$\int_{0}^{\pi} \exp\left(\cos\left(t\right)\right)\cos\left(\sin\left(t\right)\right)\,{\rm d}t=\pi$ Does anyone have a proof of the above integral? I have one proof, but I wanted to see other proofs.
 A: Hint:
By some simple change of variables we get
$$\int_0^\pi e^{\cos t}\cos\sin tdt=\frac{1}{2}\int_0^{2\pi} e^{\cos \theta}\cos \sin\theta d\theta=\mathrm{Re}\left(\frac{1}{2}\int_0^{2\pi}e^{e^{i\theta}}d\theta\right) = \pi\cdot\mathrm{Re}\left(\frac{1}{2\pi i}\int_{C}\frac{e^z}{z}dz\right),$$
where $C$ is the unit circle. Now use Cauchy integral formula.
A: We have
$\cos(\sin(t)) = \text{Real}(e^{i \sin(t)})$. Hence, the integral we want is
$$I = \text{Real}\left(\int_0^{\pi} e^{e^{it}}dt \right)$$
We have
$$e^{e^{it}} = \sum_{k=0}^{\infty} \dfrac{e^{ikt}}{k!}$$
Hence,
$$I = \sum_{k=0}^{\infty} \dfrac{I_k}{k!}$$
where $I_k = \text{Real}\left(\displaystyle \int_0^{\pi} e^{ikt}dt \right) = \pi \delta_k$. Hence, we are done.
Updated to answer TCL's claim:
I do not understand TCL's claim "I was hoping to see a proof without complex analysis, but that seems to be not possible." My proof does not rely on complex analysis at all. Writing $\cos(t)$ as Real($e^{it}$) is just a notational convenience. All we are making use of is the following identity:
$$\exp(\cos(t)) \cos(\sin(t)) = \sum_{k=0}^{\infty} \dfrac{\cos(kt)}{k!}$$
A: Since the integrand is even, this is half of the integral from $-\pi$ to $\pi.$ That, in turn is an integral over the unit circle of a meromorphic function, and thus can be done using Cauchy's integral formula. Computing the residue is a bit tedious. But then, maybe it's the same as your proof.
A: Apply Mean Value Property to the function $e^z$ at 0, we get for any $r>0$,  $$1=e^0=\frac{1}{2\pi}\int_0^{2\pi} \exp(re^{it})\,dt$$
Using the fact that the integrand is $2\pi$ periodic and that $\exp(r\cos t)\sin(r\sin(t))$ is odd and $\exp(r\cos t)\cos(r\sin(t))$ is even, we find $$1=\frac{1}{\pi}\int_0^\pi \exp(r\cos t)\cos(r\sin t)\,dt.$$
I was hoping to see a proof without complex analysis, but that seems to be not possible.
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\int_{\pi}^{2\pi}\expo{\cos\pars{t}}\cos\pars{\sin\pars{t}}\,\dd t
     =
     \int_{-\pi}^{0}\expo{\cos\pars{t}}\cos\pars{\sin\pars{t}}\,\dd t
     =
     -\int_{\pi}^{0}\expo{\cos\pars{t}}\cos\pars{\sin\pars{t}}\,\dd t}$
$\ds{=\int_{0}^{\pi}\expo{\cos\pars{t}}\cos\pars{\sin\pars{t}}\,\dd t}$

Then,
\begin{align}
&\color{#0000ff}{\large%
\int_{0}^{\pi}\expo{\cos\pars{t}}\cos\pars{\sin\pars{t}}\,\dd t} 
=\half\int_{0}^{2\pi}\expo{\cos\pars{t}}\cos\pars{\sin\pars{t}}\,\dd t
=\half\,\Re\int_{0}^{2\pi}\expo{\expo{\ic t}}\,\dd t 
\\[3mm]&=\half\,\Re\oint_{\verts{z}\ =\ 1}\expo{z}\,\pars{-\ic\,{\dd z \over z}}
=\half\,\Re\lim_{z \to 0}\pars{2\pi\ic\,z\,{-\ic\expo{z} \over z}}
=\color{#0000ff}{\Large\pi} 
\end{align}

