Proving a property of an integral

[NotEuler]

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?

 

[fresh_42]

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: $\lim_{ t \to \infty} \lim_{s \to \infty} \int_t^sf(x)\,dx = \lim_{ t \to \infty} \lim_{s \to \infty} \left(F(s)-F(t) \right) = \lim_{s \to \infty}F(s) - \lim_{t \to \infty} F(t) =0\,.$

 

[NotEuler]

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?

 

[fresh_42]

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: $\lim_{s \to \infty}F(s) = \lim_{t \to \infty}F(t) = \lim_{tree \to \infty}F(tree)\,.$

 

[NotEuler]

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.

 

[fresh_42]

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

 

[fresh_42]

$\lim_{x \to \infty} (x^3-x^2) = \lim_{x \to \infty} x^2(x-1) = \infty$ but we have $\lim_{x \to \infty} (x^3-x^3) = \lim_{\to \infty} 0=0\,.$

 

[NotEuler]

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!

 

[PeroK]

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.

 

[fresh_42]

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.

 

[fresh_42]

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.

 

[NotEuler]

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?

 

[fresh_42]

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.