# Find L{F(t)} if F(t)={(sin t, text(between ) 0<t<pi),(0,text(between ) pi<t<2pi):}

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

Please can you work this out?
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ehedem26
I presume the function is periodic. Have you tried considering
$\mathcal{L}\left[f\left(t\right)\right]={\int }_{0}^{\mathrm{\infty }}{e}^{-pt}f\left(t\right)dt=\sum _{n=0}^{\mathrm{\infty }}{\int }_{n\pi }^{\left(n+1\right)\pi }{e}^{-pt}f\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{\int }_{2n\pi }^{\left(2n+1\right)\pi }{e}^{-pt}\mathrm{sin}\left(t\right)$
But this is just
$\sum _{n=0}^{\mathrm{\infty }}\frac{\left({e}^{p\pi }+1\right)\left({e}^{\pi \left(-\left(2n+1\right)\right)p}\right)\left(p\mathrm{sin}\left(2\pi n\right)+\mathrm{cos}\left(2\pi n\right)\right)}{{p}^{2}+1}$
Since $\mathrm{sin}\left(2\pi n\right)=0$ and $\mathrm{cos}\left(2\pi n\right)=1$, this becomes
$\sum _{n=0}^{\mathrm{\infty }}\frac{\left({e}^{p\pi }+1\right)\left({e}^{\pi \left(-\left(2n+1\right)\right)p}\right)}{{p}^{2}+1}$
And in fact we can pull out terms involving p to obtain
$\frac{\left({e}^{\pi }+1\right){e}^{-p\pi }}{{p}^{2}+1}\sum _{n=0}^{\mathrm{\infty }}{e}^{-2np\pi }$
And we evaluate the sum to obtain
$\frac{\left({e}^{\pi }+1\right){e}^{-p\pi }}{{p}^{2}+1}\left(\frac{{e}^{2\pi p}}{{e}^{2\pi p}-1}\right)=\frac{\left({e}^{\pi p}+1\right)\text{csch}\left(\pi p\right)}{2\left({p}^{2}+1\right)}$
If this function is not periodic, but 0 for $t>2\pi$, then
$\mathcal{L}\left[f\left(t\right)\right]={\int }_{0}^{\mathrm{\infty }}{e}^{-pt}f\left(t\right)dt={\int }_{0}^{\pi }{e}^{-pt}\mathrm{sin}\left(t\right)dt=\frac{{e}^{-p\pi }+1}{{p}^{2}+1}$