This week's problem was correctly answered by MarkFL. You can find his solution below.
[sp]We are given the function:
$$f(t)=\begin{cases}\sin(kt) & 0\leq t< \dfrac{\pi}{k}\\ 0 & \dfrac{\pi}{k}\leq t< \dfrac{2\pi}{k}\end{cases}$$
Which is $$p=\frac{2\pi}{k}$$ periodic, and which we observe is continuous as well.
To find the Laplace transform, we begin with the definition:
$$\mathcal{L}\{f\}\equiv\int_0^{\infty} e^{-st}f(t)\,dt$$
Because $f(t)$ satisfies:
$$f(t)=f(t+p)$$ for $$0\le t$$
we may simplify our computations by partitioning the integral of the transform into integrals over subintervals of length $p$, and write:
$$]\mathcal{L}\{f\}=\sum_{j=0}^{\infty}\left(\int_{jp}^{(j+1)p} e^{-st}f(t)\,dt \right)$$
Consider the following substitution:
$$\tau=t-jp$$
Then we find:
$$e^{-st}=e^{-s(\tau+jp)}$$
and so using this and by the periodicity of $f$, we find:
$$\int_{jp}^{(j+1)p} e^{-st}f(t)\,dt=e^{-jps}\int_{0}^{p} e^{-s\tau}f(\tau)\,d\tau$$
Since the integral does not depend on $j$, we may now write:
$$\mathcal{L}\{f\}=\left(\int_{0}^{p} e^{-s\tau}f(\tau)\,d\tau \right)\sum_{j=0}^{\infty}\left(e^{-jps} \right)$$
Now, if we observe that the sum is a geometric series, we may write:
$$\sum_{j=0}^{\infty}\left(e^{-jps} \right)=\sum_{j=0}^{\infty}\left(\left(e^{-ps} \right)^j \right)=\frac{1}{1-e^{-ps}}$$
Now, for the integral, we may write (using the definition of $f$):
$$\int_{0}^{p} e^{-s\tau}f(\tau)\,d\tau=\int_{0}^{\frac{p}{2}} e^{-s\tau}\sin(k\tau)\,d\tau+\int_{\frac{p}{2}}^{p} e^{-s\tau}0\,d\tau=\int_{0}^{\frac{p}{2}} e^{-s\tau}\sin(k\tau)\,d\tau$$
At this point, we may develop a formula for:
$$I=\int_0^a e^{bx}\sin(cx)\,dx$$
Using integration by parts, we may use:
$$u=\sin(cx)\,\therefore\,du=c\cos(cx)\,dx$$
$$dv=e^{bx}\,dx\,\therefore\,v=\frac{1}{b}e^{bx}$$
Hence:
$$I=\left[\frac{1}{b}e^{bx}\sin(cx) \right]_0^a-\frac{c}{b}\int_0^a e^{bx}\cos(cx)\,dx$$
$$I=\frac{1}{b}e^{ab}\sin(ac)-\frac{c}{b}\int_0^a e^{bx}\cos(cx)\,dx$$
On the remaining integral, use integration by parts again where:
$$u=\cos(cx)\,\therefore\,du=-c\sin(cx)\,dx$$
$$dv=e^{bx}\,dx\,\therefore\,v=\frac{1}{b}e^{bx}$$
Hence:
$$I=\frac{1}{b}e^{ab}\sin(ac)-\frac{c}{b}\left(\left[\frac{1}{b}e^{bx}\cos(cx) \right]_0^a+\frac{c}{b}\int_0^a e^{bx}\sin(cx)\,dx \right)$$
$$I=\frac{1}{b}e^{ab}\sin(ac)-\frac{c}{b}\left(\frac{1}{b}e^{ab}\cos(ac)-\frac{1}{b}+\frac{c}{b}I \right)$$
$$I=\frac{1}{b}e^{ab}\sin(ac)-\frac{c}{b^2}e^{ab}\cos(ac)+\frac{c}{b^2}-\frac{c^2}{b^2}I$$
$$I\left(\frac{b^2+c^2}{b^2} \right)=\frac{\left(b\sin(ac)-c\cos(ac) \right)e^{ab}+c}{b^2}$$
$$I=\frac{\left(b\sin(ac)-c\cos(ac) \right)e^{ab}+c}{b^2+c^2}$$
Applying this formula, we may then write:
$$\int_{0}^{\frac{p}{2}} e^{-s\tau}\sin(k\tau)\,d\tau=\frac{\left(-s\sin\left(\dfrac{p}{2}k \right)-k\cos\left(\dfrac{p}{2}k \right) \right)e^{-\frac{p}{2}s}+k}{s^2+k^2}$$
And so we now have:
$$\mathcal{L}\{f\}=\frac{k-e^{-\frac{p}{2}s}\left(s\sin\left(\dfrac{p}{2}k \right)+k\cos\left(\dfrac{p}{2}k \right) \right)}{\left(s^2+k^2 \right)\left(1-e^{-ps} \right)}$$
Using $$p=\frac{2\pi}{k}$$ we have:
$$\mathcal{L}\{f\}=\frac{k-e^{-\frac{\pi}{k}s}\left(s\sin\left(\pi \right)+k\cos\left(\pi \right) \right)}{\left(s^2+k^2 \right)\left(1-e^{-\frac{2\pi}{k}s} \right)}$$
$$\mathcal{L}\{f\}=\frac{k\left(1+e^{-\frac{\pi}{k}s} \right)}{\left(s^2+k^2 \right)\left(1-e^{-\frac{2\pi}{k}s} \right)}$$
Factoring the second factor in the denominator as the difference of squares and then dividing out the resulting factor common to the numerator, we obtain:
$$\mathcal{L}\{f\}=\frac{k}{\left(s^2+k^2 \right)\left(1-e^{-\frac{\pi}{k}s} \right)}$$[/sp]
Mark did a great job solving this problem from first principles. There is a formula, though, that allows you to compute $\mathcal{L}\{f(t)\}$ for $p$-periodic functions (which Mark, in essence, derived): If $f(t)$ is $p$-periodic and piecewise continuous for $t\geq 0$, then \[\mathcal{L}\{f(t)\} = \frac{1}{1-e^{-ps}}\int_0^p e^{-st}f(t)\,dt\tag{1}\]
Below, you'll find my solution using $(1)$.
[sp]Consider the $\dfrac{2\pi}{k}$-periodic function $f(t)$, where
\[f(t) = \begin{cases}\sin(kt) & 0\leq t\leq \pi/k\\ 0 & \pi/k \leq t < 2\pi/k\end{cases}\]
By formula $(1)$, we see that
\[\mathcal{L}\{f(t)\} = \frac{1}{1-e^{-2\pi s/k}}\int_0^{2\pi/k} e^{-st}f(t)\,dt = \frac{1}{1-e^{-2\pi s/k}}\int_0^{\pi/k}e^{-st}\sin(kt)\,dt.\]
To evaluate
\[\int_0^{\pi/k} e^{-st}\sin(kt)\,dt\]
we proceed using integration by parts twice: Let $u=e^{-st}\implies \,du=-se^{-st}\,dt$ and $\,dv=\sin(kt)\,dt \implies v= -\frac{1}{k}\cos (kt)$. Therefore,
\[\begin{aligned}\int_0^{\pi/k} e^{-st}\sin(kt)\,dt &= \left.\left[-\frac{e^{-st}}{k}\cos(kt)\right]\right|_0^{\pi/k} - \frac{s}{k}\int_0^{\pi/k}e^{-st}\cos(kt)\,dt\\ &= \frac{1+e^{-\pi s/k}}{k} -\frac{s}{k}\int_0^{\pi/k} e^{-st}\cos(kt)\,dt\end{aligned}\]
We apply parts again: $u=e^{-st}\implies \,du = -se^{-st}\,dt$ and $\,dv=\cos(kt) \implies v=\frac{1}{k}\sin(kt)$. Therefore,
\[\begin{aligned}\int_0^{\pi/k}e^{-st}\sin(kt) &= \frac{1+e^{-\pi s/k}}{k}-\frac{s}{k}\left(\left.\left[ \frac{e^{-st}}{k}\sin(kt)\right]\right|_0^{\pi/k} + \frac{s}{k}\int_0^{\pi/k} e^{-st}\sin(kt)\,dt\right)\\ &= \frac{1+e^{-\pi s/k}}{k}-\frac{s^2}{k^2} \int_0^{\pi/k} e^{-st}\sin(kt)\,dt\end{aligned}\]
Solving for the integral leaves us with
\[\left(1+\frac{s^2}{k^2}\right) \int_0^{\pi/k}e^{-st}\sin(kt)\,dt = \frac{1+e^{-\pi s/k}}{k} \implies \int_0^{\pi/k}e^{-st}\sin(kt)\,dt = \frac{k}{s^2+k^2}(1+e^{-\pi s/k})\]
Therefore
\[\begin{aligned}\mathcal{L}\{f(t)\} &= \frac{1}{1-e^{-2\pi s/k}}\cdot \frac{k}{s^2+k^2}(1+e^{-\pi s/k})\\ &= \frac{k(1+e^{-\pi s/k})}{(s^2+k^2)(1+e^{-\pi s/k}) (1-e^{-\pi s/k})}\\ &= \frac{k}{(s^2+k^2)(1-e^{-\pi s/k})}\end{aligned}\][/sp]
