Cool area problem

$\begin{aligned} A &=\int_0^\infty \left | e^{-x}\sin(x) \right|dx=\int_0^\infty e^{-x} \left | \sin(x) \right|dx & & \color{#D61F06} e^{-x}>0 \\ &=\sum_{k=0}^\infty \left ( (-1)^k \int_{k \pi}^{(k+1) \pi} e^{-x}\sin(x)dx\right ) & & \color{#D61F06} k\pi<x<(k+1)\pi\Rightarrow \left | \sin(x) \right |=(-1)^k \sin(x) \\ &=\sum_{k=0}^\infty \left ( (-1)^k \left . \frac{-e^{-x}\left ( \sin(x)+\cos(x)) \right )}{2} \right |_{k\pi}^{(k+1)\pi} \right ) \\ &=\frac{-1}{2} \sum_{k=0}^\infty \left ( (-1)^k \left [ e^{-(k+1)\pi} \cos((k+1)\pi)-e^{-k\pi} \cos(k\pi) \right ] \right ) \\ &=\frac{-1}{2} \sum_{k=0}^\infty \left ( (-1)^k \left [ e^{-k\pi} e^{-\pi} (-1)^{k+1}-e^{-k\pi} (-1)^k \right ] \right ) & & \color{#D61F06} \cos(k \pi)=(-1)^k\\ &=\frac{-1}{2} \sum_{k=0}^\infty e^{-k\pi} \left (e^{-\pi} (-1)^{2k+1}-(-1)^{2k} \right ) \\ &=\frac{e^{-\pi} +1 }{2} \sum_{k=0}^\infty \left (e^{-\pi} \right )^k =\frac{e^{-\pi} +1 }{2} \cdot \frac{1 }{1-e^{-\pi}} & & \color{#D61F06} \text{geometric series}\\ A &=\frac{e^\pi+1}{2\left (e^\pi-1 \right )}\approx 0.5451657 \end{aligned}$

Let $f(x)=e^{-x}sin(x)$ .

Then the area, $A = \sum_{i=0}^\infty |\int_{i\pi}^{(i+1)\pi} f(x) dx|$ .

Calculating the integral in the summation via integration by parts gives $\int_{i\pi}^{(i+1)\pi} f(x) dx = \frac{(-1)^{i}e^{-i\pi}(e^{-\pi}+1)}{2}$ , and taking the absolute value of this gives the area as $A = \sum_{i=0}^\infty \frac{e^{-i\pi}(e^{-\pi}+1)}{2}$ . This sum is simply a geometric series, and using $S=\frac{a_0}{1-r}$ gives $\boxed{A = \frac{e^{\pi}+1}{2(e^{\pi}-1)}}$ . Note that we can also find the positive and negative areas of the graph by solving the simultaneous equations $A_1-A_2=\frac{e^{\pi}+1}{2(e^{\pi}-1)}$ and $A_1+A_2=\frac{1}{2}$ , where $A_1$ and $A_2$ are the positive and negative areas, respectively. This gives $A_1=\frac{e^{\pi}}{2(e^{\pi}-1)}$ and $A_2 = \frac{-1}{2(e^{\pi}-1)}$ .