MHB Rieman inversion formula in Laplace transform

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SUMMARY

The discussion centers on applying the Riemann inversion formula to compute the inverse Laplace transform of the function $$F(s) = \frac{e^{-5s}}{(s + 4)^{2}}$$. The integral $$\frac{1}{2i\pi}*\int_{8-i\infty}^{8+i\infty}\frac{e^{s(t-5)}}{(s+4)^2}ds$$ is evaluated to find the result as a function of $$t$$. Participants suggest using residue calculus or properties of the Laplace transform, such as the Gamma function, to solve the problem effectively.

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  • Knowledge of residue calculus in complex analysis
  • Basic concepts of the Gamma function and its applications
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lucad93
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Hello everybody! I'm sorry if it's not the right section to post in. I'm trying to solve this exercise:
$$\frac{1}{2i\pi}*\int_{8-i\infty}^{8+i\infty}\frac{e^{s(t-5)}}{(s+4)^2}ds$$
The request is to find the result in function of $$t$$
I know i must use the Riemann inversion formula, and so the request would be to anti-transform this $$\frac{e^{-5s}}{(s+4)^{2}}$$. First question: Am I right? I haven't found a transform that fit with this, how can I do?
ThankYou! :)
 
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Hi lucad93,

Welcome! (Smile) What you're trying to find here is the inverse Laplace transform of the function

$$F(s) = \frac{1}{(s + 4)^2}$$

at $t-5$. Do you have to compute this using residue calculus, or using certain properties of the Laplace transform (e.g., $\mathcal{L}(t^a)(s) = \Gamma(a+1)/s^{a+1}, a > -1$, etc.)?
 

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