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Playing around with a proof of a generalized Hardy-Littlewood Tauberian theorem (I'll state the theorem at the end of this question), I ran into the following proposition:

Let $$F_n(\alpha) = \frac{1}{n}\sum_{k=1}^{n-1} \bigg(\log\frac{n}{k}\bigg)^\alpha.$$ Then $\lim_{n\to\infty} F_n(\alpha) = \Gamma(\alpha+1)$.

I have a proof in the case $\alpha\geq 0$, which is the case I'm interested in. I'll add it as an answer in a moment. But I want to post this as a question because it's interesting, and in case others come up with different proofs and might like to add their answers.


For posterity, here's the generalized Hardy-Littlewood Tauberian theorem. Suppose $\mu$ is a measure on $[0,\infty)$ whose cumulative distribution $D_\mu(x) = \mu([0,x])$ is of bounded variation. Let $\omega(s) = \int_0^\infty e^{-x s}d\mu(x)$ be its Laplace transform. Then: $$D_\mu(x)\sim x^\alpha\ \mbox{ as $x\to\infty$ if and only if }\ \omega(s)\sim s^{-\alpha}\ \mbox{ as $s\to 0$}.$$ I was playing with the proof in Ch 8 of the second volume of Taylor's book Partial Differential Equations.

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    $\begingroup$ It's the same trick from calculus where you look at a Riemann sum and make it into an integral, except they did a small change of variables first. You can actually do this without any big machines at all. $\endgroup$ – Adam Hughes Sep 20 '16 at 23:44
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First, notice that in fact $$F_n(\alpha) = \frac{1}{n}\sum_{k=1}^n \log(n/k)^\alpha,$$ as the last term is equal to zero.

Define $f(t) = -\log(t)^\alpha$. We may reinterpret $F_n$ as a Riemann sum for $f$ over $[0,1]$ for the partition $t_k = k/n$, where $k$ ranges over $1,\ldots,n$. Then $$ F(\alpha) = \lim_{n\to\infty} F_n(\alpha) = \int_0^1 f(t)\ dt = \int_0^1 (-\log t)^\alpha\ dt $$ Now we exploit the fact that the logarithm is monotone: we perform a substitution. Let $s = -\log t$, so that $t = e^{-s}$ and $ds = -dt/t$. Therefore $dt = -e^{-s}ds$. The limits of integration $0$ and $1$ become $\infty$ and $0$, respectively. The minus sign in $ds$ reverses the limits, and voila! $$ F(\alpha) = \int_0^\infty s^\alpha e^{-s}\ ds = \Gamma(\alpha+1).$$

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    $\begingroup$ this is a good question as was your solution. FYI, you can skip the last step and go directly to the gamma function. Euler originally defined the gamma function as a factorial via an infinite product and then with the integral, $n! = \int_{0}^{1} (-\ln(t))^{n} \mathrm{d}t$ for $n \in \mathbb{Z}^{+}$. Of course your analysis recovers the more conventional integral form. $\endgroup$ – poweierstrass Oct 9 '16 at 17:48

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