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In this .pdf document, which is just a list of Putnam-style undergraduate-level problems from various sources, the third question is as I have stated it below (up to a change of notation).

Evaluate $$I=\int_{0}^{1}\log{(x)}\log{(1-x)}\,\mathrm{d}x.$$

Feel free to take a moment to try to solve this yourself, if you have never seen it before. My answer is as follows.

In the interval $(0,1)$, we may expand $\log{(1-x)}$ as a power series: \begin{eqnarray*} \log{(1-x)} & = & -\left(x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\ldots+\frac{x^{r}}{r}+\ldots\right)\\ & = & -\sum_{k=0}^{\infty}\frac{x^{k}}{k} \end{eqnarray*}

Now for any $p\in\mathbb{N}$, using integration by parts we see that $$\int_{0}^{1}x^{p}\log{x}\,\mathrm{d}x = -\frac{1}{(p+1)^{2}}.$$

Combining the above results, we have: \begin{eqnarray*} \int_{0}^{1}\log{(x)}\log{(1-x)}\,\mathrm{d}x & = & \int_{0}^{1}\log{(x)}\left[-\sum_{k=1}^{\infty}\frac{x^{k}}{k}\right]\,\mathrm{d}x\\ & = & -\sum_{k=1}^{\infty}\frac{1}{k}\left[\int_{0}^{1}x^{k}\log{x}\,\mathrm{d}x\right]\\ & = & \sum_{k=1}^{\infty}\frac{1}{k(k+1)^{2}}\\ & = & \sum_{k=1}^{\infty}\left[\frac{1}{k(k+1)}-\frac{1}{(k+1)^{2}}\right]\\ & = & \sum_{k=1}^{\infty}\frac{1}{k(k+1)}-\sum_{k=1}^{\infty}\frac{1}{(k+1)^{2}}\\ & = & 1-(\frac{\pi^{2}}{6}-1)\\ & = & 2-\frac{\pi^{2}}{6}, \end{eqnarray*} where we have evaluated one of the two series at the end via a telescoping sum, the details of which I have left out, and the value of the other series is well-known.

I found the result surprising. I performed a small sanity check by attempting to sketch the graph within the interval; Wolfram|Alpha agrees with my sketch and agrees with my answer, but, to me at least, this information is uninformative about why the result is true. To be more precise, I don't understand how or why the answer relates to the original elements of the question.

My question: Is there any reason why one would expect $\pi$ to appear in this answer? Is there some tricky change of variables or some unbeknownst-to-me complex analysis way of evaluating this integral which sheds more light on the relation between it and $\zeta(2)$?

With this in mind, I should also be precise about what I am accepting as an answer:

T's & C's: If no such "deeper relation" between question and answer is apparent to anyone at all, then I will accept "it's just a coincidence" as an answer. If a connection is apparent to somebody, but it involves mathematics that you fear may be beyond me, feel free to post it anyway if you wish, and I'll do my best to understand what you've said.

By a "deeper relation", I mean any interpretation or rephrasing of the question into terms beyond elementary calculus; other areas of mathematics, or even physical interpretations, will do. Other integrals which are surprising in a similar way may also be helpful.

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Differentiation under the integral sign gives: $$ \int_{0}^{1}\log(x)\log(1-x)\,dx = \left.\frac{\partial^2}{\partial\alpha\,\partial\beta}\int_{0}^{1}x^\alpha(1-x)^{\beta}\,dx\right|_{\alpha,\beta=0} $$ hence you just have to differentiate a beta function and $\zeta(2)$ arises as $\psi'(1)$.

Differentiation is carried on through: $$ \frac{d}{dz}\,f(z) = f(z)\cdot\frac{d}{dz}\log f(z)$$ and $\psi(z)=\frac{d}{dz}\log\Gamma(z)$.

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    $\begingroup$ I had a feeling there would be something like this, which is why I asked the question, but my course hasn't reached complex analysis yet. I am attempting to work through your hints; I see that differentiation under the integral sign gives the required result, and I see that $\Gamma$ arises due to its relation to the beta function, I'm currently puzzling over the details at the end, and where exactly $\psi^{\prime}(1)$ comes in. Thank you! $\endgroup$ – Will R Aug 29 '15 at 21:08
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    $\begingroup$ I've worked through as much as I can at the moment; in actual fact, the value is $\psi^{\prime}(2)=\frac{\pi^{2}}{6}-1$ (source: trigamma function Wikipedia page), rather than $\psi^{\prime}(1)$, but I'm sure it's not a coincidence that the two values are similar. To give full details for closure, the result is $$\mathrm{B}(1,1)[(\psi(1)-\psi(2))^{2}-\psi^{\prime}(2)]=1\cdot[(-1)^{2}-\psi^{\prime}(2)] = 2-\frac{\pi^{2}}{6}.$$ Thank you; it is likely that my adventure here will influence my future studies. $\endgroup$ – Will R Aug 29 '15 at 22:46
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    $\begingroup$ @WillR: that is not a chance. The functional equation of the digamma function gives $$\psi(x+1)=\psi(x)+\frac{1}{x},$$ hence by differentiation we get: $$\psi'(x+1)=\psi'(x)-\frac{1}{x^2}$$ and we just need to evaluate at $x=1$, then recall that $\psi'(1)=\text{Li}_2(1)=\zeta(2)$. $\endgroup$ – Jack D'Aurizio Aug 29 '15 at 23:24
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This has to do with the fact that using integration by parts you get a dilogarithm, and $\text{Li}_2(1)=\zeta(2)$, since the two defining series happen to be the same at $x=1$.

Why $\zeta(2)$ involves $\pi^2$ can be seen as a by-product of the series expansion of $\sin(x)$, as can be read about here, on the solution to the Basel problem. In short one shows that $\sin x /x$ can be factored as a product involving $n^2/\pi^2$, and comparing coefficients gives the desired sum.

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Using the integration by parts, one has \begin{eqnarray} \int_0^1\ln x\ln(1-x)dx&=&x\ln x\ln(1-x)|_0^1-\int_0^1x(\frac{\ln(1-x)}{x}-\frac{\ln x}{1-x})dx\\ &=&-\int_0^1(\ln x+\ln(1-x))dx+\int_0^1\frac{\ln x}{1-x}dx\\ &=&2-\frac{\pi^2}{6}. \end{eqnarray} Here $$ \int_0^1\frac{\ln x}{1-x}dx=-\frac{\pi^2}{6} $$ is well-known.

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  • $\begingroup$ This is very interesting; the integral whose value you state as well-known is not one that I've seen before. Can you suggest any sources where this is mentioned, e.g., where/when do you remember first hearing about it? $\endgroup$ – Will R Aug 30 '15 at 23:47
  • $\begingroup$ @WillR I know this is an old thread, but for future readers make the substitution u=1-x and then see Adam Hughes' answer which makes an incorrect substitution to convert a function into the above integral followed by the correct proof of the above integral. math.stackexchange.com/questions/1805230/… $\endgroup$ – user5389726598465 Jul 16 '17 at 0:57

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