1. ## Summation of integrals

Hey there!

I have a doubt in the notation of a problem involving series that I am trying to solve.

Are integrals and summations exchangeable? I have something like this:

$\int_0^t\sum\limits_{k=1}^\infty B_k\left(s\right)\sin\left(\frac{{k\pi}}{{L}} x\right)e^{-\alpha\left(k\pi/L\right)^2\left(t-s\right)}ds$

Is it the same as

$\sum\limits_{k=1}^\infty \int_0^t B_k\left(s\right)\sin\left(\frac{{k\pi}}{{L}} x\right)e^{-\alpha\left(k\pi/L\right)^2\left(t-s\right)}ds$

I think that for solving this is much more convenient to integrate the general term first in terms of k, and then evaluate the sum.

Any hint?

2. ## Re: Summation of integrals

Hey berni1984.

The general result for this kind of thing is Fubini's theorem. If all the terms are finite and you get the right convergence, then it should follow.

Fubini's theorem - Wikipedia, the free encyclopedia

Note that the theorem is regard to measures so the summation and integral needs to be considered in terms of the right measure on the space.

3. ## Re: Summation of integrals

Chiro,

Thanks for your quick reply! I have checked it out, but however I don't think Fubini's Theorem is exactly what this problem is about.

While I was waiting for a response, I have done something, and I would like to see if it makes sense to you.

If one expands the summation inside the integral, one gets the following: (only the first two terms are shown)

$\int_0^t\sum\limits_{k=1}^{\infty}B_k\left(s\right )\sin\left(\frac{{k\pi}}{{L}}x\right)e^{-\alpha\left(k\pi/L\right)^2(t-s)}ds=\int_0^t B_1\left(s\right)\sin\left(\frac{{1\pi}}{{L}} x\right) e^{-\alpha\left(1\pi/L\right)^2(t-s)}ds+\int_0^t B_2\left(s\right)\sin\left(\frac{{2\pi}}{{L}}x \right) e^{-\alpha\left(2\pi/L\right)^2(t-s)}ds+...$

As it can be seen, the sine part is independent from the integration variable, and it can be pulled out of the integral. Thus, after grouping, the following can be seen:

$\int_0^t\sum\limits_{k=1}^{\infty}B_k\left(s\right )\sin\left(\frac{{k\pi}}{{L}}x\right)e^{-\alpha\left(k\pi/L\right)^2(t-s)}ds=\sum\limits_{k=1}^{\infty}\left[\sin\left(\frac{{k\pi}}{{L}}x\right)\int_0^t B_k\left(s\right)e^{-\alpha\left(k\pi/L\right)^2(t-s)}ds\right]$

This may not seem a good improvement, but $B_k\left(s\right)$ is a polynomial that is not very difficult to manipulate, and with simple algebra, the exponential term can be left in terms of s only (the part dependent on t is independent of the integration variable and thus can be pulled out of the integral). The resulting term can be integrated by parts, and the final result is a series depending only on k.

Does this all make sense to you?

Thanks!!

4. ## Re: Summation of integrals

Just to explain what Fubinis theorem is intuitively: its basically a result that says that you can swap integrals and since discrete sums can be written as integrals (with a different measure than the one you are usually familiar with which is the dx) it means that you can swap the sigma with the integral provided that you have the whole expression in absolute value is finite (see the wiki for what I mean).

If you satisfy the above criteria (to use Fubinis theorem) then you can swap the sigma's and the integrals order.

5. ## Re: Summation of integrals

If $(f_n)_{n=1}^\infty$ is a sequence of Riemann integrable functions which uniformly converge with limit $f$, then $f$ is Riemann integrable and its integral can be computed as the limit of the integrals of the $f_n$ (Wikipedia, MathWorld). There are various counterexamples when convergence is not uniform and the limit is either not Riemann-integrable or the integral of the limit (sum in the case of series) does not equal the limit of the integrals. One example is here; I can post others from the book "Counterexamples in Analysis".