I need to evaluate the Laplace transform of the following integral:

$$ \phi(t)= \int_0^{t_0} K(t-x)f(x)dx $$

Note that the constant upper limit of the integral is different from the time variable, so that straightforward application of the Convolution Theorem is not feasible. I have an explicit functional form of the Laplace Transform of the Kernel $K()$, but not of the function $f()$ or its transform.

Any tips, pointers, hints or even answers would be deeply appreciated.


  • $\begingroup$ Change the order of integration (if that's justifiable, which it probably is). $\endgroup$ Oct 31, 2019 at 16:53
  • $\begingroup$ I take it you mean something like $$ \phi(t)= \int_0^{t_0} f(t-x)K(x)dx $$, but I still don't see how it would result in a tractable form, since the CT would still be inapplicable. $\endgroup$ Oct 31, 2019 at 16:56
  • 1
    $\begingroup$ No, I mean $$\int_0^{\infty} e^{-st}\int_0^{t_0} K(t-x)f(x)\,dx\,dt = \int_0^{t_0} f(x)\int_0^{\infty} K(t-x)e^{-st}\,dt\,dx\,.$$ $\endgroup$ Oct 31, 2019 at 16:59
  • $\begingroup$ Doh! Yes, my bad! I've posted an answer below. Please review it for correctness - I forgot to mention that key point that $f(x)$ is zero outside of $[0,t_0]$. I do have a follow up question that could be a bit trickier. Thanks for your help! $\endgroup$ Nov 1, 2019 at 16:22

1 Answer 1


Based on the excellent suggestion by Daniel Fischer above, I've attempted a solution below:

The Laplace Transform of

$$\phi(t)=\int_0^{t_0} K(t-x)f(x)dx$$


$$ L[\phi(t)]= \int_0^\infty e^{-st}\int_0^{t_0} K(t-x)f(x)dxdt =\int_0^{t_0}\Biggl[ \int_0^\infty e^{-st}K(t-x)dt\Biggr]f(x)dx $$

With the variable shift $y=t-x$, the inner integral can be written as

$$ \Lambda(s) = e^{-sx}\int_0^\infty e^{-sy}K(y)dy=e^{-sx}\bar K(s) $$

where the bar indicates a Laplace Transform. Substitution into the outer integral yields

$$ L[\phi(t)]=\bar K(s) \int_0^{t_0}e^{-sx}f(x)dx $$

Since $f(t)$ is zero outside of the interval $[0,t_0]$, the upper limit of the outer integral can be replaced by $t$, so that

$$ L[\phi(t)]=\bar K(s) \int_0^te^{-sx}f(x)dx=\bar K(s) \bar F(s) $$

  • $\begingroup$ When you substitute $y = t - x$, the lower integral limit changes from $t = 0$ to $y = -x$. If $K$ vanishes on $(-\infty, 0)$ we can of course omit the $[-x,0)$ part of that integral, but then, in conjunction with $f$ vanishing outside $[0,t_0]$, $\phi$ is the ordinary convolution of $K$ and $f$, so you could directly apply the convolution theorem. $\endgroup$ Nov 1, 2019 at 16:42
  • $\begingroup$ In the actual problem that I have, $f(t)$ is a piecewise continuous function over several sub-intervals, so that the actual integral that I want to break up is $$\phi(t)=\int_0^{t} K(t-x)f(x)dx=\phi(t)=\int_0^{t_0} K(t-x)f_A(x)dx +\int_{t_0}^{t} K(t-x)f_B(x)dx $$ My problem is that the Kernel $K(t)$ cannot be expressed in the form $K(t) = P(t,t_0)K(t_0)$ which would allow me to carry over $$\int_0^{t_0} K(t_0-x)f_A(x)dx $$ as a legacy term updated from one sub-interval to the next. $\endgroup$ Nov 1, 2019 at 17:13
  • $\begingroup$ A quick clarification. You wrote that I could directly apply the Convolution Theorem. Wouldn't that the same Laplace Transform as that for $$\phi(t)=\int_0^{t} K(t-x)f(x)dx$$ ? - Note the upper limit is $t$ instead of $t_0$. $\endgroup$ Nov 1, 2019 at 18:20
  • $\begingroup$ The application of the second shifting property doesn't work here. The limits of the integral don't match the step function. It also gives the wrong answer (try it with $K(t) = \exp(s_0 t)$ and $f(t) = \sin(\omega_0 t)$). $\endgroup$ Nov 1, 2019 at 18:23
  • $\begingroup$ @SharatVChandrasekhar If $K(u) = 0$ for $u < 0$ and $f(x) = 0$ for $x \notin [0,t_0]$, then $$\int_0^t K(t-x)f(x)\,dx = \int_0^{t_0} K(t-x)f(x)\,dx.$$ For $t < t_0$ we then have $K(t-x) = 0$ on $(t,t_0)$, and for $t > t_0$ we then have $f(x) = 0$ on $(t_0,t)$. $\endgroup$ Nov 1, 2019 at 18:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .