# 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!

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, 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!

• Many many thanks. – Quiet_waters Jul 23 '20 at 16:30