Showing that successive arches of $y=e^{-\alpha x}\sin\beta x$ bound areas in geometric progression with ratio $q=e^{-\alpha\pi/\beta}$ Prove that the areas $S_0$, $S_1$, $S_2$, $S_3$, $\ldots$, limited by the $x$-axis and the curve $y=e^{-\alpha x}\sin\beta x$, for $x\geq 0$, form a geometric progression of the ratio $q=e^{-\alpha\pi/\beta}$ (see image below)

Then I tried to do it by calculation: 

I don't think it works very well ...
 A: $\int_0^\infty |e^{-\alpha x}\sin\beta x| \ dx$
We break this up.
$\int_0^{\frac {\pi}{\beta}} e^{-\alpha x}|\sin\beta x| \ dx + \int_\frac {\pi}{\beta}^{2\frac {\pi}{\beta}} e^{-\alpha x}|\sin\beta x| \ dx+\cdots+\int_{n\frac {\pi}{\beta}}^{(n+1)\frac {\pi}{\beta}} e^{-\alpha x}|\sin\beta x| \ dx+\cdots $
Lets tackle the general term.
$\int_{n\frac {\pi}{\beta}}^{(n+1)\frac {\pi}{\beta}} e^{-\alpha x}|\sin\beta x| \ dx\\
u = x-n\frac{\pi}{\beta}\\
du = dx\\
\int_{0}^{\frac {\pi}{\beta}} e^{-\alpha (u+n\frac {\pi}{\beta})}|\sin\beta u+n\pi| \ du\\
e^{-\frac {\alpha n \pi}{\beta}}\int_{0}^{\frac {\pi}{\beta}} e^{-\alpha u}|\sin\beta u| \ du\\$
We can do this to each term
$\int_0^{\frac {\pi}{\beta}} e^{-\alpha x}|\sin\beta x| \ dx + e^{-\frac {\alpha \pi}{\beta}}\int_0^{\frac {\pi}{\beta}} e^{-\alpha x}|\sin\beta x| \ dx + e^{-\frac {2\alpha \pi}{\beta}}\int_0^{\frac {\pi}{\beta}} e^{-\alpha x}|\sin\beta x| \ dx+\cdots + e^{-\frac {n\alpha \pi}{\beta}}\int_0^{\frac {\pi}{\beta}} e^{-\alpha x}|\sin\beta x| \ dx+\cdots\\
\int_0^{\frac {\pi}{\beta}} e^{-\alpha x}|\sin\beta x| \ dx(1 + e^{-\frac {\alpha \pi}{\beta}} + e^{-\frac {2\alpha \pi}{\beta}}+\cdots + e^{-\frac {n\alpha \pi}{\beta}}+\cdots)$
A: Examined below is the $k$-th term in the series.
$$I_k = \int_{\frac{k\pi}{\beta}}^{\frac{(k+1)\pi}{\beta}}
e^{-\alpha x}\sin{\beta x}dx$$
Integrate-by-parts twice to get
$$I_k = -\frac{\beta^2}{\alpha^2}I_k + \frac{\beta}{\alpha^2}
(e^{-\frac{\alpha\pi}{\beta}}+1)e^{-\frac{k\alpha\pi}{\beta}}$$
Thus, 
$$\frac{I_{k+1}}{I_{k}}= e^{-\frac{\alpha\pi}{\beta}}$$
