How to find the inverse Laplace transform of $F(s) = {se^{-\frac{s}{2}} + \pi e^{-s} \over (s^2 + \pi^2)}$? How to find the inverse Laplace transform of
$$
F(s) = {se^{-\frac{s}{2}} + \pi e^{-s} \over (s^2 + \pi^2)}
$$
Can't use partial fractions, it isn't in any of the standard forms either. I have no clue how to approach this question. Any hints please?
 A: Given that :
$$L^{-1}\left\{\frac{e^{-\frac{s}{2}}s+\pi e^{-s}}{s^2+\pi ^2}\right\}$$
Expand :
$$=L^{-1}\left\{\frac{e^{-\frac{s}{2}}s}{s^2+\pi ^2}+\frac{\pi e^{-s}}{s^2+\pi ^2}\right\}$$
$$\mathrm{Use\:the\:linearity\:property\:of\:Inverse\:Laplace\:Transform:}$$
$$\mathrm{For\:functions\:}f\left(s\right),\:g\left(s\right)\mathrm{\:and\:constants\:}a,\:b:$$
$$L^{-1}\left\{a\cdot f\left(s\right)+b\cdot g\left(s\right)\right\}=a\cdot L^{-1}\left\{f\left(s\right)\right\}+b\cdot L^{-1}\left\{g\left(s\right)\right\}$$
We will get :
$$=L^{-1}\left\{\frac{se^{-\frac{1}{2}s}}{s^2+\pi ^2}\right\}+\pi L^{-1}\left\{\frac{e^{-s}}{s^2+\pi ^2}\right\}$$
Solve first part :
$$L^{-1}\left\{\frac{se^{-\frac{1}{2}s}}{s^2+\pi ^2}\right\}$$
Notice  that :
$$\mathrm{Apply\:inverse\:transform\:rule:\quad if\:}L^{-1}\left\{F\left(s\right)\right\}=f\left(t\right)$$
$$\mathrm{\:then}\:L^{-1}\left\{e^{-as}F\left(s\right)\right\}=H\left(t-a\right)f\left(t-a\right))$$
$$\mathrm{Where\:}\text{H}\left(t\right)\mathrm{\:is\:Heaviside\:step\:function}$$
$$\mathrm{For\:}\frac{se^{-\frac{1}{2}s}}{s^2+\pi ^2}:\quad F\left(s\right)=\frac{s}{s^2+\pi ^2},\:\quad a=\frac{1}{2}$$
We will get :
$$=\text{H}\left(t-\frac{1}{2}\right)L^{-1}\left\{\frac{s}{s^2+\pi ^2}\right\}\left(t-\frac{1}{2}\right)$$
$$=\text{H}\left(t-\frac{1}{2}\right)\cos \left(\sqrt{\pi ^2}\left(t-\frac{1}{2}\right)\right))$$
Solve the second part :
$$L^{-1}\left\{\frac{e^{-s}}{s^2+\pi ^2}\right\}$$
Notice  that :
$$\mathrm{Apply\:inverse\:transform\:rule:\quad if\:}L^{-1}\left\{F\left(s\right)\right\}=f\left(t\right)$$
$$\mathrm{\:then}\:L^{-1}\left\{e^{-as}F\left(s\right)\right\}=H\left(t-a\right)f\left(t-a\right))$$
$$\mathrm{Where\:}\text{H}\left(t\right)\mathrm{\:is\:Heaviside\:step\:function}$$
$$\mathrm{For\:}\frac{e^{-s}}{s^2+\pi ^2}:\quad F\left(s\right)=\frac{1}{s^2+\pi ^2},\:\quad a=1$$
We will get :
$$=\text{H}\left(t-1\right)L^{-1}\left\{\frac{1}{s^2+\pi ^2}\right\}\left(t-1\right)$$
$$\text{H}\left(t-1\right)\frac{1}{\sqrt{\pi ^2}}\sin \left(\sqrt{\pi ^2}\left(t-1\right)\right)$$
Combine first and second part :
$$=\text{H}\left(t-\frac{1}{2}\right)\cos \left(\sqrt{\pi ^2}\left(t-\frac{1}{2}\right)\right)+\pi \text{H}\left(t-1\right)\frac{1}{\sqrt{\pi ^2}}\sin \left(\sqrt{\pi ^2}\left(t-1\right)\right)$$
Refine and we will get :
$$=\text{H}\left(t-\frac{1}{2}\right)\cos \left(\sqrt{\pi ^2}\left(t-\frac{1}{2}\right)\right)+\text{H}\left(t-1\right)\sin \left(\sqrt{\pi ^2}\left(t-1\right)\right)$$
A: From the property of Laplace transform: time shifting,
$$
\boxed{
\mathcal{L}\{g(t-a)\theta(t-a)\}(s) = e^{-as}G(s)}
$$
where $a>0$, $\theta(t)$ is the Heaviside step function. Let
$$
F(s) = e^{-\frac{s}{2}}\underbrace{\frac{s}{s^2+\pi^2}}_{G_1(s)} + e^{-s}\underbrace{\frac{\pi}{s^2+\pi^2}}_{G_2}
$$
Then
$$
\mathcal{L}\left\{g_1\left(t-\frac{1}{2}\right) \theta\left(t-\frac{1}{2}\right) + g_2(t-1)\theta(t-1)\right\}(s) =e^{-\frac{s}{2}} G_1(s) + e^{-s}G_2(s)
$$
Note that $$
\mathcal{L}\{\cos \omega t\}(s) = \frac{p}{p^2+\omega^2},\quad \mathcal{L}\{\sin \omega t\}(s) = \frac{\omega}{p^2+\omega^2}
$$
So you can obtain $g_1(t)$ and $g_2(t)$:
$$
g_1(t) = \mathcal{L}^{-1}\left\{\frac{s}{s^2+\pi^2}\right\}=\cos \pi t,\quad g_2(t) = \mathcal{L}^{-1}\left\{\frac{\pi}{s^2+\pi^2}\right\}=\sin \pi t
$$
Thus, the inverse Laplace transform of $F(s)$ is:
$$
\boxed{
\begin{align}
\mathcal{L}^{-1}\{F(s)\}&= \mathcal{L}^{-1}\left\{\frac{s e^{-\frac{s}{2}}+\pi e^{-s}}{s^2+\pi^2}\right\}\\[.2cm]
&=g_1\left(t-\frac{1}{2}\right) \theta\left(t-\frac{1}{2}\right) + g_2(t-1)\theta(t-1)\\[.2cm]
&=\boxed{\left[\theta\left(t-\frac{1}{2} \right)- \theta(t-1)\right]
\sin \pi t} \end{align}
}
$$
