Is there a way to exactly solve this integral?

Question

Is there a way to determine $\theta_c$ in the following integral equation:

$$\int_{-\theta_c}^{\theta_c} d\theta \, \exp (a \cos\theta)=1,$$

where $a$ belongs to positive reals?

Answer 1

For $a>0$, the integrand has a maximum at $\theta=0$. Writing the series for $\cos$ around zero, we find$^\dagger$

$$ e^a \int_{-\theta_0}^{\theta_0} d\theta \ \exp( -a \theta^2 /2)=e^a \sqrt{\frac{2\pi}{a}} \operatorname{erf}\left(\theta_0 \sqrt{\frac{a}{2}} \right)=1$$

Rearranging

$$\operatorname{erf}\left(\theta_0 \sqrt{\frac{a}{2}} \right)=e^{-a} \sqrt{\frac{a}{2\pi}} $$

Apply inverse $\operatorname{erf}$

$$ \theta_0= \sqrt{\frac{2}{a}} \operatorname{erf}^{-1} \left(e^{-a} \sqrt{\frac{a}{2\pi}}\right) $$

The argument of $\operatorname{erf}^{-1}$ above goes to zero exponentially fast as $a\rightarrow\infty$. The series for $\operatorname{erf}^{-1}(z)$ around $z=0$ is

$$ \operatorname{erf}^{-1}(z)=\frac{\sqrt{\pi}}{2}z+O(z^2) ,\ z\rightarrow0$$

We are left with

$$ \theta_0 \sim \frac{e^{-a}}{2} \ \ , \ \ a\rightarrow\infty$$

Here is a plot of the original integral: $\int_{-\theta_0}^{\theta_0}d\theta \exp(a \cos\theta)$, numerically evaluated with $\theta_0=e^{-a}/2$. We see that 'large $a$' means modest values of $a>1$.

$\dagger$ The justification of what we are doing is the same as the justification for Laplace's method.