Sfdwhf

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.

For the main problem, a bit of algebra:
begin{align*}e^{vcosthetacos t}cosh(vsinthetasin t) &= e^{vcosthetacos t}left(e^{vsinthetasin t}+e^{-vsinthetasin t}right)\
&= frac12left(e^{vcosthetacos t+vsinthetasin t}+e^{vcosthetacos t-vsinthetasin t}right)\
&= frac12left(e^{vcos(theta-t)}+e^{vcos(theta+t)}right)end{align*}
Now we integrate that:
begin{align*}int_0^{pi}e^{vcosthetacos t}cosh(vsinthetasin t),dt &= frac12int_0^{pi}e^{vcos(theta-t)}+e^{vcos(theta+t)},dt\
&=frac12left(int_0^{pi}e^{vcos(theta+t)},dt+int_{-pi}^{0}e^{vcos(theta+s)},dsright)\
&=frac12int_{-pi}^{pi}e^{vcos(theta+t)},dt = frac12int_{-pi-theta}^{pi-theta}e^{vcos s},dsend{align*}
Flipping $theta-t$ to $theta+s$ gives us an integral over the other half of the period - and it's the same function, so we just write it as one integral. Then, in that final integral of $e^{vcos s}$ over one full period, it doesn't matter where that period is; from $-pi$ to $pi$ is the same as from $-pi-theta$ to $pi-theta$.

That leaves us with the Bessel function identity, that the average value of $e^{vcos t}$ over a full period is $I_0(cos v)$. For this, since Bessel functions are defined by a differential equation, we differentiate (under the integral sign):
begin{align*}I(v) &= frac1{2pi}int_0^{2pi}e^{vcostheta},dtheta\
I'(v) &= frac1{2pi}int_0^{2pi}costhetacdot e^{vcostheta},dtheta\
I''(v) &= frac1{2pi}int_0^{2pi}cos^2thetacdot e^{vcostheta},dtheta\
I'(v) &= frac1{2pi}left[sinthetacdot e^{vcostheta}right]_{theta=0}^{theta=2pi} +frac1{2pi}int_0^{2pi}sinthetacdot vsinthetacdot e^{vcostheta},dtheta\
I'(v) &= frac{v}{2pi}int_0^{2pi}sin^2thetacdot e^{vcostheta},dthetaend{align*}
The first three lines are $I$ and its derivatives, calculated the obvious way. Then, in the next two, we apply integration by parts to transform the $I'$ integral into a form that works better with the others. Then, from $cos^2+sin^2=1$, we get $vI''(v)+I'(v)-vI(v)=0$, the modified Bessel equation of order zero. Together with the initial condition $I(0)=1$ (since the average value of $1$ is $1$) and $I'(0)=0$, this gives that $I(v)=I_0(v)$. Done.

A brief note on the bonus question: we can apply the same identities, but we run into trouble when we try to fold over and transform the $e^{vcos(theta-t)}$ term into an integral over $[-pi,0]$. The way the $t$ factor transforms, we end up with
$$frac12int_{-pi}^{pi}|t|cdot e^{vcos(theta+t)},dt$$
Multiplying by a triangle wave isn't going to come out cleanly. I might look at Fourier series next, but not in this answer.

Note that
$$begin{align*}
I_theta &= frac14int_0^{2pi}e^{vcosthetacos t}left(e^{vsinthetasin t}+e^{-vsinthetasin t}right) mathrm dt\&=frac14int_0^{2pi}e^{vcos(t-theta)} mathrm dt+frac14int_0^{2pi}e^{vcos(t+theta)} mathrm dt.
end{align*}$$ Because $tmapsto e^{vcos t}$ is $2pi$-periodic, we have
$$
int_0^{2pi}e^{vcos(t-theta)} mathrm dt=int_0^{2pi}e^{vcos(t+theta)} mathrm dt=int_0^{2pi}e^{vcos t} mathrm dt
$$ so it follows that
$$
I_theta = frac 12int_0^{2pi}e^{vcos t} mathrm dt =int_0^pi e^{vcos t} mathrm dt.
$$