How would one go about Laplace Transforming $\max(0, \sin(t))$?

Question

The function described in the Image below is meant with $\max(0, \sin(t))$ I'm guessing you need to do a series expansion of sorts or some clever arithmetic with Euler's identity. But I am unable to get something concrete.

Answer 1

Whilst the the methods in the other answers are correct too, I find this one easiest to understand.

You can represent the function in the question as the following: \begin{equation} max(0,\sin{t}) = \frac{1}{2}\sin{t}+\frac{1}{2}|\sin{t}| \end{equation} $|\sin{t}|$ can be transformed to $\frac{e^{\pi s} + 1}{e^{\pi s} - 1} \cdot \frac{1}{s^2 + 1}$ using the "Rule of Periodicity". (See the References below if you don't know what that is).

As $\sin{t}$ is knowingly $\frac{1}{s^2+1}$ therefore the whole function transforms to:

\begin{equation} \frac{e^{\pi s}}{e^{\pi s} - 1}\cdot \frac{1}{s^2+1} \end{equation}

REF1: https://www.intmath.com/laplace-transformation/5-transform-periodic-function.php

REF2: https://books.google.at/books?id=NRin1j9FESsC&pg=PA28&lpg=PA28&dq=laplace+%22max+0+sin%22&source=bl&ots=71ovKnZh2U&sig=ACfU3U3VFdMkKtPkq1B9EDlsnY4DF5RH2w&hl=de&sa=X&ved=2ahUKEwiqrv2W5efpAhVi-yoKHWJdDSAQ6AEwAnoECAoQAQ#v=onepage&q=laplace%20%22max%200%20sin%22&f=false

Answer 2

We want the integral \begin{align} I = \int_0^\infty e^{-tx} \max\{\sin(x),0\} dx, \end{align} note that \begin{align} \max\{\sin(x),0\} &= \begin{cases} \sin(x), & 2n\pi\leq x\leq (2n+1)\pi\\0, &(2n+1)\pi \leq x \leq (2n+2)\end{cases}, \end{align} for each natural $n$. We can (ignoring convergence issues) split the integral up into a bunch of intervals \begin{align} I = \sum_{n=0}^\infty \int_{2n\pi}^{(2n+1)\pi} e^{-tx} \sin(x) dx, \end{align} since the bits between $(2n+1)\pi$ and $(2n+2)\pi$ are zero. The integral in each interval is easy if you express $\sin$ as a sum of exponentials and the infinite sum is just a geometric series once you've done the integral in each little interval.

Answer 3

Here's an attempt using your observation that $max(0, \sin(t))=0$ for $[(2k+1) \pi, (2k+2) \pi]$, where $k=0,1,2,...$.

This means that we can split the Laplace transform into a discrete sum of integrals \begin{equation} \mathcal{I} \equiv \mathcal{L}[max(0,\sin(t))] = \int_0^{\infty}max(0,\sin(t)) e^{-st} dt = \sum_{k=0}^{\infty} \mathcal{I}_k, \end{equation}

where \begin{equation} \mathcal{I}_k = \int_{2k\pi}^{(2k+1) \pi } \sin(t) e^{-st} dt. \end{equation}

Then we have (you should double check this)

\begin{equation} \mathcal{I}_k = \frac{1}{2} \left( e^{-\pi s} -1 \right) e^{-2k\pi s}. \end{equation}

Finally using the formula for geometric progression we have that \begin{equation} \mathcal{I}(s) = \frac{1}{2} \left(e^{-\pi s} -1 \right) \frac{1}{1 - e^{-2\pi s}}. \end{equation}

Hopefully I didn't make any mistakes in writing this.