Understanding the double integral $\int_0^\infty\int_0^t f(x)g(t-x)dxdt$ I am dealing with the integral $\int_0^\infty\int_0^t f(x)g(t-x)dxdt$ and $g(x)=0$ if $x<0$.
I need to arrive to $\int_0^\infty f(x)dx\int_0^\infty g(t)dt$.
Using Fubini, I have:
$$\int_0^\infty\int_0^t f(x)g(t-x)dxdt=$$
$$\int_0^t f(x)\int_0^\infty g(t-x)dtdx=$$
$$\int_0^t f(x)\int_{-x}^\infty g(t)dtdx=$$
$$\int_0^t f(x)dx\int_{0}^\infty g(t)dt.$$
How can I proceed to get limit $\infty$ at first integral?
Many thanks!
 A: Well, you applied Fubini's theorem incorrectly.  We have
$$\begin{align}
\int_0^\infty \int_0^t f(x) g(t-x)\,dx\,dt&=\int_0^\infty \int_x^\infty f(x) g(t-x)\,dt\,dx\\\\
&=\int_0^\infty f(x)\int_x^\infty  g(t-x)\,dt\,dx\\\\
&=\int_0^\infty f(x)\int_0^\infty  g(t)\,dt\,dx\\\\
&=\left(\int_0^\infty f(x)\,dx\right)\left(\int_0^\infty g(x)\,dx\right)
\end{align}$$
as was to be shown!

NOTE:
Alternatively, we could have exploited the fact that $g(x)=0$ for $x<0$.  Then noting that $g(t-x)=0$ for $t<x$, we see that
$$\begin{align}
\int_0^\infty f(x)g(t-x)\,dx&=\int_0^t f(x) g(t-x)\,dx+\int_t^\infty f(x)\underbrace{g(t-x)}_{=0}\,dx\\\\
&=\int_0^t f(x) g(t-x)\,dx\tag1
\end{align}$$
Using $(1)$, we find that
$$\begin{align}
\int_0^\infty \int_0^t f(x) g(t-x)\,dx\,dt&=\int_0^\infty \int_0^\infty f(x) g(t-x)\,dx\,dt\\\\
&=\int_0^\infty f(x) \int_0^\infty g(t-x)\,dt\,dx\\\\
&=\int_0^\infty f(x) \int_{-x}^\infty g(t)\,dt\,dx\\\\
&=\int_0^\infty f(x) \int_0^\infty g(t)\,dt\,dx\\\\
&=\left(\int_0^\infty f(x)\,dx\right)\left(\int_0^\infty g(x)\,dx\right)
\end{align}$$
as expected!
