I want to show that

$$f(x) =\sum_{k=1}^\infty\frac{x^{2k}}{k!}\ln{k}$$ asymptotically approaches

$$2e^{x^2}\ln x$$

in the limit of $x \to \infty$. Any help would be appreciated.

The things I tried, which didn't work:

1) I tried to follow the same approach as here, i.e. to express the derivative $f'(x)$ in terms of some function $g(x)$, such that $f'(x) = 2x g(x) f(x)$ and then try to estimate the behaviour of $g(x)$ using the identity:

$$\lim_{x\to \infty} \frac{a_n x^n}{b_n x^n} =\lim_{n\to \infty} \frac{a_n}{b_n}.$$

It is possible to show that $g(x) = 1 + o(1)$, however in the next step I faced the original problem of finding $$\lim_{x \to \infty} \frac{2e^{x^2}\ln x}{f(x)}$$

2) I tried to expand $\ln k $ into different series representations, such as:

$$\ln k = 2\sum_n \frac{1}{2n-1} \left(\frac{k-1}{k+1} \right)^{2n-1}, $$ or $$\ln k = \sum_n \frac{1}{n} \left(\frac{k-1}{k} \right)^{n}, $$ to try to gather the desired asymptotic expression. There are closed forms of the resulting summations, however it was hard for me to evaluate them at arbitrary n.

3) I tried to look at the integral representation of $f(x)$: $$ f(x)=\oint_{C} \ln(-\rm{i}\,a)e^{-{\rm i}\,a \ln(x)-\ln[\Gamma(1-{\rm i}\, a)]}\coth(\pi a)\frac{d a}{2{\rm i}},$$ where $C$ is around the positive part of the imaginary axis. The $\coth$ is a series of delta-functions along the imaginary axis, therefore the sum of residues of the remaining expression under the integral gives the original summation.

The function: $$I(a,x) = \ln(a)e^{a \ln(x)-\ln[\Gamma(1+ a)]}$$ has a maximum along $a$. One can get the position $a_0$ of this maximum from $\partial_a I=0$ and try to approximate $I$ by a Gaussian (in $a$) centered around $a_0$. The result of this looks reasonable numerically, however it is still hard to see how it might lead to the simple $2e^{x^2}\ln x$ behaviour.

  • $\begingroup$ One observation is that you can substitute $t=x^2$, in which case the problem is to show that $g(t)=\sum_{k=1}^\infty \frac{t^k}{k!}\ln k \sim e^t \ln t$ as $t\to+\infty$. So the squares are in that respect superfluous. As far as how to show this, you might try considering the product $e^{-t}g(t)$ and see how the resulting product of power series behaves. $\endgroup$ Dec 10, 2017 at 18:39
  • $\begingroup$ @Semiclassical: the most reasonable approach, indeed. The interesting part is that $e^{-t}g(t)$ has a manageable integral representation. $\endgroup$ Dec 10, 2017 at 18:50
  • 3
    $\begingroup$ A straight forward modification of the argument here gives us the result. A possibly interesting way of rephrasing the problem, we are trying to estimate $\mathbb{E}(\log N_x)$ for large $x$ where, $N_x$ is a poisson random variable with parameter $x$ (use Jensen's inequality to estimate the same). $\endgroup$
    – r9m
    Dec 10, 2017 at 19:19
  • 1
    $\begingroup$ Related: math.stackexchange.com/a/2510830/44121 $\endgroup$ Dec 10, 2017 at 19:22

1 Answer 1


$$ e^{-x^2} = \sum_{k\geq 0}\frac{x^{2k}}{k!}(-1)^k $$

$$\begin{eqnarray*} e^{-x^2}f(x) &=& \sum_{m\geq 1}\frac{x^{2m}}{m!}\sum_{n=1}^{m}\binom{m}{n}(-1)^{m-n}\log(n)\\&=&\sum_{m\geq 1}\frac{x^{2m}}{m!}\int_{0}^{\infty}\sum_{n=1}^{m}\binom{m}{n}(-1)^{m-n}\frac{e^{-z}-e^{-nz}}{z}\,dz\\&=&\sum_{m\geq 1}\frac{x^{2m}(-1)^{m+1}}{m!}\int_{0}^{+\infty}\frac{e^{-z}-1+(1-e^{-z})^m}{z}\,dz\\&=&\sum_{m\geq 1}\frac{x^{2m}(-1)^{m}}{m!}\int_{0}^{1}\frac{u^m-u}{(1-u)\log(1-u)}\,du\\&=&\int_{0}^{1}\frac{-1+e^{-u x^2}+u-e^{-x^2} u}{(1-u)\log(1-u)}\,du\\&=&\int_{0}^{1}\frac{u-e^{-x^2} \left(-1+e^{u x^2}+u\right)}{u\log u}\,du\end{eqnarray*}$$ and the claim now follows from Laplace method, since $$ \int_{0}^{1}\frac{u^k-1}{\log u}\,du = \log(k+1).$$

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ Dec 10, 2017 at 20:51

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