# Interesting integral $\int_0^\pi \frac{\sin(x)}{x} e^{x \cot(x)} \ \mathrm{d}x = \pi$

I recently received the following integral result which I've been trying to prove, from a friend who himself got it from the famous M.L. Glasser as a potential new identity: $$\boxed{\int_0^\pi \frac{\sin(x)}{x} e^{x \cot(x)} \ \mathrm{d}x = \pi}$$ Numerically this checks out, but I've made virtually no progress on this, due to how bizarre the integrand is. The following are some results I have:

• The Lobachevsky identity doesn't work in this case as we'd need an integral from $$0$$ to $$\infty$$, and the only $$2\pi$$-length interval the integrand converges on is $$(-\pi, \pi)$$.
• I haven't found any complex analytic ways of proceeding, but given the size and complexity of possible functions and contours, this doesn't really mean much.
• Series expanding the exponential is bad since term-by-term integration yields divergent values, and so the sum is not well-behaved.
• One thing I have noticed is that $$\frac{\sin(x)}{x} e^{x \cot(x)} = \frac{\sin(x)}{x} \exp\left (\frac{x}{\sin(x)} \cos(x)\right )$$ so DUTIS may be a viable option, but I didn't make much progress from here.

Any help with proving this amazing result would be great!

The integral is a special case of a conjecture by Nuttall (1985) $$\int_0^\pi\left[\frac{\sin x}xe^{x\cot x}\right]^\nu\,dx=\frac{\pi\nu^\nu}{\Gamma(\nu+1)},\quad\nu\ge0$$ that was later proven by many including Glasser.
In the linked paper, Bouwkamp (1986) proved the more generalised identity $$\int_0^\pi\frac{\cos(\nu(f(x)\sin x-x))+\frac{f'(x)}{f(x)}\sin(\nu(f(x)\sin x-x))}{(f(x)e^{-f(x)\cos x})^\nu}\,dx=\frac{\pi\nu^\nu}{\Gamma(\nu+1)}$$ for any positive, continuously differentiable $$f(x)$$ on $$[0,\pi)$$ such that $$f(\pi^-)\to\infty$$.
This follows directly from the Hankel contour integral $$\frac i2\int_\infty^{(0^+)}(-s)^{-\nu-1}e^{-\nu s}\,ds=\frac{\pi\nu^\nu}{\Gamma(\nu+1)}$$ with path $$s=-f(x)e^{ix}$$. Choosing $$f(x)=x/\sin x$$ recovers Nuttall's identity.