Decoding Infinite Series with Double Integrals

[Screwdriver]

Homework Statement

Check out matt grime's post in this thread (it's the last one):

https://www.physicsforums.com/showthread.php?p=470773#post470773"

How exactly did he know that the sum could be represented as that double integral? Also, is there a method of converting sums like that to integrals (double or otherwise) for summands other than $n^{-2}$ such as $n^{-7}$ or something?

Homework Equations

$$\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xy}dxdy=\sum_{n=1}^{\infty }\frac{1}{n^2}$$

The Attempt at a Solution

Come to think of it, I don't even really see how that helps you, because the series expansion for the $y$ integral (after computing the $x$ integral) is the derivative of the series you're trying to find, so integrating it just brings you back to where you started. The only thing I was able to note was that:

$$\sum_{n=0}^{\infty} (xy)^n=\frac{1}{1-xy}$$

 

[micromass]

What exactly are you asking?? Are you asking why

$$\int_0^1\int_0^1\frac{dxdy}{1-xy}=\sum{\frac{1}{n^2}}$$

Well, you're on the right way. Just notice that

$$\frac{1}{1-xy}=\sum{(xy)^{n-1}}$$

So by monotone convergence, we can switch the limit and the integral: So

$$\int_0^1\int_0^1\frac{dxdy}{1-xy}=\sum\int_0^1\int_0^1 (xy)^{n-1}dxdy$$

So you only need to calculate $\int_0^1\int_0^1 (xy)^{n-1}dxdy$ now...

 

[Screwdriver]

Well what I'm really asking is how to convert the sum to an integral. Like If I just saw $\sum_{n=1}^{\infty }\frac{1}{n^2}$ somewhere, I would start with this:

$$\sum_{n=1}^{\infty} (xy)^{n-1}=\frac{1}{1-xy}$$

But how did I know to start with that? It appears to have no relation to $\sum_{n=1}^{\infty }\frac{1}{n^2}$ ... and then all of a sudden there's a double integral with non-infinite limits.

 

[lineintegral1]

Recall the (real) geometric series,

$\frac {1}{1-x}=\sum_{n=0}^{\infty}x^n$

This is true provided, of course, that $|x|<1$.

Anytime you see a rational expression of this form, a light should go off in your head regarding the geometric series. It should be easy to see now how to prove the equivalence of your double integral and $\zeta(2)$.

Not sure if this clarifies anything; let us know if you have more questions.

 

[Screwdriver]

Recall the (real) geometric series,

$\frac {1}{1-x}=\sum_{n=0}^{\infty}x^n$

This is true provided, of course, that $|x|<1$.

Anytime you see a rational expression of this form, a light should go off in your head regarding the geometric series. It should be easy to see now how to prove the equivalence of your double integral and $\zeta(2)$.

Not sure if this clarifies anything; let us know if you have more questions.

It's not showing that the integral and sum are equivalent that's my problem, it's coming up with the integral in the first place . Let me try to rephrase this:

Given a sum, I want to know how to write down a computable integral whose value is the value of that sum.

 

[Screwdriver]

First and last bump for this thread.