Moving a limit inside an integral

In summary, the grad student mentioned that you cannot move a limit inside of an integral without meeting certain conditions, but did not specify what those conditions were. I was under the impression that this was unrestricted (and the particular theorem we were looking at worked fine if we did this). However, the limit doesn't exist, so it is inadmissible to interchange the limiting operations of integration and sequence limiting.
  • #1
strangequark
38
0
A grad student mentioned the other day that you cannot move a limit inside of an integral without meeting certain conditions, unfortunately, he didnt say what those condition were... I was under the impression that this was unrestricted (and the particular theorem we were looking at worked fine if we did this)...


Can anyone name these conditions for me?
 
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  • #2
A typical case is the following:
Let fe(x)=0, |x|>=e>0, f(x)=1/(2e), |x|<e.

Thus, we have:
[tex]I_{e}=\int_{-\infty}^{\infty}f_{e}(x)dx=1, \lim_{e\to{0}}I_{e}=1[/tex]

However, the following limit doesn't EXIST, [tex]\lim_{e\to{0}}f_{e}(x)[/tex]
Thus, it is inadmissible to interchange the limiting operations of integration and sequence limiting.
 
  • #3
From my notes from an Analysis course:

Let [itex]V \subset \mathbb{R}^n, a, b \in \mathbb{R}, a < b[/itex]. Further, assume that the function [itex]f: V \times [a, b] \to \mathbb{R}[/itex] is continuous on [itex]V \times [a, b] \subset \mathbb{R}^{n + 1}[/itex]. Then the function [itex]I: V \to \mathbb{R}[/itex], defined by
[tex]I(x) := \int_a^b f_x(t) \, \mathrm{d}t = \int_a^b f(x, t) \, \mathrm{d}t,
\qquad f_x: t \mapsto f(x, t)[/tex]
is continuous.

This means that for each [itex]\xi \in V[/itex],
[tex]
\lim_{x \to \xi} \int_a^b f(x, t) \, \mathrm{d}t =
\lim_{x \to \xi} I(x) =
I(\xi) =
\int_a^b f(\xi, t) \, \mathrm{d}t =
\int_a^b \left( \lim_{x \to \xi} f(x, t) \right) \, \mathrm{d}t[/tex],
using the continuity of [itex]x \mapsto f(x, t)[/itex] in the last step.

The proof is not hard, but not easy to type out since it relies on all sorts of definitions and earlier results (e.g. on interchange of limits and results on Riemann integration).
 
  • #4
CompuChip said:
From my notes from an Analysis course:

Let [itex]V \subset \mathbb{R}^n, a, b \in \mathbb{R}, a < b[/itex]. Further, assume that the function [itex]f: V \times [a, b] \to \mathbb{R}[/itex] is continuous on [itex]V \times [a, b] \subset \mathbb{R}^{n + 1}[/itex]. Then the function [itex]I: V \to \mathbb{R}[/itex], defined by
[tex]I(x) := \int_a^b f_x(t) \, \mathrm{d}t = \int_a^b f(x, t) \, \mathrm{d}t,
\qquad f_x: t \mapsto f(x, t)[/tex]
is continuous.

This means that for each [itex]\xi \in V[/itex],
[tex]
\lim_{x \to \xi} \int_a^b f(x, t) \, \mathrm{d}t =
\lim_{x \to \xi} I(x) =
I(\xi) =
\int_a^b f(\xi, t) \, \mathrm{d}t =
\int_a^b \left( \lim_{x \to \xi} f(x, t) \right) \, \mathrm{d}t[/tex],
using the continuity of [itex]x \mapsto f(x, t)[/itex] in the last step.

The proof is not hard, but not easy to type out since it relies on all sorts of definitions and earlier results (e.g. on interchange of limits and results on Riemann integration).

CompuChip, can you please elaborate on the arguments behind the proof (a sketch would be enough). Thanks!
 
  • #5
The monotone convergence theorem and dominated convergence theorem form measure theory can often be used to pass the limit inside an integral, rephrasing convergence in terms of convergence of sequences when necessary.
 
  • #6
Sorry, there is no sequence or series in the equatiion below:

$\underset{x \rightarrow \infty}\lim
\int_{0}^{\infty} f(x,y)dy = \int_{0}^{\infty} \underset{x \rightarrow \infty}\lim \Bigl[
f(x,y) dy$

assuming f is continuous everywhere in $[a,\infty[ \times [0,\infty$ and the limit inside the integral on the above equation exists and is finite. According to Compuchip the above statement holds and my question what are the mathematical arguments.

Thanks,

zrezki
 
  • #7
strangequark said:
A grad student mentioned the other day that you cannot move a limit inside of an integral without meeting certain conditions, unfortunately, he didnt say what those condition were... I was under the impression that this was unrestricted (and the particular theorem we were looking at worked fine if we did this)...


Can anyone name these conditions for me?

You we under a bad impression.
The integral is not important.
The issue is interchanging two limits.
Or more precisely when interchanged limits exist and are equal.
say U and V are limits and f a function for what U,V and f is
UVf=VUf
It is a hard question to answer in general.
In simple cases a sufficient condition is used.
Limits can be interchanged when uniform or close to it.
That is to say the inside limit must not change rapidly as the outer limit is approched.
 
  • #8
Still heuristic and I need more rigourous arguments. This is a well-formed problem to which should exist a clear answer.

Thanks
 
  • #9
^This problem is not well formed and no clear answer exist. You did not specify what rigorous means. In particular the type of integration has not been specified nor the conditions on the function to be integrated, nor the nature of the limit being taken.

Here is an example of a well-formed problem to which a clear answer exists.

let I indicate (proper) Riemann integration with respect to x with limits a and b
let L indicate a limit operation with respect to t

let the function f(x,t) be continuous in x for the range a<=x<=b for all t in the approach,
and if f(x,t) approaches a limiting function F(x) uniformly in x for this range, then:
ILf and LIf exist and are equal

notice that one can extent this if desired and even then all cases are not settled

we will have (by previous theorem F is continuous hence integrable
ILf=IF exist
by uniform convergence for epsilon>0 there exist tepsilon
|F(x)-f(x,t)|<epsilon for t beyond tepsilon, a<=x<=b
so
|IF(x)-If(x,t)|=|I[F(x)-f(x,t)]|<=I|F(x)-f(x,t)|<=|b-a|epsilon
hence
LIf exist and
ILf=LIf
 
  • #10
All the ifs you mentionned were contained in my previous reply. The only thing I forgot is that the integral is Riemann one. Thar's what I meant by well formed problem. Following your reasonning in the first part, all we need to move the limit inside the integral is unform convergence of f with respect to x. Is that true ?
 
  • #11
Incredible amounts of messy answers!

The comment from GSpeight seems to be the only one that answers the original question properly. The Lebesgue's dominated convergence theorem is the standard tool for changing the order of integration and limit.

Warning: Please ignore confusing nonsensical comments like this

The dominated convergence theorem does not hold for the Riemann integral, and its power and utility are one of the primary theoretical advantages of Lebesgue integration.

The value of the Riemann integral is the same as the value of the Lebesgue integral, so of course the dominated convergence can be used for Riemann integrals!
 
  • #12
I couldn't see how may I going to apply the dominated convergence theorem to check if this is true:

$\underset{x \rightarrow \infty}\lim
\int_{0}^{\infty} f(x,y)dy = \int_{0}^{\infty} \underset{x \rightarrow \infty}\lim \Bigl[
f(x,y) \Bigl}dy$

assuming f is continuous everywhere in $[a,\infty[ \times [0,\infty$ and the limit inside the integral on the above equation exists and is finite.

Thanks
 
  • #13
zrezki said:
I couldn't see how may I going to apply the dominated convergence theorem to check if this is true:

$\underset{x \rightarrow \infty}\lim
\int_{0}^{\infty} f(x,y)dy = \int_{0}^{\infty} \underset{x \rightarrow \infty}\lim \Bigl[
f(x,y) \Bigl}dy$

assuming f is continuous everywhere in $[a,\infty[ \times [0,\infty$ and the limit inside the integral on the above equation exists and is finite.

Thanks

This hypothesis is not true. For example set

[tex]
f(x,y) = \left\{\begin{array}{ll}
0,\quad &y<x\\
\sin(y-x),\quad & 0\leq y-x\leq \pi\\
0,\quad &x+\pi < y\\
\end{array}\right.
[/tex]

[tex]
\int\limits_0^{\infty} f(x,y) dy = \int\limits_{x}^{x+\pi} \sin(y-x)dy = 2\quad\quad\forall\; x\in [0,\infty[
[/tex]

[tex]
\lim_{x\to\infty} f(x,y) = 0,\quad\quad \forall\; y\in [0,\infty[
[/tex]

The dollar signs $ don't work over here. Instead use [ tex] and [ /tex].
 
Last edited:
  • #14
This answers my question. Thanks.
 

What is the rule for moving a limit inside an integral?

The rule for moving a limit inside an integral is known as the "limit theorem for integrals." It states that if a function f(x) is continuous on [a, b] and if the limit of f(x) as x approaches c exists for all c in [a, b], then the limit of the integral from a to b of f(x) as x approaches c is equal to the integral from a to b of the limit of f(x) as x approaches c.

Why is it important to be able to move a limit inside an integral?

Moving a limit inside an integral is important because it allows us to solve more complicated integrals. It also helps us to understand the behavior of a function as it approaches a certain point or value.

What are the steps for moving a limit inside an integral?

The steps for moving a limit inside an integral are as follows: 1. Use the limit theorem for integrals to rewrite the integral with the limit inside.2. Evaluate the integral as usual.3. Take the limit of the resulting expression as x approaches the given value.

What are some common mistakes when moving a limit inside an integral?

Some common mistakes when moving a limit inside an integral include forgetting to use the limit theorem for integrals, incorrectly evaluating the integral, and forgetting to take the limit at the end. It is important to carefully follow the steps and pay attention to the limits of integration and the given value of x.

Can a limit always be moved inside an integral?

No, a limit cannot always be moved inside an integral. The limit theorem for integrals only applies if the function is continuous on the interval and if the limit of the function as x approaches the given value exists. If these conditions are not met, then the limit cannot be moved inside the integral.

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