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Consider the following integral (originating from the product of the Laplace transforms of $f$ and $g$):

$$\int_0^\infty \int_0^\infty f(u)\ g(v) e^{-s(u+v)}\ du\ dv.$$

For this integral, the variables are to be changed from $(u, v)$ to $(u, t)$ so that $v = t - u$.

According to this video (~at 25:00), the result is

$$\int_0^\infty \int_0^t f(u)\ g(t - u) e^{-st}\ du\ dt.$$

The integrand is found easily, but what about the limits? In the video Prof. Mattuck gives a geometric argument, but I wonder how to the limits can be found in a more systematic way, like in the standard substitution rule

$$\int_{g(a)}^{g(b)} f(x)\ dx = \int_a^b f(g(t)) g'(t)\ dt.$$

In particular, where does the $t$ in the upper limit of the inner integral come from?

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2 Answers 2

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Using Fubini in a generous way we can argue as follows: The given double integral can be written as $$J:=\int_0^\infty \left( f(u)e^{-s u}\int_0^\infty g(v)e^{-sv}\ dv\right)\ du\ .$$ The substitution $v:=t-u$ $(u\leq t<\infty)$ transforms the inner integral into $$\int_u^\infty g(t-u) e^{s(u-t)}\ dt\ ,$$ and using Fubini again we obtain $$J=\int_0^\infty\int_u^\infty f(u)g(t-u) e^{-st} dt\ du=\int_B f(u)g(t-u) e^{-st} {\rm d}(u,t)\ ,$$ where $B$ is the the domain $$B:=\{(u,t)\ |\ u\geq 0,\ t\geq u\}=\{(u,t)\ |\ t\geq 0,\ 0\leq u\leq t\}\ .$$ (Note that you need some "geometry" in order to verify this.) Using Fubini with respect to the second representation of $B$, i.e., making $u$ the inner variable, we get $$J=\int_0^\infty\int_0^u f(u)g(t-u) e^{-st} du\ dt\ ,$$ as claimed by Prof. Mattuck.

A second way to arrive at the same result is the following:

(As the given integral is over an unbounded domain we have to be a little careful.)

We are told to replace the given independent variables $(u,v)$ by new variables $(x,t)$ (here $x$ is later to be renamed by $u$), using the formulas $${\bf w}:\quad(x,t)\mapsto \cases{u:= x\cr v:=t-x\cr}\ .$$

enter image description here

In this way the triangle $\Delta:=\{(x,t)\ |\ 0\leq t\leq T,\ 0\leq x\leq t\}$ in the $(x,t)$-plane is mapped onto the triangle $\Delta':=\{(u,v)\ |\ u\geq 0,\ v\geq0,\ u+v\leq T\}$ in the $(u,v)$-plane. (One can verify this by checking where the vertices go.) The Jacobian of this map is of absolute value $1$. Therefore one has $$\eqalign{\int\nolimits_{\Delta'} f(u)g(v)e^{-s(u+v)}\ {\rm d}(u,v)&= \int\nolimits_\Delta f(x)g(t-x)e^{-s\,t}\ {\rm d}(x,t)\cr &=\int_0^T e^{-st} \int_0^t f(x)g(t-x)\ dx\ dt\ .\cr}$$ On the right hand side the outer integration is with respect to the variable $t$, and the inner integration is over segments $t={\rm const.}$ parallel to the $x$-axis. (At the end one may replace the letter $x$ with $u$, if so desired.)

You will now have to check whether the assumptions made about $f$ and $g$ allow letting $T\to\infty$ in the last formula. If yes, you arrive at Prof. Mattuck's claim.

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Thanks for your answer and the nice picture! What I find a bit hard to understand is how to come up with the reasoning about the triangles... by what rationale do you get from the double integral to the triangles? –  koletenbert Nov 6 '12 at 10:37
One more thought: I think the main point of my question is how to change variables for a double integral (improper or not). I was under the impression that there exists a simple substition formula (like in the single integral case), but it seems the "general" way to do it is to change it to a surface integral (thereby introducing geometric reasoning)... is that correct? –  koletenbert Nov 6 '12 at 10:44
@koletenbert: See my edit. –  Christian Blatter Nov 6 '12 at 11:34
Thanks! I like the Fubini version, that makes it clearer (to me, at least ;-)) –  koletenbert Nov 6 '12 at 22:58

$\int_0^\infty \int_0^\infty f(u)\ g(v) e^{-s(u+v)}\ du\ dv$=$\int_0^\infty f(u)e^{-su}du\int_0^\infty\ g(v) e^{-sv}\ dv$ then you can consider the Laplace transformation.

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