2

$$ \lim_{n \rightarrow \infty} \int_{a}^b\ \sum_{k=1}^{n}f_k(x) \mathrm dx= \sum_{k=1}^{\infty}\int_{a}^b f_k(x) \mathrm dx $$

Is this generaly true ? Integral is a sum , two sums can interchange, right?

I ve faced this in a proof of a theorem that says that integral of a uniformly convergent sum is equal to the sum of integral. $\int \sum g_n = \sum \int g_n$ ( $ \sum g_n $ converges uniformly )

The problem i am facing is that the stament in the title is used to prove the previous theorem.

I dont think this is a duplicate, this question is about a partial sum that changes order with an integral not an infinite

It is said the answer below is incorrect, can someone explain why

Community
  • 1
Milan
  • 1,631

2 Answers2

1

This is true (only after my edits) directly by the linearity of the integral and the definition of infinite sum.

What you wrote is correct - An integral and a (finite) sum can be interchanged with no further conditions. This is exactly the linearity of the integral.

MOMO
  • 1,264
  • 6
  • 20
  • Yeah it should be f sub k – Milan Jan 26 '19 at 19:22
  • That looks like a cheat. If i had sum that does to infinity i couldnt do it, but if the limit that goes to infinity is in front then i can :) – Milan Jan 26 '19 at 19:27
  • 1
    Please 1) do not accept a wrong answer, and 2) do not change the question such that you make an already posted answer wrong, or both. – lcv Jan 26 '19 at 19:29
  • I don't know what changes have occurred, but this answer is wrong. You cannot exchange the limit with the integral unless you know the sum of the functions converges uniformly. ONLY for finite sums, in general, can you interchange sum and integral. – Ted Shifrin Jan 26 '19 at 19:34
  • 1
    @TedShifrin this is a finite sum there is just a limit in front of it ( if this makes sense) – Milan Jan 26 '19 at 19:42
1

Consider the calculation attempt $$\lim_{n\to\infty}\int_a^b\sum_{k=1}^nf_k(x)dx=\lim_{n\to\infty}\sum_{k=1}^n\int_a^bf_k(x)dx=\sum_{k=1}^\infty\int_a^bf_k(x)dx.$$The first $=$ sign works provided each $\int_a^bf_k(x)dx$ is finite, since then$$\int_a^b\sum_{k=1}^nf_k(x)dx=\lim_{n\to\infty}\sum_{k=1}^n\int_a^bf_k(x)dx$$and both sides have the same $n\to\infty$ behaviour (which might involve a limit not existing).

The second $=$ sign is true by the definition of the summation operator $\sum_{k=1}^\infty$ as a limit of partial sums. In other words:

  • If each $\int_a^bf_k(x)dx$ is finite then $\int_a^b\sum_{k=1}^nf_k(x)dx$ and $\sum_{k=1}^n\int_a^bf_k(x)dx$ are equal and have the same limit or each have no limit; and
  • $\lim_{n\to\infty}\sum_{k=1}^n\int_a^bf_k(x)dx$ is either $\sum_{k=1}^\infty\int_a^bf_k(x)dx$ if that exists, or non-existent if it doesn't.
J.G.
  • 115,835
  • Why cant this be done to every infinte sum in integral, just changing it to a partial sum with a limit to infinity ? – Milan Jan 27 '19 at 09:56
  • 1
    @MilanStojanovic I'm not sure which counterexample you have in mind, but when at least one $\int_a^b f_k(x) dx$ isn't finite we have problems. There are also cases where $\lim_{n\to\infty}\int_a^b\sum_{k=1}^n f_k(x)dx\ne\int_a^b\lim_{n\to\infty}\sum_{k=1}^n f_k(x)dx=\int_a^b\sum_{k=1}^\infty f_k(x)dx$. – J.G. Jan 27 '19 at 09:58
  • 1
    @MilanStojanovic You can do this in every integral. What you cannot do is interchange the limit and the integral, that why the sum in the integral is not infinite. – MOMO Jan 27 '19 at 17:51