$\int_{0}^{\pi} \exp\left(\cos\left(t\right)\right)\cos\left(\sin\left(t\right)\right)\,{\rm d}t=\pi$

Question

Does anyone have a proof of the above integral? I have one proof, but I wanted to see other proofs.

Answer 1

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.

Answer 2

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: $$\exp(\cos(t)) \cos(\sin(t)) = \sum_{k=0}^{\infty} \dfrac{\cos(kt)}{k!}$$

Answer 3

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.

Answer 4

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.

Answer 5

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{\pi}^{2\pi}\expo{\cos\pars{t}}\cos\pars{\sin\pars{t}}\,\dd t = \int_{-\pi}^{0}\expo{\cos\pars{t}}\cos\pars{\sin\pars{t}}\,\dd t = -\int_{\pi}^{0}\expo{\cos\pars{t}}\cos\pars{\sin\pars{t}}\,\dd t}$ $\ds{=\int_{0}^{\pi}\expo{\cos\pars{t}}\cos\pars{\sin\pars{t}}\,\dd t}$

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}