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: $$ \mathcal{L}\{f(t)\}(s)=\int_0^\infty e^{-st}f(t)dt $$


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

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.


$$\int_0^\infty \int_0^{\infty}e^{-st}x(t-\tau)d\tau dt = \int_0^\infty \int_0^{\infty}e^{-st}x(t-\tau) \frac{e^{-\tau t}}{e^{-\tau t}}d\tau dt $$ 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. $$\int_0^{\infty}x(t-\tau)d\tau = 1 \ast x(t) $$

  • 1
    $\begingroup$ The theory of distributions allows to say that $\mathcal{L}\left\{\int_0^{\infty}x(t-\tau)d\tau\right\}(s) = \int_0^\infty e^{-\tau s} X(s)d\tau$ $\endgroup$
    – reuns
    Commented Jun 27, 2020 at 19:46
  • $\begingroup$ @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 ! $\endgroup$
    – Gadorcha
    Commented Jun 27, 2020 at 21:56

1 Answer 1


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: $$\int_0^{+\infty}x(t-\tau)d\tau = \frac{1}{s}X(s)$$.

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    $\begingroup$ So $\mathcal{L}\{ x(t-\tau) \}(s) = \mathcal{L}\{ x(t) \}(s) = X(s)$ ? $\endgroup$
    – Gadorcha
    Commented Jun 27, 2020 at 23:12
  • $\begingroup$ No. $\mathcal{L}\{x(t-t_0)\} = X(s)e^{-st_0}$. It is a known property of the laplace transform. $\endgroup$
    – sted
    Commented Jun 27, 2020 at 23:50
  • $\begingroup$ 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! $\endgroup$
    – Gadorcha
    Commented Jun 28, 2020 at 0:37

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