Finding which function has this Fourier series I was attempting to solve
$$\int_0^\infty e^{-t}|\sin(t)|dt$$
Without Laplace transforms, which I know is the standard way to solve it.
I exploited how $|\sin(x)|$ has a period of $\pi$, so I generalized the integral into a limit and sum. The $(-1)^k$ comes from the fact that one would subtract off the negative area in order to add the same area the absolute value function would have accumulated for when the ordinary sin function would be negative. Thus I arrived at:
$$\lim_{n\to \infty} \Bigl(\sum_{k=0}^n (-1)^k \int_k^{k+\pi} e^{-x}\sin(x)dx\Bigr) (1)$$
Then, I made a substitution $u=x-k$, then integrated by parts:
$$\Bigl(\frac{1}{2} + \frac{1}{2e^\pi}\Bigr)\lim_{n\to \infty}\sum_{k=0}^n \frac{(-1)^k}{e^k}(\sin(k) + \cos(k))$$
Here, I’m not sure if I have reached an unfortunate dead end, or if there is perhaps some identity I can benefit from; maybe a Fourier series, since there are sinusoids involved. 
Thanks!
edit:
The bounds on the integral are not supposed to be $\int_k^{k+\pi}$. Rather, $(1)$ should be:
$$\lim_{n\to \infty} \Bigl(\sum_{k=0}^n (-1)^k \int_{k\pi}^{(k+1)\pi} e^{-x}\sin(x)dx\Bigr) $$
 A: With
$$F(t) = \frac{-e^{-t}(\cos(t) + \sin(t))}{2} \tag{1}\label{eq1A}$$
then
$$\begin{equation}\begin{aligned}
\frac{dF(t)}{dt} & = \frac{e^{-t}(\cos(t) + \sin(t))}{2} + \frac{-e^{-t}(-\sin(t) + \cos(t))}{2} \\
& = \frac{e^{-t}(\cos(t) + \sin(t) + \sin(t) - \cos(t))}{2} \\
& = \frac{2e^{-t}\sin(t)}{2} \\
& = e^{-t}\sin(t)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note I determined \eqref{eq1A} by twice using integration by parts of your integral, both times using $u = e^{-t}$, with $dv = \sin(t)dt$ for the first time and $dv = \cos(t)dt$ the second time. Also, you have
$$\cos(t) + \sin(t) = \sqrt{2}\cos\left(t - \frac{\pi}{4}\right) \tag{3}\label{eq3A}$$
so \eqref{eq1A} can be equivalently expressed as
$$F(t) = \frac{-e^{-t}\cos\left(t - \frac{\pi}{4}\right)}{\sqrt{2}} \tag{4}\label{eq4A}$$
I'll leave it to you to finish the rest of the problem by using \eqref{eq4A} when integrating the sections where $|\sin(t)| = \sin(t)$, i.e., $t \in [2n\pi,(2n+1)\pi]$, and the negative of \eqref{eq4A} in sections where $|\sin(t)| = -\sin(t)$, i.e., $t \in [(2n+1)\pi,(2n+1)\pi]$, for $n \in \mathbb{N} \cup \{0\}$, and finally summing the resulting infinite geometric series.
