Double integral with infinite limits

[NotEuler]

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?

 

[PeroK]

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

 

[NotEuler]

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

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]

Here's an example of where my guess works:

##f(y) = \frac{1}{(y+1)^3}##
Then:
##\int_0^{\infty } \left(\int_x^{\infty } f(y) \, dy\right) \, dx = \int_0^{\infty } \ \frac{1}{2 (x+1)^2} \, dx = 1/2 ##

And also:
##\int_0^{\infty } x f(x) \, dx = 1/2##

So my guess gives the correct answer with this example (and many others).

 

[NotEuler]

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.

Now I am wondering if my counterexample really is a counterexample.

##\int_0^{\infty } \left(\int_x^{\infty } f(y) \, dy\right) \, dx## and ##\int_0^{\infty } x f(x) \, dx## both diverge with ##f(y) = \frac{1}{(y+1)^2}##, so I suppose from that perspective the conjecture still holds?

Either ##\int_0^{\infty } \left(\int_x^{\infty } f(y) \, dy\right) \, dx = \int_0^{\infty } x f(x) \, dx## or both of them diverge to infinity and are still 'equal' in that sense.

I'll stop talking to myself now...

 

[renormalize]

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

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

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]

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.

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.

 

[pasmith]

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?

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]

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].)

Thanks pasmith, looks great. One question: Is it always valid to take the limit of the two R:s in the last integral simultaneously? What I mean is that in the integral ##\int_0^{\infty } \left(\int_x^{\infty } f(y) \, dy\right) \, dx## it seems like the two infinities are 'separate'. In your solution the two R:s go to infinity at the same time. Can there be situations where these two are different?

 

[Office_Shredder]

I think you can ignore how the two R's go to infinity at the same time because you're integrand is non-negative. You usually only have to worry about stuff like that when your infinite sum/ integral does not converge absolutely.

 

[vanhees71]

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?

The idea is to change to order of the integrations. Let's rewrite the integral in the physicists' notation first, which is more clear concerning the order of integrations:
$$I=\int_0^{\infty} \mathrm{d} x \int_x^{\infty} \mathrm{d} y f(y).$$
You integrate over the "upper triangle" of the plane ##[0,\infty] \times [0,\infty]##. So changing the order of integrations you get
$$I=\int_0^{\infty} \mathrm{d} y \int_0^y \mathrm{d} x f(y)=\int_0^{\infty} \mathrm{d} y y f(y).$$
Now you can call the integration variable anything you like. So renaming the ##y## to ##x## leads to
$$I=\int_0^{\infty} \mathrm{d} x x f(x).$$
Of course, all this is valid only, if the integrals exist/converge ;-)).

 

[mathman]

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.

In general, the switching can be justified. You can make upper limits finite and let to infinity.