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Find the inverse Laplace transform of the giveb function by using the

convolution theorem.$$F(x) = \frac{s}{(s+1)(s^2+4)}$$

If I use partial fractions I get: $$\frac{s+4}{5(s^2+4)} – \frac{1}{5(x+1)}$$

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which gives me Laplace inverses:

$$\frac{1}{5}(\cos2t + \sin2t) -\frac{1}{5} e^{-t}$$

But the answer is:

$$f(t) = \int^t_0 e^{-(t -\tau)}\cos(2\tau) d\tau$$

How did they get that?

- Convolution product on the linear dual of the polynomial algebra
- Laplace transform of $f(t^2)$
- Translations AND dilations of infinite series
- Convolution is uniformly continuous and bounded
- Convolution of a function with itself
- Evaluating improper integrals using laplace transform
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- Inverse Laplace transform of fraction $F(s) = \large\frac{2s+1}{s^2+9}$
- Sufficient conditions to conclude that $\lim_{a \to 0^{+}} \int_{0}^{\infty} f(x) e^{-ax} \, dx = \int_{0}^{\infty} f(x) \, dx$
- Inequality for the p norm of a convolution

Related techniques (I), (II). Using the fact about the Laplace transform $L$ that

$$ L(f*g)=L(f)L(g)=F(s)G(s)\implies (f*g)(t)=L^{-1}(F(s)G(s)) .$$

In our case, given $ H(s)=\frac{1}{(s+1)}\frac{s}{(s^2+4)}$

$$F(s)=\frac{1}{s+1}\implies f(t)=e^{-t},\quad G(s)=\frac{s}{s^2+4}\implies g(t)=\cos(2t).$$

Now, you use the convolution as

$$ h(t) = \int_{0}^{t} e^{-(t-\tau)}\cos(2\tau) d\tau . $$

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