# I'd like to compute the Laplace transform of the following

I'd like to compute the Laplace transform of the following function:
\[f(t)=\begin{cases}0, & \text{if }\ 0 \le t
Could someone please provide some pointers?

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Tionant

Hint: by definition
$$\displaystyle{L}{\left({f}\right)}{\left({s}\right)}\:={\int_{{0}}^{{\infty}}}{f{{\left({t}\right)}}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}={\int_{{0}}^{{\pi}}}{0}\cdot{e}^{{-{s}{t}}}{\left.{d}{t}\right.}+{\int_{{\pi}}^{{\infty}}}{\sin{{t}}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}={\int_{{\pi}}^{{\infty}}}{\sin{{t}}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}$$
Using $$L(f)(s) := \int_0^{\infty} f(t)e^{-st} dt=\int_0^{\pi} 0 \cdot e^{-st} dt+\int_{\pi}^{\infty} \sin te^{-st}dt=\int_{\pi}^{\infty}\sin te^{-st} dt$$ you can arrive at the answer.

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breisgaoyz

You can also use the shifting property of the Laplace transform. Your function is
$$\displaystyle{f{{\left({t}\right)}}}={\sin{{\left({t}\right)}}}{u}{\left({t}-\pi\right)}=-{\sin{{\left({t}-\pi\right)}}}{u}{\left({t}-\pi\right)}={g{{\left({t}-\pi\right)}}}$$
with $$g(t)=-\sin(t)u(t)$$. The transform of $$g(t)$$ is a standard result that can be found in any Laplace transform table:
$$\displaystyle{G}{\left({s}\right)}=-{\frac{{{1}}}{{{s}^{{2}}+{1}}}}$$
and by the shifting property
$$\displaystyle{F}{\left({s}\right)}={e}^{{-\pi{s}}}{G}{\left({s}\right)}=-{\frac{{{e}^{{-\pi{s}}}}}{{{s}^{{2}}+{1}}}}$$