# Find inverse Laplace Transform having s-shifting and t-shifting. (5.3-56)

Please check my work. Did I calculate the following inverse Laplace Transform correctly? Our Laplace Transform is...

$$\mathcal{L}\{f(t)\}=\frac{e^{-\pi s}}{s^2+2s+2}$$

My solution:

Recognizing the exponential factor as a t-shift we pull it to the side. We can also rewrite the denominator as the sum of two squares...

$$\mathcal{L}\{f(t)\}=e^{-\pi s}\cdot \frac{1}{(s+1)^2+1}$$

This reveals that we also have an s-shift. The rest is just pattern matching...

$$\mathcal{L^{-1}}\{f(t)\}=e^{-t}\cdot sin(t-\pi)\cdot u(t-\pi)$$

Therefore...

$$f(t)=-e^{-t}\cdot sin(t)\cdot u(t-\pi)$$

I obtained a slightly different answer then yours. In this post of yours, I defined the Inverse Laplace transform. The poles of this function occur at $s=-1\pm i$. Using the same definition, we have \begin{align} \mathcal{L}^{-1}\{F(s)\} &= \int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{s(t-\pi)}}{s^2+2s+2}ds\\ &=\lim_{s\to -1+i}(s+1-i)\frac{e^{s(t-\pi)}}{s^2+2s+2}+\lim_{s\to -1-i}(s+1+i)\frac{e^{s(t-\pi)}}{s^2+2s+2} \end{align} Again, we have to contend with convergence of the exponential. As in the linked post, we take $s(t-\pi)<0$ so we need $\mathcal{U}(t-\pi)$. \begin{gather} \mathcal{L}^{-1}\{F(s)\} = \frac{e^{-(t-\pi)}e^{i(t-\pi)}}{2i}\mathcal{U}(t-\pi)+\frac{e^{-(t-\pi)}e^{-i(t-\pi)}}{-2i}\mathcal{U}(t-\pi)\\ =e^{-(t-\pi)}\mathcal{U}(t-\pi)\sin(t-\pi) \end{gather}