TITLE: A peculiar integral identity
QUESTION [8 upvotes]: Here I was, innocently trying to solve this daunting-looking integral
$$\int_0^\pi e^{v \cos \theta \cos t} \cosh(v \sin \theta \sin t) dt $$
when the inner beauty behind this beast slowly started to disclose itself.

Of course, the first thing I did, was checking if WolframAlpha can help, futilely.
Next on my lookup table was Gradshteyn and Ryzhik. Yet again, the god of integrals had no mercy upon me and I was left to my own device.

To get a first impression I plotted the function, that needs to be integrated, that is for $f_{\theta}(t) = e^{\cos \theta \cos t} \cosh(\sin \theta \sin t)$, we get the following plots:

Okay, nothing too special about that. So I proceeded by trying to numerically evaluate the integral itself. And then, something strange happened...

Turns out the integral is invariant under $\theta$!
Even better, we have

$$\int_0^\pi e^{v \cos \theta \cos t} \cosh(v \sin \theta \sin t) dt = \int_0^\pi e^{v \cos t} dt = \pi I_0(v)\quad \forall \theta \in [-\pi,\pi]\; ,$$

where for the first equality I just set $\theta = 0$ and the second equality is a known identity of the modified Bessel function of the first kind. Now, I only stumbled upon this identity numerically, and I was wondering if someone can share some analytical wisdom regarding this. Put into a question:

Does someone know, why this identity holds?


Bonus
I now face the same integral but with an additional linear term, that is
$$\int_0^{\pi} tf_{\theta}(t)dt \; ,$$
with $f_{\theta}$ as defined above. I am hoping that the techniques that illuminate the identity above will also shed some light at this new integral, which by the way is not constant in $\theta$ anymore.

REPLY [2 votes]: Note that
$$\begin{align*}
I_\theta &= \frac14\int_0^{2\pi}e^{v\cos\theta\cos t}\left(e^{v\sin\theta\sin t}+e^{-v\sin\theta\sin t}\right) \mathrm dt\\&=\frac14\int_0^{2\pi}e^{v\cos(t-\theta)} \mathrm dt+\frac14\int_0^{2\pi}e^{v\cos(t+\theta)} \mathrm dt.
\end{align*}$$ Because $t\mapsto  e^{v\cos t}$ is $2\pi$-periodic, we have
$$
\int_0^{2\pi}e^{v\cos(t-\theta)} \mathrm dt=\int_0^{2\pi}e^{v\cos(t+\theta)} \mathrm dt=\int_0^{2\pi}e^{v\cos t} \mathrm dt
$$ so it follows that
$$
I_\theta = \frac 12\int_0^{2\pi}e^{v\cos t} \mathrm dt =\int_0^\pi e^{v\cos t} \mathrm dt.
$$