Laplace Inverse.

Question

Let $\mathcal{L}[f(x)](s)$ be the Laplace Transformation of a function $f(x).$

I know $\mathcal{L}^{-1}[\frac{1}{s+10}] = \frac{1}{s-(-10)} = e^{-10x}$

But how would I calculate $\mathcal{L}^{-1}$ for the expression $\frac{1}{s+10}\mathcal{L}[f(x)]?$ , where $f(x)$ is arbitary.

Now I'm confused with how to proceed further. I don't think I can take the fraction inside and write this: $\mathcal{L}[\frac{f(x)}{s+10}]$. It feels plain wrong.

So I used the definition and came to this point: $ \mathcal{L}^{-1} [\int_{0}^{\infty}\frac{e^{-sx}f(x)}{s+10} dx]$

Does this have anything to do with convolution?

Answer 1

$$\mathcal{L}[\frac{f(x)}{s+10}]$$ First you should only have functions of $s$ not $x$. And you take inverse Laplace Transform not laplace Transform : $$\mathcal{L^{-1}}[\frac{F(s)}{s+10}]$$ Use the convolution theorem. $$\mathcal{L^{-1}}\left (\frac{F(s)}{s+10}\right)= \int_0^t e^{-10(t-\tau)}f(\tau) d\tau$$ Where $F(s)$ is the Laplace Transform of $f(x)$. $F(s)$ is not given ? Then keep the integral form.