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I want to prove that $$\iint\limits_{x^2+y^2<1} u(x,y) dxdy=\pi$$ where $u(x,y)=e^x \cos y$.

There is a theorem which says that if $u\in C^2(\Omega)$ and $\nabla^2 u=0$ in a domain $\Omega\subseteq \mathbb{R}^n$, then for any ball $B=B_R(v)$ with $\overline{B}\subset\Omega$, $$u(v)=\frac{1}{\omega_n R^n}\int_B u dx$$ where $\omega_n$ is the volume of the ball of radius 1 in $\mathbb{R}^n$. The double integral above can be seen as a particular case of the theorem, since $\omega_2=\pi$, $R=1$ and $u(0,0)=1$. It's also clear that $\nabla^2 u=0$ (It's the real part of $e^z,z\in\mathbb{C}$).

I want to prove it without using this mean value theorem. In a standard way, I get to the integral $$2\int_{-1}^1 e^x\sin \sqrt{1-x^2}dx$$ which seems crazy. Numerically it seems to be $\pi$ efectively.

How could I calculate the integral?

LeviathanTheEsper
  • 3,163

2 Answers2

4

Write

\begin{align*} 2 \int_{-1}^{1} e^x \sin \sqrt{1-x^2} \, dx &= 2 \Im \int_{-1}^{1} e^{x+i\sqrt{1-x^2}} \, dx \\ &= 2 \Im \int_{0}^{\pi} e^{e^{i\theta}}\sin\theta \, d\theta \qquad (x = \cos\theta) \\ &= \Im \int_{0}^{\pi} i e^{e^{i\theta}} (e^{-i\theta} - e^{i\theta}) \, d\theta. \end{align*}

Now expanding the double exponential term using the Taylor series, we can write

\begin{align*} 2 \int_{-1}^{1} e^x \sin \sqrt{1-x^2} \, dx &= \sum_{n=0}^{\infty} \frac{1}{n!} \Im \int_{0}^{\pi} i (e^{(n-1)i\theta} - e^{(n+1)i\theta}) \, d\theta. \end{align*}

But notice that we have the following sifting property:

$$ \Im \int_{0}^{\pi} i e^{ik\theta} \, d\theta = \begin{cases} \pi, & k = 0 \\ 0, & k \neq 0 \end{cases}. $$

Therefore we get

$$ 2 \int_{-1}^{1} e^x \sin \sqrt{1-x^2} \, dx = \frac{1}{0!}(0 - 0) + \frac{1}{1!}({\color{red}\pi} - 0) + \frac{1}{2!}(0 - 0) + \cdots = \pi. $$

Sangchul Lee
  • 167,468
4

It is not that crazy. It is equivalent to $$ I = 2\cdot\text{Im}\int_{0}^{1}\exp\left(x+i\sqrt{1-x^2}\right)+\exp\left(-x+i\sqrt{1-x^2}\right)\,dx $$ or to $$ I = 2\cdot\text{Im}\int_{0}^{\pi/2}\left[\exp(e^{i\theta})+\exp(-e^{-i\theta})\right]\sin(\theta)\,d\theta$$ or, by exploiting $e^z=\sum_{n\geq 0}\frac{z^n}{n!}$, to $$ I = 2\cdot\sum_{n\geq 0}\frac{1}{n!}\int_{0}^{\pi/2}\left[\sin(n\theta)-(-1)^n \sin(n\theta)\right]\sin(\theta)d\theta $$ that simplifies to: $$ I = 4\cdot\sum_{m\geq 0}\frac{1}{(2m+1)!}\int_{0}^{\pi/2}\sin((2m+1)\theta)\sin(\theta)\,d\theta $$ where only the contribute given by $m=0$ is non-zero. It follows that: $$ I = 4\int_{0}^{\pi/2}\sin^2(\theta)\,d\theta = \color{red}{\pi}.$$

Jack D'Aurizio
  • 353,855