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Definite Triple Integral to Series

  • Thread starter Elysian
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  • #1
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Homework Statement



Does the triple integral
[itex]\int^{1}_{0}\int^{1}_{0}\int^{1}_{0}\frac{1}{1+x^2 y^2 z^2}[/itex] = [itex]\sum^{∞}_{n=0}\frac{1}{(2n+1)^3}[/itex]

Homework Equations





The Attempt at a Solution


I've not a single clue on what to do with this problem. I figured maybe I could find a decent conversion of variables and find the Jacobian and switch variables to make this easier but tht didn't seem to work.

I don't know why but I think the series expansion of arcsin or arctan is needed here, but I'm not exactly sure how to incorporate it in. I don't even know how to evaluate this triple integral..
 

Answers and Replies

  • #2
jbunniii
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Homework Statement



Does the triple integral
[itex]\int^{1}_{0}\int^{1}_{0}\int^{1}_{0}\frac{1}{1+x^2 y^2 z^2}[/itex] = [itex]\sum^{∞}_{n=0}\frac{1}{(2n+1)^3}[/itex]
Hint: the integrand is ##\leq 1##, so does that give you an upper bound for the left hand side? Notice also that the first term of the series on the right hand side is ##1##, so that gives you a lower bound for the right hand side.
 
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Hint: the integrand is ##\leq 1##, so does that give you an upper bound for the left hand side? Notice also that the first term of the series on the right hand side is ##1##, so that gives you a lower bound for the right hand side.
Ohh I think I get what you mean

The upper bound for the LHS is 1, and the first term of the right hand side is 1 and then adding constants, so they can't be equal?

There's only one intersection of their ranges so it doesnt work out then I guess?
 
  • #4
jbunniii
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The upper bound for the LHS is 1, and the first term of the right hand side is 1 and then adding constants, so they can't be equal?
That's right, the LHS is ##\leq 1##, and the RHS is ##> 1##, so they can't be equal.
 

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