# Triple Integral

1. Aug 3, 2005

### amcavoy

I need to prove the following:

$$\int_{0}^1\int_{0}^1\int_{0}^1\frac{1}{1-xyz}dxdydz=\sum_{n=1}^{\infty}\frac{1}{n^3}$$

Or, as a generalization:

$$\int_{0}^1\cdots\int_{0}^1\frac{1}{1-\prod_{k=1}^mx_k}\prod_{k=1}^mdx_k=\sum_{n=1}^{\infty}\frac{1}{n^m}$$

...if there is such a generalization.

I don't know where to begin, any suggestions?

Thanks a lot for your help.

Last edited: Aug 3, 2005
2. Aug 3, 2005

### MalleusScientiarum

You can make a Taylor expansion of the denominator since it is convergent for every point in the domain of integration.

3. Aug 4, 2005

### amcavoy

Alright, a friend showed me how to do this for m=3. Does this work?

$$\sum_{n=1}^{\infty}\frac{1}{n^3}=\sum_{n=0}^{\infty}\frac{1}{(n+1)^3}$$

and since $$\int_{0}^1x^kdx=\frac{1}{k+1}$$, the sum can be rewritten as follows:

$$\sum_{n=0}^{\infty}\frac{1}{(n+1)^3}=\sum_{n=0}^{\infty}\int_{0}^1\int_{0}^1\int_{0}^1(xyz)^kdxdydz=\int_{0}^1\int_{0}^1\int_{0}^1\frac{1}{1-xyz}dxdydz$$.

This works? Or, is the proof more in-depth? I would like to know whether the needs to be a justification for evaluating the outside sum first.

Thanks again.

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