Double integral with infinite limits

PeroK said:

Isn't ##f(x) = \dfrac 1 {x^3}## a counterexample to your conjecture?

NotEuler said:

It doesn't look like the integral ##\int_0^{\infty } f(y) \, dy## is finite with that function? So the conditions are not met.

But actually ##f(y) = \frac{1}{(y+1)^2}## does seem to be a counterexample: the inner integral converges, but the outer one does not. So I suppose the conjecture isn't true in general. Could it still be true if both integrals converge? At least numerical examples suggest so, and a similar construction using infinite sums seems to give a similar answer to my guess.

NotEuler said:

I now need to calculate (or simplify) the double integral:

$$\int_0^{\infty } \left(\int_x^{\infty } f(y) \, dy\right) \, dx$$

Now, I have a conjecture that this can be written as

$$\int_0^{\infty } x f(x) \, dx$$

renormalize said:

Replace ##f## by ##F'## (where ##F## is the antiderivative of ##f##) in both integrals, integrate-by-parts in the second integral, and then compare it to the first.

NotEuler said:

I have the following problem and am almost sure of the answer but can't quite prove it:

##f(y)## is nonnegative, and I know that ##\int_0^{\infty } f(y) \, dy## is finite.

I now need to calculate (or simplify) the double integral:

$$\int_0^{\infty } \left(\int_x^{\infty } f(y) \, dy\right) \, dx$$

Now, I have a conjecture that this can be written as

$$\int_0^{\infty } x f(x) \, dx$$How could I go about proving such a thing?

pasmith said:

Let [itex]R > 0[/itex] and integrate [itex]xf(x)[/itex] by parts: [tex]
\begin{split}
\int_0^R xf(x)\,dx &= \left[ x\int_0^x f(y)\,dy\right]_0^R - \int_0^R \int_0^x f(y)\,dy\,dx \\
&= \int_0^R 1\,dx \int_0^R f(y)\,dy - \int_0^R \int_0^x f(y)\,dy\,dx \\
&= \int_0^R \left( \int_0^R f(y)\,dy - \int_0^x f(y)\,dy\right)\,dx \\
&= \int_0^R \int_x^R f(y)\,dy\,dx. \end{split}[/tex] Now take the limit [itex]R \to \infty[/itex]. (Note that not all of these steps are valid if you start with [itex]R = \infty[/itex].)

NotEuler said:

I have the following problem and am almost sure of the answer but can't quite prove it:

##f(y)## is nonnegative, and I know that ##\int_0^{\infty } f(y) \, dy## is finite.

I now need to calculate (or simplify) the double integral:

$$\int_0^{\infty } \left(\int_x^{\infty } f(y) \, dy\right) \, dx$$

Now, I have a conjecture that this can be written as

$$\int_0^{\infty } x f(x) \, dx$$How could I go about proving such a thing?

NotEuler said:

Ah yes, I think I see at least partly. If I write ##F(x)=-\int_x^{\infty } f(y) \, dy##, then ##\int_0^{\infty } x f(x) \, dx = \int_0^{\infty } x F'(x) \, dx = [xF(x)]_0^∞ - \int_0^{\infty } F(x) \, dx##.

##[xF(x)]_0^∞## goes to 0 at the lower limit if ##F(x)## converges, but I am not quite sure how I can justify it going to zero at the upper limit. On the other hand, the last term ##- \int_0^{\infty } F(x) \, dx = \int_0^{\infty } \left(\int_x^{\infty } f(y) \, dy\right) \, dx ##. So, apart from the upper limit of ##[xF(x)]_0^∞## I think I get this. Many thanks!I had another idea, but I am not sure what the conditions are for it to be valid. The integration limits specify a triangle to the right of the y-axis and above the liny y=x. So can I then change the order of integration as follows:
##\int_0^{\infty } \left(\int_x^{\infty } f(y) \, dy\right) \, dx = \int_0^{\infty } \left(\int_0^{y } f(y) \, dx\right) \, dy = \int_0^{\infty }\left([xf(y)]_0^y \right)\, dy = \int_0^{\infty }\left(yf(y) \right)\, dy = \int_0^{\infty }xf(x) \, dx##

The new integration limits seem to specify exactly the same area, but again I am not sure exactly how to justify this.