Take the 2-minute tour ×  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?  share|improve this
question add comment  2 Answers  up vote 2 down vote accepted
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^{-s