# How to calculate $\int_{0}^ 1 \int_{0}^ 1 \frac{1}{(1-xy)^ p} dxdy$ with $p>0$.

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!

• You should have $p(p+1)\cdots (p+n-1)$, and not $p(p+1)\cdots (p+n)$. – Batominovski Nov 21 '18 at 22:02
• @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? – Dora y Diego Nov 21 '18 at 22:23

## 1 Answer

This is a well-known integral for $$0 and involves the Harmonic number:

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

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