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How to find the Laplace transform of $ |\sin (t)| $?

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up vote 8 down vote accepted

$$ \mathcal{L}_s\left(\left| \sin t \right|\right) = \int_0^\infty \left|\sin(t)\right|\mathrm{e}^{-s t} \mathrm{d} t = \sum_{n=0}^\infty \int_{\pi n}^{\pi (n+1)} \left|\sin(t)\right|\mathrm{e}^{-s t} \mathrm{d} t = \sum_{n=0}^\infty \mathrm{e}^{-s \pi n} \int_{0}^{\pi} \sin(t)\mathrm{e}^{-s t} \mathrm{d} t = \frac{1}{1-\exp(-\pi s)} \int_{0}^{\pi} \sin(t)\mathrm{e}^{-s t} \mathrm{d} t $$ The latter integrate is easy to evaluate as follows: $$ \int_{0}^{\pi} \sin(t)\mathrm{e}^{-s t} \mathrm{d} t = \Im \int_{0}^{\pi} \mathrm{e}^{i t -s t} \mathrm{d} t = \Im\left(\frac{\exp(i \pi - \pi s) - 1}{i - s}\right) = \frac{1+\exp(-\pi s)}{1+s^2} $$ Thus $$ \mathcal{L}_s\left(\left| \sin t \right|\right) = \frac{1+\exp(-\pi s)}{1+s^2} \cdot \frac{1}{1-\exp(-\pi s)} $$

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+1. Pretty good !!!. – Felix Marin Jul 22 '14 at 5:41

Here is a short derivation. First note that $|\sin t|$ has a period of $\pi$. Then $$\mathcal{L}(|\sin t|)$$ $$= \frac{1}{1 - e^{-\pi s}}\quad \int_0^{\pi} e^{-st} |\sin t|\, dt$$ $$= \frac{1}{1 - e^{-\pi s}}\quad \int_0^{\pi} e^{-st} \sin t \,dt$$ $$= \frac{1}{1 - e^{-\pi s}}\quad \left. \left(-e^{-st} \frac{s \,\sin t + \cos t}{s^2 + 1} \quad \right|^{\pi}_0 \right )$$ $$= \frac{1}{1 - e^{-\pi s}}\quad \left ( \frac{e^{-\pi s}}{s^2 + 1}\;\; + \frac{1}{s^2 + 1} \right )$$ $$= \frac{1 + e^{-\pi s}}{1 - e^{-\pi s}} \quad \cdot \frac{1}{s^2 + 1}$$

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Here is another way:

First, I make a variable change so I need only work with integers

$$\int_{0}^{\infty}e^{-st}|\sin(t)|dt=\pi\int_{0}^{\infty}e^{-\pi st}|\sin(\pi t)|dt.$$ Now, lets remove the absolute values. The above is

$$\pi\sum_{n=0}^{\infty}\left(\int_{2n}^{2n+1}e^{-\pi st}\sin(\pi t)dt-\int_{2n+1}^{2n+2}e^{-\pi st}\sin(\pi t)dt\right) $$

Now, use the fact that $$\pi\int e^{-\pi st}\sin(\pi t)dt=-\frac{e^{-\pi st}\left(s\sin(\pi t)+\cos(\pi t)\right)}{s^{2}+1} +C$$ to piece together a solution.

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There is another simple way using step function as follows.
The period is $\pi$ and absolute ($\sin x$) can be written as $$\sin(x)(u(x)-u(x-\pi))=\sin(x)u(x) - \sin(x)u(x-\pi)$$ and $\sin(x)=\sin(\pi-x)= - \sin(x-\pi)$ because $\sin$ is a odd function.
$f(x) = \sin(x)u(x) + \sin(x-\pi)u(x-\pi)$ now we can transform $f(s) = \frac1{1+s^2} + \frac{e^{-s\pi}}{s^2+1}$ using $f(x-a)u(x-a)$ ' s lap trans $e^{-sa}\mathcal L(f(x))$.

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