A problematic integral: $\int_0^{2\pi} e^{-2\pi i\lambda\cos(t)}\,dt$

Question

Is there a special trick to calculate this integral? $$\int_0^{2\pi} e^{-2\pi i\lambda\cos(t)}\,dt$$ for $\lambda>0$.

Answer 1

The Bessel function of the zeroth order can be represented by the integral: $$J_0 (z) = \frac{1}{\pi} \int_{0}^\pi e^{i z \cos(t)} dt$$ Since the $i\sin(z\cos x)$ part of the exponent is zero over the integral until $2\pi$, only the $\cos(z \cos x))$ remains. This term remains identical through $x\to x+\pi$, so that: $$\int_{0}^{2\pi} e^{-2\pi i \lambda \cos(t)} dt = 2\pi J_0(2\pi\lambda)$$ As for the original integral, for a proof that this is indeed a solution of Bessel's ODE, see here.