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Question:$$\int\limits_0^1\mathrm dx\,\frac {\log\log\frac 1x}{(1+x)^2}=\frac 12\log\frac {\pi}2-\frac {\gamma}2$$

I’ve had some practice with similar integrals, but this one eludes me for some reason. I first made the transformation $x\mapsto-\log x$ to get rid of the nested log. Therefore$$\mathfrak{I}=\int\limits_0^{\infty}\mathrm dx\,\frac {e^{-x}\log x}{(1+e^{-x})^2}$$ The inside integrand can be rewritten as an infinite series to get$$\mathfrak{I}=\sum\limits_{n\geq0}(n+1)(-1)^n\int\limits_0^{\infty}\mathrm dx\, e^{-x(n+1)}\log x$$The inside integral, I thought, could be evaluated by differentiating the gamma function to get$$\int\limits_0^{\infty}\mathrm dt\, e^{-t(n+1)}\log t=-\frac {\gamma}{n+1}-\frac {\log(n+1)}{n+1}$$ However, when I simplify everything and split the sum, neither sum converges. If we consider it as a Cesaro sum, then I know for sure that$$\sum\limits_{n\geq0}(-1)^n=\frac 12$$Which eventually does give the right answer. But I’m not sure if we’re quite allowed to do that especially because in a general sense, neither sum converges.

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By the dominated convergence theorem we have $$ \mathfrak{I} = \lim_{r \nearrow 1} I(r) \, ,$$ where for $r \in (0,1)$ we have defined $$ I(r) = \int\limits_0^1 \mathrm{d} x\,\frac {\log\log\frac 1x}{(1+r x)^2} \, . $$ With this regularisation interchanging summation and integration is actually justified and your calculations lead to $$ I(r) = - \gamma \sum \limits_{n=0}^\infty (-r)^n - \sum \limits_{n=0}^\infty (-r)^n \log(1+n) \equiv I_1 (r) + I_2(r) \, . $$ The first sum is easy: $$ I_1(r) = - \frac{\gamma}{1+r} \, , $$ so $\lim_{r \nearrow 1} I_1(r) = - \frac{\gamma}{2}$ .

For the second sum we can write \begin{align} I_2(r) &= \frac{1}{r} \sum_{n=1}^\infty (-r)^n \log(n) \\ &= \frac{1}{2r} \sum_{k=1}^\infty [2 r^{2k} \log(2k) - r^{2k-1} \log(2k-1) - r^{2k+1} \log(2k+1)] \\ &= \frac{1}{2r} \sum_{k=1}^\infty r^{2k} \left[\log\left(\frac{4k^2}{4k^2-1}\right) + (1-r) \log(2k+1) - \frac{1}{r} (1-r) \log(2k-1)\right] \\ &= \frac{1}{2r} \sum_{k=1}^\infty r^{2k} \left[\log\left(\frac{4k^2}{4k^2-1}\right) + (1-r)^2 \log(2k+1)\right] \, . \end{align} The second term can be estimated by \begin{align} \frac{(1-r)^2}{2r} \sum_{k=1}^\infty r^{2k} \log(2k+1) &\leq \frac{(1-r)^2}{2r^2} \sum_{n=1}^\infty \sqrt{n} r^{n} \\ &= \frac{(1-r)^2}{2r^2} \operatorname{Li}_{-1/2} (r) \\ &= \frac{\sqrt{\pi}}{4 r^2} \sqrt{1-r} + \mathcal{O} \left((1-r)^{3/2}\right) \end{align} as $r \nearrow 1$. The asymptotic behaviour of the polylogarithm can be deduced from the series given here (the second one below 2.). Now we can use the monotone convergence theorem and Wallis' product to find $$ \lim_{r \nearrow 1} I_2 (r) = \frac{1}{2} \sum_{k=1}^\infty \log\left(\frac{4k^2}{4k^2-1}\right) = \frac{1}{2} \log \left(\prod_{k=1}^\infty \frac{4k^2}{4k^2-1}\right) = \frac{1}{2} \log \left(\frac{\pi}{2}\right) \, . $$

Therefore $$ \mathfrak{I} = \frac{1}{2} \left[\log \left(\frac{\pi}{2}\right) - \gamma\right]$$ as claimed.

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  • $\begingroup$ Confirmed by Mathematica: $\frac{1}{2} \left(\log \left(\frac{\pi }{2}\right)-\gamma \right)$ $\endgroup$ – David G. Stork Jul 16 '18 at 21:51

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