Laplace Transform of an integral - Convolution Theorem not feasible

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.

Thanks

• Change the order of integration (if that's justifiable, which it probably is). Oct 31, 2019 at 16:53
• 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. Oct 31, 2019 at 16:56
• No, I mean $$\int_0^{\infty} e^{-st}\int_0^{t_0} K(t-x)f(x)\,dx\,dt = \int_0^{t_0} f(x)\int_0^{\infty} K(t-x)e^{-st}\,dt\,dx\,.$$ Oct 31, 2019 at 16:59
• 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! Nov 1, 2019 at 16:22

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

The Laplace Transform of

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

is

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

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

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

• 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. Nov 1, 2019 at 16:42
• 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. Nov 1, 2019 at 17:13
• 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 $t_0$. Nov 1, 2019 at 18:20
• 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 $f(t) = \sin(\omega_0 t)$). Nov 1, 2019 at 18:23
• @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 $(t_0,t)$. Nov 1, 2019 at 18:30