# Laplace transform of the improper integral of a function

i need to solve this ODE using Laplace transform. $$\theta(t)$$ is the Heaviside step function. $$x$$ and $$x'$$ both have a Laplace Transform.

$$x'(t) = \sin(2t)\theta(t-\pi)+\int_0^{\infty}x(t-\tau)d\tau~~~~, ~~~~~x(0)=0$$

I'm having trouble with the improper integral. By definition know that: $$\mathcal{L}\{f(t)\}(s)=\int_0^\infty e^{-st}f(t)dt$$

So

$$\mathcal{L}\left\{\int_0^{\infty}x(t-\tau)d\tau\right\}(s)=\int_0^\infty e^{-st}\int_0^{\infty}x(t-\tau)d\tau dt$$

Since $$e^{-st}$$ does not depend of $$\tau$$

$$\mathcal{L}\left\{\int_0^{\infty}x(t-\tau)d\tau\right\}(s)=\int_0^\infty \int_0^{\infty}e^{-st}x(t-\tau)d\tau dt$$

And using a Laplace Transform Table the equation would be:

$$sX(s) = e^{-\pi s}\frac{2}{s^2+4} + \int_0^\infty \int_0^{\infty}e^{-st}x(t-\tau)d\tau dt ~~~~,~~~~X(s)=\mathcal{L}\{x(t)\}(s)$$

As I said before, I am having a lot of trouble with the improper integral... don't really know what to do so i can find $$X(s)$$ and then do the inverse transform. Really blocked here.

Thanks a lot.

EDIT:

$$\int_0^\infty \int_0^{\infty}e^{-st}x(t-\tau)d\tau dt = \int_0^\infty \int_0^{\infty}e^{-st}x(t-\tau) \frac{e^{-\tau t}}{e^{-\tau t}}d\tau dt$$ If I change switch $$dt$$ and $$d \tau$$ $$\int_0^\infty e^{-\tau t} \int_0^{\infty}e^{-s(t-\tau)}x(t-\tau) e^{-\tau t} dt d\tau ≟ \int_0^\infty e^{-\tau t} \mathcal{L}\{x(t-\tau)\}(s) d\tau$$ Now, how can I take this to an expression with $$X(s)$$?

EDIT2: Someone told me I could use the convolution product. This solves my problem but i'd really like to know if I could've done something with what I tried in the first place. $$\int_0^{\infty}x(t-\tau)d\tau = 1 \ast x(t)$$

• The theory of distributions allows to say that $\mathcal{L}\left\{\int_0^{\infty}x(t-\tau)d\tau\right\}(s) = \int_0^\infty e^{-\tau s} X(s)d\tau$ Commented Jun 27, 2020 at 19:46
• @reuns thanks for your answer! I edited the question since I couldn't demonstrate what you told me, but i'm sure what you said is correct ! Commented Jun 27, 2020 at 21:56

So, you want to calculate the laplace transform of $$\int_0^{+\infty}x(t-\tau)d\tau$$.

As you have written, this is equal to $$\int_0^{+\infty} e^{-st} \int_0^{+\infty}x(t-\tau)d\tau dt$$

We can rearrange $$\int_0^{+\infty} \int_0^{+\infty}e^{-st}x(t-\tau)dtd\tau$$

The term $$\int_0^{+\infty}e^{-st}x(t-\tau)dt$$ is the Laplace transform of $$x(t-\tau)$$, therefore,

$$\int_0^{+\infty} \int_0^{+\infty}e^{-st}x(t-\tau)dtd\tau = \int_0^{+\infty} X(s) e^{-s\tau} d\tau = X(s) \int_0^{+\infty}e^{-s\tau}d\tau$$

And because $$\int_0^{+\infty}e^{-s\tau}d\tau = \frac{1}{s}$$, the answer is: $$\int_0^{+\infty}x(t-\tau)d\tau = \frac{1}{s}X(s)$$.

• So $\mathcal{L}\{ x(t-\tau) \}(s) = \mathcal{L}\{ x(t) \}(s) = X(s)$ ? Commented Jun 27, 2020 at 23:12
• No. $\mathcal{L}\{x(t-t_0)\} = X(s)e^{-st_0}$. It is a known property of the laplace transform.
– sted
Commented Jun 27, 2020 at 23:50
• Oh I see. The table I have says that the Laplace Transform of $\theta(t-t_0) f(t-t_0)$ is $F e^{-t_0s}$. Your expression seems more useful! Commented Jun 28, 2020 at 0:37