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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!

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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!

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  • $\begingroup$ Many many thanks. $\endgroup$ – Quiet_waters Jul 23 '20 at 16:30

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