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What is the value of $$\int_{[0,1]^2} \!\!\!dx\,dy\,\mathcal{P} \frac{\log(x)}{x-y},$$ where $\mathcal{P}$ denotes Cauchy's principal value. I solved this once for a homework, but I can neither remember the answer nor reproduce it right now... (bad memory)

Any help or comment is highly welcome.

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what do you mean by principal value? you have infinite values along $y=x$ and $\{0\}\times[0,1]$. how are you approaching those? –  yoyo Mar 7 '11 at 17:16
The logarithmic singularity at ${0}\times [0,1]$ seems to be integrable. So only the pole at $x=y$ poses problems. Cauchy's princial value (en.wikipedia.org/wiki/Cauchy_principal_value) denotes the limit when this singularity is approached symetrically... –  Fabian Mar 7 '11 at 17:32
Mathematica tells me that the result is $\pi^2/6$ [looks like $\zeta(2)$]. Any thoughts why (or if) this is correct? –  Fabian Mar 7 '11 at 17:34
Can you not take the Hilbert transform of $1_{[0,1]}$ and then compute the (principal value) integral of $\log(x)$ times the result of that? –  Jonas Teuwen Mar 7 '11 at 21:30
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1 Answer

up vote 2 down vote accepted

Let $f(x,y)=\frac{\log x}{x-y}$. Since the principal value does not seem well-defined here, let's integrate $f(x,y)+f(y,x)= \frac{\log (x/y)}{x-y}$ (which is positive) on the triangle defined by $0<y<x<1$. Now $\int_0^x \frac{\log (x/y)}{x-y} dy = -\int_0^1 \frac{\log u}{1-u} du$ (change of variable $y=xu$), which does not depend on $x$, and so this is equal to the first integral (integrating for $x$ between $0$ and $1$).

Now $-\int_0^{1-\epsilon} \frac{\log u}{1-u} du = \int_0^{1-\epsilon} \sum_{ n \geq 1} \frac{u^{n-1}}{n} du = \sum_{n \geq 1} \frac{(1-\epsilon)^n}{n^2}$ and letting $\epsilon$ go to $0$ gives $\zeta(2)=\frac{\pi^2}{6}$.

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Why you think the principal value is not well defined? –  Fabian Mar 9 '11 at 18:22
The examples at the end of the Wikipedia page illustrate this fact, for a given integral you can get different values if you choose your epsilons differently. But most of the time if you make "reasonable" choices the result is the same: for example in your case if we integrate on a closed symetric (with respect to the diagonal) domain not containing the diagonal and if this domain "converges" to the whole of $[0,1]^2$. If we integrate wrt $y$ first, excluding $[x-\epsilon,x+\epsilon]$ (then $\epsilon \rightarrow 0$) we still find $\zeta(2)$. But other choices would give different values. –  Plop Mar 9 '11 at 19:12
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