Easy way of computing $\iint\limits_{B_R(0)} (e^y+e^{-y})\cos(x) d(x,y)$

Question

So I was asked to compute the integral

$$\iint\limits_{B_R(0)} (e^y+e^{-y})\cos(x) \, d(x,y), \qquad B_R(0):=\{(x,y)\in\mathbb{R}^2\mid x^2+y^2\leq R\}$$

the explicit way and the apparent "easy" way where no calculation is needed.

I did the first part but I fail to see the easy way to compute this.

Can anyone give me a hint? Thanks

Answer 1

Consolidating several comments, the integrand is $2\Re\cos(x+iy)$, so a change to polar coordinates lets us apply the mean value property viz.$$2\int_0^Rrdr\int_0^{2\pi}d\theta\cos(re^{i\theta})=2\int_0^Rrdr\int_0^{2\pi}d\theta1=2\pi R^2.$$As an alternative to the MVP, we can expand $\cos z$ as its Taylor series, using $\int_0^{2\pi}e^{ik\theta}d\theta=2\pi\delta_{k0}$ for $k\in\Bbb Z$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\iint_{x^{2} + y^{2} < R^{2}} \pars{\expo{y} + \expo{-y}}\cos\pars{x}\,\dd x\,\dd y} \\[5mm] = &\ \iint_{\large\mathbb{R}^{2}} \pars{\expo{y} + \expo{-y}}\cos\pars{x} \bracks{R^{2} - x^{2} - y^{2} > 0}\,\dd x\,\dd y \\[5mm] = &\ \iint_{\large\mathbb{R}^{2}} \pars{\expo{y} + \expo{-y}}\cos\pars{x} \bracks{\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{\pars{R^{2} - x^{2} - y^{2}}s} \over s} \,{\dd s \over 2\pi\ic}}\dd x\,\dd y \\[5mm] = &\ \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{R^{2}s} \over s} \overbrace{\bracks{\int_{-\infty}^{\infty}\cos\pars{x}\expo{-sx^{2}}\dd x}}^{\ds{\root{\pi}\,{\expo{-1/\pars{4s}} \over \root{s}}}}\ \times \\[2mm] &\ \phantom{\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{R^{2}s} \over s}\,\,} \underbrace{\bracks{\int_{-\infty}^{\infty}\expo{-sy^{2} + y}\dd y + \int_{-\infty}^{\infty}\expo{-sy^{2} - y}\dd y}} _{\ds{2\root{\pi}\,{\expo{1/\pars{4s}} \over \root{s}}}} {\dd s \over 2\pi\ic} \\[5mm] = &\ -\ic\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} {\expo{R^{2}s} \over s^{2}}\,\dd s = -\ic\bracks{2\pi\ic\,\lim_{s \to 0}\totald{\expo{R^{2}s}}{s}} = \bbx{\large 2\pi R^{2}} \\ & \end{align}