Question

Given the curve $\displaystyle \large{y=e^{-x}\sin x \ \ (x\geq 0)$ let the area enclosed by the curve and the x-axis (above the x-axis) be S0, S1, S2, ... , Sn, ... in order from y-axis. Find $\displaystyle \large{ \lim_{n \to \infty} \sum_{k=0}^{n} S_k}$

Answer 1

The given curve crosses the x-axis whenever x is a multiple of π, and it lies above the x-axis between consecutive even and odd multiples of π. So the regions with area S₀, S₁, S₂, ... are the sets

$R_0 = \left\{(x, y) : 0 \le x \le \pi \text{ and } 0 \le y \le e^{-x}\sin(x)\right\}$

$R_1 = \left\{(x, y) : 2\pi \le x \le 3\pi \text{ and } 0 \le y \le e^{-x}\sin(x)\right\}$

$R_2 = \left\{(x, y) : 4\pi \le x \le 5\pi \text{ and } 0 \le y \le e^{-x}\sin(x)\right\}$

and so on, with

$R_k = \left\{(x, y) : 2k\pi \le x \le (2k+1)\pi \text{ and } 0 \le y \le e^{-x}\sin(x)\right\}$

for natural number k.

The areas themselves are then given by the integral

$S_k = \displaystyle \int_{2k\pi}^{(2k+1)\pi} \int_0^{e^{-x}\sin(x)} dy \, dx = \int_{2k\pi}^{(2k+1)\pi} e^{-x}\sin(x) \, dx$

Integrate by parts twice. Take

$u = e^{-x} \implies du = -e^{-x} \, dx$

$dv = \sin(x) \, dx \implies v = -\cos(x)$

so that

$\displaystyle \int e^{-x}\sin(x) \, dx = -e^{-x}\cos(x) - \int e^{-x}\cos(x) \, dx$

then

$u = e^{-x} \implies du = -e^{-x} \, dx$

$dv = \cos(x) \, dx \implies v = \sin(x)$

so that

$\displaystyle \int e^{-x}\cos(x) \, dx = e^{-x}\sin(x) + \int e^{-x}\sin(x) \, dx$

Overall, we find

$\displaystyle \int e^{-x}\sin(x) \, dx = -e^{-x}\cos(x) - e^{-x}\sin(x) - \int e^{-x}\sin(x) \, dx$

or

$\displaystyle \int e^{-x}\sin(x) \, dx = -\frac12 e^{-x} (\cos(x)+\sin(x)) + C$

Using the antiderivative and the fundamental theorem of calculus, we compute the k-th area to be

$\displaystyle S_k = -\frac12 e^{-(2k+1)\pi} (\cos((2k+1)\pi)+\sin((2k+1)\pi)) + \frac12 e^{-2k\pi} (\cos(2k\pi)+\sin(2k\pi))$

$\displaystyle S_k = \frac12 e^{-2k\pi} \cos(2k\pi) \left(e^{-\pi} + 1\right)$

$\displaystyle S_k = \frac{e^{-\pi}+1}2 e^{-2k\pi}$

Since $\left|e^{-2\pi}\right|<1$, the sum we want is a convergent geometric sum. As n goes to ∞, we have

$\displaystyle \lim_{n\to\infty} \sum_{k=0}^n S_k = \frac{e^{-\pi}+1}2 \cdot \frac1{1 - e^{-2\pi}} = \boxed{\frac{e^\pi+1}{4\sinh(\pi)}}$