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This is Hewitt and Stromberg's Real and Abstract Analysis Problem 21.22. My approach was to compare with this question

How to derive $\int_0^1 \int_0^1 \frac{1}{1-xy} \,dy\,dx = \sum_{n=1}^{\infty}\frac{1}{n^{2}}$

and use the series expansion of $\frac{1}{(1-xy)^ p}$ to calculate this integral. I found this expansion to be

$$\frac{1}{(1-xy)^ p}=\sum_{n=0}^ \infty \frac{p \cdots (p+n-1)}{n!} (xy)^ n.$$

However, I got this integral is

$$\lim_{r \to 1} \int_0^ r \int_0^ r (\sum_{n=0}^ \infty \frac{p \cdots (p+n-1)}{n!} (xy)^ n) dxdy = \sum_{n=0}^ \infty \frac{p\cdots (p+n-1)}{n!(n+1)^ 2}.$$

But, I don't seem to find if this series converges or diverges, much less what is the value of the integral. Also it says to calculate the following integrals,

$$\int_{0}^ 1 \int_{0}^ 1 \frac{1}{(1-xy)^ p} dydx, \int_{0}^ 1 \int_{0}^ 1 \Bigg|\frac{1}{(1-xy)^ p}\Bigg| dxdy, \int_{0}^ 1 \int_{0}^ 1 \Bigg|\frac{1}{(1-xy)^ p}\Bigg| dydx$$

and to compare with Fubini's Theorem.

Have I done something wrong?

Thanks a lot!

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  • $\begingroup$ You should have $p(p+1)\cdots (p+n-1)$, and not $p(p+1)\cdots (p+n)$. $\endgroup$ – Batominovski Nov 21 '18 at 22:02
  • $\begingroup$ @Batominovski I had it with the $p+n-1$ but I didn't seem to conclude anything so I thought I could just use $p+n$. I've edited now, how can I know if it converges? $\endgroup$ – Dora y Diego Nov 21 '18 at 22:23
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This is a well-known integral for $0<p<1$ and involves the Harmonic number:

$$\frac{H_{1-p}}{1-p}$$

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  • $\begingroup$ I do not see why you had a downvote. $\to +1$. $\endgroup$ – Claude Leibovici Nov 22 '18 at 6:53
  • $\begingroup$ @ClaudeLeibovici: Nor did I. Thanks. $\endgroup$ – David G. Stork Nov 22 '18 at 7:15

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