# Double integral with infinite limits

• A
• NotEuler
In summary, the conversation discusses a conjecture about the equivalence of two double integrals, one involving a function ##f(y)## and the other involving the function ##xf(x)##. The conversation provides a counterexample to the conjecture and discusses possible conditions for the conjecture to hold, as well as proposing a proof for its validity.
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?

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

PeroK said:
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.

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 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.

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...

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

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

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.
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.

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?

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

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

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?

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.

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?
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 ;-)).

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

## 1. What is a double integral with infinite limits?

A double integral with infinite limits is a type of integral that involves calculating the area between a function and the x- and y-axes over an infinite region. This means that the limits of integration for both the x and y variables are infinity.

## 2. How is a double integral with infinite limits different from a regular double integral?

The main difference between a double integral with infinite limits and a regular double integral is the range of integration. In a regular double integral, the limits of integration are finite values, while in a double integral with infinite limits, the limits are infinity. This makes the calculation more complex and requires special techniques to solve.

## 3. What are some common applications of double integrals with infinite limits?

Double integrals with infinite limits are commonly used in physics, engineering, and economics to calculate the total area under a curve or surface that extends to infinity. They are also used in probability and statistics to calculate the probability of an event occurring within a certain range.

## 4. How do you solve a double integral with infinite limits?

Solving a double integral with infinite limits involves using special techniques such as improper integration, polar coordinates, or change of variables. It is important to carefully set up the integral and understand the properties of the function being integrated in order to find the correct solution.

## 5. Can a double integral with infinite limits have a finite value?

Yes, a double integral with infinite limits can have a finite value. This occurs when the function being integrated approaches zero as the limits of integration approach infinity. In this case, the integral can be evaluated using improper integration techniques to find a finite result.

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