# Laplace transform of convolution with modified limits

I have an expression such as $\int_0^{x+l}y(z)g(x-z) dz$ and I want to evaluate its Laplace transform w.r.t $x$ in terms of the Laplace transform of $y(x)$. I know that I can substitute $t=x+l$, and coerce it into the standard from to get the Laplace transform w.r.t $t$, but I need its transform w.r.t $x$.

Motivation:

I would like to solve an integral equation: $y(x) = f(x) + \int_0^{x+l} y(z) g(x-z) dz$.

If the integral limit had been to $x$, we would have had $y(x) = f(x) + \int_0^{x} y(z) g(x-z) dz$. This leads to $Y(s) = \frac{F(s)}{1+K(s)}$.

• May I ask what made this equation to you? – mrs Jun 28 '12 at 7:48
• @BabakSorouh, as you may guess, I have a certain process where I am trying to calculate some quantity along an axis, the $f(x)$ is an external component and $y(z)g(x-z)$ are local interactions. The applications are related to my research, but it could occur anywhere from EM fields, heat conduction, or chemical and biological processes. – highBandWidth Jun 28 '12 at 17:00
• eqworld.ipmnet.ru/en/solutions/ie/ie0217.pdf might be helpful. – doraemonpaul Jul 11 '12 at 1:15

## 1 Answer

Hint:

$y(x)=f(x)+\int_0^{x+l}y(z)g(x-z)~dz$

$y(x)=f(x)+\int_x^{-l}y(x-t)g(t)~d(x-t)$

$y(x)=f(x)-\int_x^{-l}y(x-t)g(t)~dt$

$y(x)=f(x)+\int_{-l}^xy(x-t)g(t)~dt$