Proving a property of an integral

In summary, the limit ##\lim_{t\to\infty} \int_t^{\infty } f(x) \, dx## is always 0 if the infinite integral converges and undefined if the integral does not have a finite value. This is due to the fact that the limit can be rewritten as ##\lim_{t\to\infty} \lim_{s\to\infty} (F(s)-F(t))##, which is equal to ##0## if the limit of the antiderivative ##F(s)## exists and undefined if it does not. The order of operations does not matter in this case.
  • #1
NotEuler
55
2
I'm pondering something about properties of integrals. What can we say about the following limit?

##\lim_{t\to\infty} \int_t^{\infty } f(x) \, dx##

On one hand, the 'gap' from the lower to upper integration limit diminishes, so that would suggest the limit is always 0.
But what if f is an ever-increasing function? Does the limit exist in that case? Or does the question even make sense, if the integral ##\int_t^{\infty } f(x) \, dx## does not have a finite value?

What would be a formal way to approach a question like this?
 
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  • #2
You actually have two limits: ##\lim_{t \to \infty} \int_t^\infty f(x)\,dx = \lim_{ t \to \infty} \lim_{s \to \infty} \int_t^sf(x)\,dx##. And if we assume ##F(x)## as antiderivative of ##f(x)## we have: ##\lim_{ t \to \infty} \lim_{s \to \infty} \int_t^sf(x)\,dx = \lim_{ t \to \infty} \lim_{s \to \infty} \left(F(s)-F(t) \right) = \lim_{s \to \infty}F(s) - \lim_{t \to \infty} F(t) =0\,.##
 
  • #3
Thanks fresh_42, that's great! Seems so clear now in hindsight.

And what if ##\lim_{s \to \infty}F(s)## is infinite? Would that mean that the limit I was originally after is not well defined?
 
  • #4
NotEuler said:
Thanks fresh_42, that's great! Seems so clear now in hindsight.

And what if ##\lim_{s \to \infty}F(s)## is infinite? Would that mean that the limit I was originally after is not well defined?
Their actual value isn't of interest, as they are simply identical terms: ##\lim_{s \to \infty}F(s) = \lim_{t \to \infty}F(t) = \lim_{tree \to \infty}F(tree)\,.##
 
  • #5
That's interesting.. I vaguely recall being taught that subtracting infinity from infinity doesn't have a well specified value. So something different is going on here?
Say, if f(x)=x and F(x)=x^2/2.
 
  • #6
NotEuler said:
That's interesting.. I vaguely recall being taught that subtracting infinity from infinity doesn't have a well specified value. So something different is going on here?
Say, if f(x)=x and F(x)=x^2/2.
That's correct and I do not subtract infinity from infinity. I subtract two expressions which are literally identical. There is no possibility to distinguish ##\lim_{t \to \infty} F(t)## from ##\lim_{s \to \infty} F(s)##. They have to be equal, and the difference between equal expressions is zero.

Here is a nice example about what can happen "at infinity":
https://www.physicsforums.com/threads/if-0-999-1-does-0-00-1-0.945434/#post-5983309
 
  • #7
##\lim_{x \to \infty} (x^3-x^2) = \lim_{x \to \infty} x^2(x-1) = \infty## but we have ##\lim_{x \to \infty} (x^3-x^3) = \lim_{\to \infty} 0=0\,.##
 
  • #8
fresh_42 said:
That's correct and I do not subtract infinity from infinity. I subtract two expressions which are literally identical. There is no possibility to distinguish ##\lim_{t \to \infty} F(t)## from ##\lim_{s \to \infty} F(s)##. They have to be equal, and the difference between equal expressions is zero.

Here is a nice example about what can happen "at infinity":
https://www.physicsforums.com/threads/if-0-999-1-does-0-00-1-0.945434/#post-5983309
So do I understand this correctly that in some cases, the order of operations matters, but not in others? In my example it doesn't so it's justifiable to write:
##\lim_{s \to \infty}(lim_{t \to \infty}F(t)-F(s))=lim_{s \to \infty}(F(s)-F(s))=lim_{s \to \infty}(0)=0##

Does this sound about right? Thanks very much for your help!
 
  • #9
fresh_42 said:
You actually have two limits: ##\lim_{t \to \infty} \int_t^\infty f(x)\,dx = \lim_{ t \to \infty} \lim_{s \to \infty} \int_t^sf(x)\,dx##. And if we assume ##F(x)## as antiderivative of ##f(x)## we have: ##\lim_{ t \to \infty} \lim_{s \to \infty} \int_t^sf(x)\,dx = \lim_{ t \to \infty} \lim_{s \to \infty} \left(F(s)-F(t) \right) = \lim_{s \to \infty}F(s) - \lim_{t \to \infty} F(t) =0\,.##

Assuming the integral converges.
 
  • #10
PeroK said:
Assuming the integral converges.
I don't think so. I only used the definitions for the abbreviations used and ended up with 2 identical expressions. They are 2 words within the formal language that we use, hence their difference - however defined - is necessarily zero, or the empty word if you like.
 
  • #11
NotEuler said:
So do I understand this correctly that in some cases, the order of operations matters, but not in others? In my example it doesn't so it's justifiable to write:
##\lim_{s \to \infty}(lim_{t \to \infty}F(t)-F(s))=lim_{s \to \infty}(F(s)-F(s))=lim_{s \to \infty}(0)=0##

Does this sound about right? Thanks very much for your help!
No.

There is no order anywhere.

##\int_t^\infty f(x)\,dx## is defined as ##\lim_{s \to \infty} \int_t^s f(x)\,dx##.
Now I assume that ##f(x) ## is integrable and ##F(x)## the antiderivative.
Then ##\int_t^\infty f(x)\,dx = \lim_{s \to \infty} (F(s)-F(t))\,.##

Your question was: What is ##\lim_{t\to \infty} \int_t^\infty f(x)\,dx\,\,?##
So by the previous use of definition and the fundamental theorem of integration we have
\begin{align*}
\lim_{t\to \infty} \int_t^\infty f(x)\,dx &= \lim_{t\to \infty} \left( \lim_{s \to \infty} (F(s)-F(t)) \right)\\
&= \lim_{t\to \infty} \left( \left( \lim_{s \to \infty} F(s) \right) - F(t) \right) \\
&= \left( \lim_{s \to \infty} F(s) \right) - \left( \lim_{t\to \infty} F(t) \right) \\
&= \left( \lim_{t \to \infty} F(t) \right) - \left( \lim_{t\to \infty} F(t) \right) \\
&= 0
\end{align*}

I see that you and @PeroK are right. I used ##\lim_{n \to \infty} (c+a_n)=c+\lim_{n\to \infty}a_n## which is only correct if ##c## is a finite number. Thus I need that ##\lim_{x\to \infty} F(x)## converges, i.e. exists.
 
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  • #12
Thanks for the interesting discussion.

So in the end, would you say the limit
##\lim_{t\to\infty} \int_t^{\infty } f(x) \, dx##
is always 0 if the infinite integral converges, and undefined otherwise?
 
  • #13
NotEuler said:
Thanks for the interesting discussion.

So in the end, would you say the limit
##\lim_{t\to\infty} \int_t^{\infty } f(x) \, dx##
is always 0 if the infinite integral converges,...
Yes, with the above calculation. If the antiderivative converges at infinity, then the calculation can be applied.
...and undefined otherwise?
I wouldn't go so far. If the calculation fails, then this won't allow a general conclusion. It is more relevant to ask whether the expression makes sense at all in such a case. E.g. if we allow ##f(x)## to also depend on ##t##, then we get more examples and possible outcomes for ##\lim_{t\to \infty} \int_t^\infty f(x,t)\,dx\,.##

If not, it looks as if you are right (in sloppy notation):
##\lim_{t \to \infty}\int_t^\infty f(x)\,dx= \lim_{t\to \infty} (F(\infty) -F(t)) = \lim_{t\to \infty} (\infty -\underbrace{F(t)}_{finite}) = \lim_{t\to \infty} \infty =\infty##
However, a real proof is something else! I wouldn't bet that this notation will survive a closer look. If it comes to infinity, the usual definitions change: We do not approach a given point anymore, instead we increase above all boundaries. And then a limit from an expression which already is infinitely large doesn't make much sense anymore, i.e. it has first to be said how it has to be understood.
 

1. What is the purpose of proving a property of an integral?

Proving a property of an integral allows us to establish the validity of a mathematical statement or theorem related to integrals. It helps us to understand the behavior and characteristics of integrals and their applications in various fields of science and mathematics.

2. How do you prove a property of an integral?

To prove a property of an integral, we use mathematical techniques such as substitution, integration by parts, and the fundamental theorem of calculus. We also use algebraic manipulations and logical reasoning to demonstrate the validity of the property.

3. What are some common properties of integrals?

Some common properties of integrals include linearity, commutativity, and additivity. Linearity states that the integral of a sum is equal to the sum of the integrals. Commutativity states that the order of integration does not affect the value of the integral. Additivity states that the integral of a function over a union of intervals is equal to the sum of the integrals over each individual interval.

4. Why is it important to prove properties of integrals?

Proving properties of integrals helps us to better understand the fundamental concepts and principles of integration. It also allows us to manipulate and simplify integrals to solve complex problems in various fields such as physics, engineering, and economics. Additionally, proving properties of integrals helps to avoid errors and ensure the accuracy of calculations involving integrals.

5. Can properties of integrals be applied to other mathematical concepts?

Yes, many properties of integrals can be applied to other mathematical concepts such as derivatives, limits, and series. This is because integrals and these concepts are closely related and share similar properties and rules. Understanding the properties of integrals can also help in understanding and solving problems involving these other mathematical concepts.

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