin\phi+b\cos\phi$" is small, though), it becomes impossible to compute it using the Bessel function numerically.

I would rather not do it the hard way, is there another way to compute this efficiently (in terms of some functions) ?

Thanks so much! Any help would be greatly appreciated!

Answer 1

So basically we can compute a scaled Bessel function. MATLAB, Pythons scipy package, etc., could do this.

Answer 2

The following is just a trial and it is not guaranteed correct. (Please don't trust my calculation too much as I am very tired and may commit careless mistake.)

Note that we can write $a\sin\phi+b\cos\phi=r\sin(\phi+\phi_{0})$, where $r=\sqrt{a^{2}+b^{2}}$, and $\phi_{0}\in[0,2\pi)$.

The actual value of $\phi_{0}$ is not important. Note that the integrand is periodic with period $2\pi$. Therefore \begin{eqnarray*} & & \int_{0}^{2\pi}\exp(-C)\exp(a\sin\phi+b\cos\phi)d\phi\\ & = & \int_{\phi_{0}}^{2\pi+\phi_{0}}\exp(-C)\exp(r\sin\phi)d\phi\\ & = & \int_{0}^{2\pi}\exp(-C)\exp(r\sin\phi)d\phi. \end{eqnarray*} The last equality holds because $\int_{\phi_{0}}^{2\pi+\phi_{0}}=\int_{0}^{2\pi}+\int_{2\pi}^{2\pi+\phi_{0}}-\int_{0}^{\phi_{0}}=\int_{0}^{2\pi}$, with the second and the third integrals cancel each other. Note that $\exp(x)=1+x+\frac{x^{2}}{2!}+\cdots$ and the series converges uniformly on $[-r,r]$. Therefore, we can integrate termwisely and get \begin{eqnarray*} & & \int_{0}^{2\pi}\exp(r\sin\phi)d\phi\\ & = & \sum_{k=0}^{\infty}\frac{r^{k}}{k!}\int_{0}^{2\pi}\sin^{k}\phi d\phi. \end{eqnarray*} If $k$ is odd, $\int_{0}^{2\pi}\sin^{k}\phi d\phi=0$ because $\int_{\pi}^{2\pi}\sin^{k}\phi d\phi=-\int_{0}^{\pi}\sin^{k}\phi d\phi$. If $k$ is even, $\int_{0}^{2\pi}\sin^{k}\phi d\phi$ is well-known (by reduction method and integration-by-part). For, let $I_{k}=\int_{0}^{2\pi}\sin^{k}\phi d\phi$, then for $n\geq1$, \begin{eqnarray*} I_{2n} & = & \int_{0}^{2\pi}-\sin^{2n-1}\phi d\cos\phi\\ & = & -\sin^{2n-1}\phi\cos\phi|_{0}^{2\pi}+\int_{0}^{2\pi}(2n-1)\sin^{2n-2}\phi\cos^{2}\phi d\phi\\ & = & (2n-1)\int_{0}^{2\pi}\left(\sin^{2n-2}\phi-\sin^{2n}\phi\right)d\phi\\ & = & (2n-1)(I_{2n-2}-I_{2n}). \end{eqnarray*} Hence $I_{2n}=\frac{2n-1}{2n}I_{2n-2}$.

Answer 3

Notice that $$ \int e^{ a \sin \phi + b \cos \phi} \frac{d\phi}{2\pi} = I_0(c), \quad c=\sqrt{a^2+b^2} $$ Since $I_0(c)$ grows like (see Wikipedia) $$ I_0(c)\sim \frac{e^c}{\sqrt{2\pi c}} +\cdots $$ You can compute your integral as $$ e^{-C}I_0(c)\sim \frac{e^{c-C}}{\sqrt{2\pi c}} +\cdots $$