My mathematical physics test

A simple solution: We can deduce that the result of the integral is the same for every real $k$ . So, if we take $k=0$ , then:

$\frac { 1 }{ \pi } \int _{ 0 }^{ \pi }{ { e }^{ k\cos { \theta } }·\cos { (k\sin { \theta } ) } } d\theta =$

$\quad \frac { 1 }{ \pi } \int _{ 0 }^{ \pi }{ { e }^{ 0·\cos { \theta } }·\cos { (0·\sin { \theta } ) } } d\theta =$

$\frac { 1 }{ \pi } \int _{ 0 }^{ \pi }{ { e }^{ 0 }·\cos { (0) } } d\theta =\quad \frac { 1 }{ \pi } \int _{ 0 }^{ \pi }{ 1 } d\theta \quad =$

$\quad \frac { 1 }{ \pi } (\pi -0)=1$

Thus, the solution is $\boxed{1}$ .

Cauchy's Integral Formula implies that $I_1(z, 0) = 2\pi i$ .

Moreover, if we consider the parametrization of $C :f(\theta) = e^{i\theta}, \theta \in [0, 2\pi]$ to solve $I_1(z, 0)$ by definition, we obtain the following:

$\displaystyle I_1(z, 0) = i\int\limits_0^{2\pi} \frac{ e ^ {k e ^{ i \theta }}}{e^{ i\theta}} e^{i \theta} d\theta = i\int\limits_0^{2\pi} e^{ ke^{i\theta}} d\theta = i\int\limits_0^{2\pi} e^{k\cos(\theta)} e^{ik\sin(\theta)} d\theta$

$\displaystyle I_1(z, 0) = i\int\limits_0^{2\pi} e^{k\cos(\theta)}(\cos(k\sin(\theta)) + i\sin(k\sin(\theta))) d\theta$

$\displaystyle I_1(z, 0) = i\int\limits_0^{2\pi} e^{k\cos(\theta)}\cos(k\sin(\theta))d\theta - \int\limits_0^{2\pi} e^{k\cos(\theta)}\sin(k\sin(\theta))d\theta$ .

Since $I_1(z, 0) = 2\pi i$ , it follows that $\displaystyle 2\pi = \int\limits_0^{2\pi} e^{k\cos(\theta)}\cos(k\sin(\theta))d\theta$ .

Let $g(\theta) = e^{k\cos(\theta)}\cos(k\sin(\theta))$ . Notice that $g(\pi + \theta) = g(\pi - \theta), \forall \theta \in \mathbb{R}$ .

Therefore $\displaystyle \int\limits_0^{2\pi} g(\theta) d\theta = 2 \int\limits_0^{\pi} g(\theta) d\theta = 2\pi \Rightarrow \frac{1}{\pi} \int\limits_0^{\pi} g(\theta) d\theta = 1$ .