the area between $e^{-x}\sin x\quad (x\ge0)$ and x-axis

Question

I encountered this problem in a test,and I didn’t work it out back then. what I know is the area $$S=\int_{0}^{\infty}e^{-x}|\sin x|dx$$ and I suppose the indefinite integral $$\int e^{-x}\sin x\ dx= -\frac{1}{2}e^{-x}(\sin x+\cos x)+C $$ should help to solve this problem,but that’s all I got,No further progress for now.

Answer 1

Hint:

Let $f(x)=e^{-x}$, $g(x)= \sin(x)$ and $h(x)=\cos(x)$. Note that $f'=-f$, $f''=f$, $g'=h$ and $g''=-g$

With this,

$$(fg)'=f'g+fg'=-fg+fh \\(fh)'=f'h+fh'=-fh-fg$$

Adding, $$(fg+fh)'=-2fg$$

Integrating, $$ fg+fh+C=\int (fg+fh)' = -2\int fg$$

Finally, we have $$\int e^{-x}\sin(x)dx = -\frac{1}{2}e^{-x}(\sin(x)+\cos(x))+C$$

Answer 2

One approach is relate the integral to a simpler one $\int_0^\infty e^{-x}\sin x dx$ which can be evaluated in many ways. The following uses a complex exponential function first mentioned by Jykri Lahtonen in comment.

$$\begin{align} \int_0^\infty e^{-x}|\sin x| dx &= \sum_{k=0}^\infty \int_{k\pi}^{(k+1)\pi} e^{-x}|\sin(x)|dx\\ &= \sum_{k=0}^\infty e^{-k\pi} \int_0^{\pi} e^{-x} \sin x dx\\ &= \frac{1+e^{-\pi}}{1-e^{-\pi}}\sum_{k=0}^\infty (-1)^k e^{-k\pi} \int_0^{\pi} e^{-x} \sin x dx\\ &= \frac{1+e^{-\pi}}{1-e^{-\pi}}\sum_{k=0}^\infty \int_{k\pi}^{(k+1)\pi} e^{-x} \sin x dx\\ &= \frac{1+e^{-\pi}}{1-e^{-\pi}}\int_0^\infty e^{-x}\sin x dx\\ &= \coth\frac{\pi}{2}\cdot\Im\left[\int_0^\infty e^{-(1-i)x} dx\right]\\ &= \coth\frac{\pi}{2}\cdot\Im\left[\frac{1}{1-i}\right]\\ &= \frac12 \coth\frac{\pi}{2} \end{align} $$

Answer 3

we know that $$\sin x\ge0,x\in(2k\pi,(2k+1)\pi)$$ $$\sin x\le0,x\in((2k+1)\pi,(2k+2)\pi)$$ $$k=0,1,2,\dots$$ So $$S=\int_{0}^{\infty}e^{-x}\vert\sin x\vert dx\\ =\sum_{k=0}^{\infty}\int_{2k\pi}^{(2k+1)\pi}e^{-x}\sin x\ dx-\sum_{k=0}^{\infty}\int_{(2k+1)\pi}^{(2k+2)\pi}e^{-x}\sin x\ dx$$ Let $$I=\int e^{-x}\sin x\ dx\\ =-\int \sin x\ de^{-x}\\ =-e^{-x}\sin x+\int e^{-x}\ d\sin x\\ =-e^{-x}\sin x+\int e^{-x}\cos x\ dx\\ =-e^{-x}\sin x-\int \cos x\ de^{-x}\\ =-e^{-x}\sin x-e^{-x}\cos x+\int e^{-x}\ d\cos x\\ =-e^{-x}(\sin x+\cos x)-\int e^{-x}\sin x\ dx\\ =-e^{-x}(\sin x+\cos x)-I$$ Hence $$I=-\frac{1}{2}e^{-x}(\sin x+\cos x)+C$$ Then $$S=\sum_{k=0}^{\infty}-\frac{1}{2}e^{-x}(\sin x+\cos x)\Big |_{2k\pi}^{(2k+1)\pi}-\sum_{k=0}^{\infty}-\frac{1}{2}e^{-x}(\sin x+\cos x)\Big |_{(2k+1)\pi}^{(2k+2)\pi}\\ =\sum_{k=0}^{\infty}\frac{1}{2}\left[e^{-(2k+1)\pi}+e^{-2k\pi}\right]+\sum_{k=0}^{\infty}\frac{1}{2}\left[e^{-(2k+2)\pi}+e^{-(2k+1)\pi}\right]\\ =\sum_{k=0}^{\infty}e^{-(2k+1)\pi}+\frac{1}{2}\sum_{k=0}^{\infty}e^{-2k\pi}+\frac{1}{2}\sum_{k=0}^{\infty}e^{-(2k+1)\pi}\\ =\sum_{k=0}^{\infty}e^{-(2k+1)\pi}+\frac{1}{2}\sum_{k=0}^{\infty}e^{-2k\pi}+\frac{1}{2}\sum_{k=1}^{\infty}e^{-2k\pi}\\ =\sum_{k=0}^{\infty}e^{-(2k+1)\pi}+\frac{1}{2}\sum_{k=0}^{\infty}e^{-2k\pi}+\frac{1}{2}\sum_{k=0}^{\infty}e^{-2k\pi}-\frac{1}{2}\\ =\sum_{k=0}^{\infty}e^{-(2k+1)\pi}+\sum_{k=0}^{\infty}e^{-2k\pi}-\frac{1}{2}\\ =\sum_{k=0}^{\infty}\left[e^{-2k\pi}+e^{-(2k+1)\pi}\right]-\frac{1}{2}\\ =\sum_{n=0}^{\infty}e^{-n\pi}-\frac{1}{2}\\ =\frac{1}{1-e^{-\pi}}-\frac{1}{2}$$ $$\left(\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}\Rightarrow \sum_{n=0}^{\infty}(e^{-\pi})^n=\frac{1}{1-e^{-\pi}}\right)$$