Hi I am trying to integrate $$ \mathcal{I}:=\iint \limits_{{x,y \ \in \ [0,1]}} \frac{\log(1-x)\log(1-y)}{1-xy}dx\,dy=\int_0^1\int_0^1 \frac{\log(1-x)\log(1-y)}{1-xy}dx \,dy $$ A closed form does exist. I tried to write \begin{align} \mathcal{I} &=\int_0^1 \log(1-y)dy\int_0^1 \log(1-x)\frac{dx}{1-xy} \\ &= \int_0^1\log(1-y)dy \ \int_0^1 \sum_{n\geq0}(xy)^n\, \ln(1-x) \ dx \\ &= \sum_{n\geq 0}\frac{1}{n+1}\int_0^1 \log(1-y) y^n\, dy \\ &= \sum_{n\geq 0}\frac{1}{n+1}\int_0^1 \log(1-y)y^n\, dy = ? \end{align} I was able to realize that $$ \mathcal{I}=\sum_{n\geq 1}\left(\frac{H_n}{n}\right)^2=\frac{17\zeta_4}{4}=\frac{17\pi^4}{360},\qquad \zeta_4=\sum_{n\geq 1} n^{-4} $$ however this does not help me solve the problem. How can we calculate $\mathcal{I}$? Thanks.
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How does the last expression not help you solve the problem? $\dfrac{17\pi^4}{360}$ is a closed form, is it not? – May 26 '14 at 23:13
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1@SanathDevalapurkar Obviously it is a closed form...I am trying to prove this, knowing a closed form exists is not a proof. – Jeff Faraci May 26 '14 at 23:14
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1Duplicate of this. – Lucian May 27 '14 at 02:27
3 Answers
First notice that
$$ \int_{0}^{1} x^{n} \log(1-x) \ dx = -\int_{0}^{1} x^{n} \sum_{k=1}^{\infty} \frac{x^{k}}{k} \ dx $$
$$ = -\sum_{k=1}^{\infty} \frac{1}{k} \int_{0}^{1} x^{n+k} \ dx = -\sum_{k=1}^{\infty} \frac{1}{k(n+k+1)}$$
$$ = - \frac{1}{n+1} \sum_{k=1}^{\infty} \left(\frac{1}{k} - \frac{1}{n+k+1} \right)$$
$$ = -\frac{H_{n+1}}{n+1}$$
Then
$$ \int_{0}^{1} \int_{0}^{1} \frac{\log(1-x) \log(1-y)}{1-xy} \ dx \ dy $$
$$ =\sum_{n=0}^{\infty} \int_{0}^{1} x^{n} \log(1-x) \ dx \int_{0}^{1} y^{n} \log(1-y) \ dy $$
$$ = \sum_{n=0}^{\infty} \left( \frac{H_{n+1}}{n+1} \right)^{2}$$
A closed form for the sum $\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$
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Are you missing a minus sign? $$ \log(1-x)=-\sum_{k=1}^\infty \frac{x^k}{k} $$ Also you reduced them to sums but how can we evaluate the sums? Thanks – Jeff Faraci May 26 '14 at 23:52
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Yes, I was missing a minus sign. Thanks. I know that sum has been evaluated on here. Let me do a search. – Random Variable May 26 '14 at 23:56
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http://math.stackexchange.com/questions/554003/a-closed-form-for-the-sum-sum-n-1-infty-left-frach-nn-right2 – Random Variable May 26 '14 at 23:58
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There is another way to evaluate that harmonic sum, and that is to integrate the generating function for $(H_{n})^{2}$ twice. That's what I have in my notes. Maybe I'll add it to the other thread. – Random Variable May 27 '14 at 00:12
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Please let me know if you add it to the other thread in case I do not realize instantly. Thanks – Jeff Faraci May 27 '14 at 00:48
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1Here's even a better approach. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1917270#p1917270 – Random Variable May 27 '14 at 01:19
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Consider the double integral \begin{align} I = \int_{0}^{1} \int_{0}^{1} \frac{\ln(1-x) \ \ln(1-y)}{1-xy} \ dx dy \end{align} which, upon expansion into series form, becomes \begin{align} I &= \sum_{n=0}^{\infty} \ \int_{0}^{1} x^{n} \ln(1-x) dx \ \int_{0}^{1} y^{n} \ln(1-y) dy \\ &= \sum_{n=0}^{\infty} \left( \int_{0}^{1} x^{n} \ \ln(1-x) \ dx \right)^{2}. \end{align} Without much difficulty it can be shown that \begin{align} \int_{0}^{1} x^{n} \ \ln(1-x) \ dx &= \left. \partial_{\mu} B(n+1, \mu+1) \right|_{\mu = 1} \\ &= B(n+1,1) \left( \psi(1) - \psi(n+2) \right) \\ &= \frac{H_{n+1}}{n+1}. \end{align} Now, returning to the primary integral, it is seen that \begin{align} I = \sum_{n=0}^{\infty} \left( \frac{H_{n+1}}{n+1}\right)^{2} = \sum_{n=1}^{\infty} \left( \frac{H_{n}}{n} \right)^{2} = \frac{17 \zeta(4)}{4}. \end{align} Hence \begin{align} \int_{0}^{1} \int_{0}^{1} \frac{\ln(1-x) \ \ln(1-y)}{1-xy} \ dx dy = \frac{17 \zeta(4)}{4}. \end{align}
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Do you have proof of $$ \sum_{n=1}^{\infty} \left( \frac{H_{n}}{n} \right)^{2} = \frac{17 \zeta(4)}{4}, $$ Thanks for your help – Jeff Faraci May 26 '14 at 23:54
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1This series and a few related can be found in the article http://www.ams.org/journals/proc/1995-123-04/S0002-9939-1995-1231029-X/ – Leucippus May 26 '14 at 23:58
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Thanks a lot my friend. I checked off the answer because you only because it was first. But this is very nice...Thanks, my one other question is, what do you mean "without much difficulty it can be shown...", I don't see the immediate connection with the Beta function since this is where I got stuck – Jeff Faraci May 27 '14 at 00:03
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Nice answer. I considered using the Beta function, but I changed my mind at the last minute. – Random Variable May 27 '14 at 00:16
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1@Integrals : practice and experience is where most of it begins, but in this case it was more "I didn't want to type all the steps". – Leucippus May 27 '14 at 01:03
Wow, RV and L are pretty fast. Well, even though their ways are more efficient, I worked out yet another way so I may as well post it.
$$\int_{0}^{1}\log(1-y)\int_{0}^{1}\frac{\log(1-x)}{1-xy}dxdy$$
Let $u=xy, \;\ \frac{du}{x}=dy$
$$\int_{0}^{x}\frac{\log(1-u/x)}{1-u}du\int_{0}^{1}\frac{\log(1-x)}{x}dx$$
The left integral w.r.t u is a classic dilog: $\int_{0}^{x}\frac{\log(1-u/x)}{1-u}du=-Li_{2}(\frac{x}{1-x})$
Thus, $$\int_{0}^{1}\frac{Li_{2}(\frac{x}{1-x})\log(1-x)}{x}dx$$
But, $$-Li_{2}(\frac{x}{1-x})=\sum_{n=1}^{\infty}\frac{H_{n}}{n}x^{n}$$
giving:
$$\int_{0}^{1}\log(1-x)\sum_{n=1}^{\infty}\frac{H_{n}}{n}x^{n-1}dx$$
Also, $\int_{0}^{1}x^{n-1}\log(1-x)dx=\frac{H_{n}}{n}$
So, we finally obtain:
$$\sum_{n=1}^{\infty}\frac{H_{n}^{2}}{n^{2}}=\frac{17\pi^{4}}{360}$$
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Nice work. Random Variable posted a link where you prove this sum, so thanks for this. +1 – Jeff Faraci May 27 '14 at 00:04