# Laplace Transform of this function

Find $L\{F(t)\}$ if

$$F(t) = \begin{cases} \sin t & \text{between }0 < t < \pi \\ 0 & \text{between } \pi < t < 2\pi \end{cases}$$

Really stumped by this one. Please can you work this out?

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I'm stumped too... – David Mitra Jan 20 '13 at 2:10
Do you wish to find the Laplace transform of $F$? – David Mitra Jan 20 '13 at 2:11
yes, I do! The laplace transform of F(t) – user58944 Jan 20 '13 at 2:17
You need to specify $F$ for all $t\ge 0$, or say it is periodic. As stated problem is incomplete. – Maesumi Jan 20 '13 at 6:04

I presume the function is periodic. Have you tried considering

$$\mathcal{L}[f(t)] = \int_{0}^{\infty} e^{-pt}f(t) dt = \sum_{n=0}^{\infty}\int_{n\pi}^{(n+1)\pi} e^{-pt}f(t) = \sum_{n=0}^{\infty}\int_{2n\pi}^{(2n+1)\pi} e^{-pt}\sin(t)$$

But this is just

$$\sum_{n=0}^{\infty} \frac{(e^{p\pi} + 1)(e^{\pi(-(2n+1))p})(p\sin(2\pi n) + \cos (2 \pi n))}{p^2 + 1}$$

Since $\sin (2\pi n) = 0$ and $\cos (2 \pi n) = 1$, this becomes

$$\sum_{n=0}^{\infty} \frac{(e^{p\pi} + 1)(e^{\pi(-(2n+1))p})}{p^2 + 1}$$

And in fact we can pull out terms involving $p$ to obtain

$$\frac{(e^{\pi} + 1)e^{-p\pi}}{p^2 +1}\sum_{n=0}^{\infty} e^{-2n p \pi}$$

And we evaluate the sum to obtain

$$\frac{(e^{\pi} + 1)e^{-p\pi}}{p^2 +1}\left(\frac{e^{2\pi p}}{e^{2 \pi p} - 1} \right) = \frac{(e^{\pi p} +1)\mbox{csch} (\pi p)}{2(p^2 + 1)}$$

If this function is not periodic, but $0$ for $t> 2\pi$, then

$$\mathcal{L}[f(t)] = \int_{0}^{\infty} e^{-pt}f(t) dt = \int_{0}^{\pi} e^{-pt}\sin(t) dt = \frac{e^{-p\pi} + 1}{p^2 + 1}$$

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Thank you for your help, I think the answer is expected in the s domain – user58944 Jan 20 '13 at 2:13
wow! Thank you for taking the time to do all of this. You're a great help! What is the csch though? – user58944 Jan 20 '13 at 2:32
The function $\mbox{csch}$ is the hyperbolic cosecant. It's defined by $\mbox{csch}z = \frac{2}{e^{z} - e^{-z}}$ – Isaac Solomon Jan 20 '13 at 2:33
OK, I think this might be a more complex solution than I should have! Thanks for guiding me through this anyway. I'll try and get it into the form of s – user58944 Jan 20 '13 at 2:39
This is in the form of $s$, I've just written $p$ for $s$, because it's what I'm used to. – Isaac Solomon Jan 20 '13 at 2:40