How do I even start doing this?

>>8389622
i have no idea

>>8389622
Wolfram alpha

Why did you snapshot an equation about an iterared integral

u = 1 - xyz
du = -dxdydz

File: zeta of three.png (12KB, 671x285px) Image search: [Google]

12KB, 671x285px

>>8389622
it checks out.

Geometric series

>>8389622
Try from right to left using complex residues then use some symmetry argument.

[eqn] \int_0^1 \int_0^1 \int_0^1 \sum_{k=0}^\infty (xyz)^k dx dy dz = \sum_{k=0}^\infty \int_0^1 \int_0^1 \int_0^1 (xyz)^k dx dy dz [/eqn]
[eqn] = \sum_{k=0}^\infty \frac{1}{(k+1)^3} [/eqn]
[eqn] = \sum_{n=1}^\infty \frac{1}{n^3} [/eqn]

You can pull the series out of the integral because Lebesgue's monotone convergence theorem.

>>8389622
[eqn]\frac{1}{1-xyz} = \sum_{n=0}^{\infty}(xyz)^n \iff |xyz| \leq 1[/eqn]

brainlet detected lol

>>8389724
noice

>>8389622
That's a statement, not a question. What are you supposed to do?

>>8389622
Sorry, there is no known way to put that in terms of Pi and a constant.

>he thinks you can just DO a triple integral

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41KB, 549x591px

>>8389622
Is this a troll thread? It's easy to show that the integral evaluates to Apery's constant.

>>8389724
Kudos sir!

>>8389733
>>[math]\leq 1[/math]

>>8391998

Implying the right hand side is constant

>>8391467
Povide a proof for the given identity.

>>8389666
/thread

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44KB, 480x600px

>>8389724
>Lebesgue's monotone convergence theorem
mfw he's this right

>>8389724
Thank you, on a side note. Did you major in math? I wish to become better, any tips?

>>8392400
Read, Try Ex, Repeat

>>8389622
Take a triple derivative to get rid of the triple integral.