Prove that $\int_{0}^{\pi}e^{\cos(\theta)}\cos(\sin(\theta))\,d\theta = \pi$

Question

I have to prove that

$$\int_{0}^{\pi}e^{\cos(\theta)}\cos(\sin(\theta))\,d\theta = \pi$$

by considering $$\int_{\gamma} \frac{e^z}{z}$$ where $\gamma(t) = e^{2{\pi}it}, 0 \leq t \leq 1$.

I have calculated that $$\int_{\gamma} \frac{e^z}{z} = 2{\pi}i$$

but I am not quite sure how to use the latter equality to prove the desired equality. Any help would be appreciated.

Answer 1

Consider $$\int_{0}^{\pi}e^{\cos(\theta)}\cos(\sin(\theta))\,d\theta+i\int_{0}^{\pi}e^{\cos(\theta)}\sin(\sin(\theta))\,d\theta =\int_{0}^{\pi}e^{cos(\theta)+i\sin(\theta)}\,d\theta.$$ This last integral can be easily computed using $$\int_{\gamma} \frac{e^z}{z}dz,$$ (just use the definition). Then you take the real part to conclude.

Answer 2

Complex analysis is not really needed: we just need to recall that $$ \forall n\in\mathbb{Z}\setminus\{0\},\qquad \text{Re}\left(\int_{0}^{\pi} e^{ni\theta}\,d\theta\right) = 0,$$ hence:

$$\begin{eqnarray*} I = \text{Re}\left(\int_{0}^{\pi} e^{\cos\theta+i\sin\theta}\,d\theta\right)&=&\text{Re}\left(\int_{0}^{\pi}\exp(e^{i\theta})\,d\theta\right)\\&=&\sum_{n\geq 0}\frac{1}{n!}\cdot\text{Re}\left(\int_{0}^{\pi}e^{ni\theta}\,d\theta\right)\\&=&\frac{\pi}{0!}=\color{red}{\pi}.\end{eqnarray*} $$

Answer 3

If you want to use complex analysis, note that from $$I=\int_{\gamma}\frac{e^{z}}{z}dz=2\pi i\tag{1} $$ we have, taking $z=e^{i\theta}=\cos\left(\theta\right)+i\sin\left(\theta\right)$ \begin{align*} I= & i\int_{-\pi}^{\pi}e^{\cos\left(\theta\right)+i\sin\left(\theta\right)}d\theta\\ = & i\int_{-\pi}^{\pi}e^{\cos\left(\theta\right)}\left(\cos\left(\sin\left(\theta\right)\right)+i\sin\left(\sin\left(\theta\right)\right)\right)d\theta\\ = & i\int_{-\pi}^{\pi}e^{\cos\left(\theta\right)}\cos\left(\sin\left(\theta\right)\right)d\theta-\int_{-\pi}^{\pi}e^{\cos\left(\theta\right)}\sin\left(\sin\left(\theta\right)\right)d\theta \end{align*} and now, since \begin{align*} \int_{-\pi}^{\pi}e^{\cos\left(\theta\right)}\sin\left(\sin\left(\theta\right)\right)d\theta= & \int_{0}^{\pi}e^{\cos\left(\theta\right)}\sin\left(\sin\left(\theta\right)\right)d\theta+\int_{-\pi}^{0}e^{\cos\left(\theta\right)}\sin\left(\sin\left(\theta\right)\right)d\theta\\ = & \int_{0}^{\pi}e^{\cos\left(\theta\right)}\sin\left(\sin\left(\theta\right)\right)d\theta-\int_{0}^{\pi}e^{\cos\left(\theta\right)}\sin\left(\sin\left(\theta\right)\right)d\theta\\ = & 0 \end{align*} we have $$I=i\int_{-\pi}^{\pi}e^{\cos\left(\theta\right)}\cos\left(\sin\left(\theta\right)\right)d\theta=2i\int_{0}^{\pi}e^{\cos\left(\theta\right)}\cos\left(\sin\left(\theta\right)\right)d\theta $$ and from $(1)$ the final result.