Problem 7 IMC 2015 - Integral and Limit

For the integral in question: $$\large\lim_{A\to\infty} \, \frac{1}{A} \, \int_{1}^{A} A^{\frac{1}{x}}\,dx$$ the following is considered: \begin{align} A^{\frac{1}{x}} = e^{\frac{1}{x} \, \ln A} = 1 + \frac{\ln A}{x} + \sum_{k=2}^{\infty} \frac{\ln^{k}A}{k! \, x^{k}} \end{align} for which \begin{align} \int_{1}^{A} A^{\frac{1}{x}} \, dx &= [ x ]_{1}^{A} - \ln A \, [ \ln x ]_{1}^{A} + \sum_{k=2}^{\infty} \frac{\ln^{k}A}{k! \, (1-k)} \, [ x^{1-k} ]_{1}^{A} \\ &= A - 1 + \ln A - \sum_{k=2}^{\infty} \frac{1}{k! \, (k-1)} \, \left(\frac{A \, \ln^{k}A}{A^{k}} - \frac{\ln^{k}A}{1} \right). \end{align} Now dividing by $A$ leads to \begin{align} \frac{1}{A} \, \int_{1}^{A} A^{\frac{1}{x}} \, dx &= 1 - \frac{1}{A} + \frac{\ln A}{A} - \sum_{k=2}^{\infty} \frac{1}{k! \, (k-1)} \, \left(\frac{ \ln^{k}A}{A^{k}} - \frac{\ln^{k}A}{A} \right). \end{align} In order to evaluate the neccessary limits the following components are needed: \begin{align} \lim_{A \to \infty} \frac{\ln A}{A} &\to 0 \\ \lim_{A \to \infty} \frac{\ln^{k}A}{A^{k}} &= \left( \lim_{A \to \infty} \frac{\ln A}{A} \right)^{k} \to 0 \\ \lim_{A \to \infty} \frac{\ln^{k}A}{A} &= k \, \lim_{A \to \infty} \frac{\ln^{k-1}A}{A} = \cdots = \frac{k!}{1!} \, \lim_{A \to \infty} \frac{\ln A}{A} = 0. \end{align} With these limiting values it is evident that \begin{align} \lim_{A \to \infty} \, \frac{1}{A} \, \int_{1}^{A} A^{\frac{1}{x}} \, dx &= \lim_{A \to \infty} \left[ 1 - \frac{1}{A} + \frac{\ln A}{A} - \sum_{k=2}^{\infty} \frac{1}{k! \, (k-1)} \, \left(\frac{ \ln^{k}A}{A^{k}} - \frac{\ln^{k}A}{A} \right) \right] \\ &= 1 - \lim_{A \to \infty} \frac{1}{A} - \sum_{k=2}^{\infty} \frac{1}{k! \, (k-1)} \, \lim_{A \to \infty} \left(\frac{ \ln^{k}A}{A^{k}} - \frac{\ln^{k}A}{A} \right) \\ &= 1. \end{align}

The key to showing that this limit converges to $1$ is to show that the function $A^{1/x}$ decreases quickly towards $1$ as $x$ increases beyond $1$, so that the integral isn't significantly larger than $\int_1^A 1\,dx \sim A$. To this end we will focus on showing that the difference (which is clearly positive) is small enough:

$$\int_1^A (A^{1/x} - 1)\,dx = o(A).$$

This can be done be with three piecewise-constant approximations:

1) We first consider $x$ close to $1$, where we don't have an upper bound better than $A^{1/x} \le A$. So to get a contribution of $o(A)$ we need to cut it off at $x = 1 + o(1)$. We want $A^{1/x}$ to shrink down to a decent amount, so it behooves us to choose the $o(1)$ to be very slowly shrinking, say at $x = 1 + 2/\sqrt{\ln A}$. Then the contribution from this piece is $$\int_1^{1 + 2/\sqrt{\ln A}} (A^{1/x} - 1)\, dx \le \int_1^{1 + 2/\sqrt{\ln A}} A\, dx = \frac{2A}{\sqrt{\ln A}} = o(A).$$

2) Now we consider $x \ge 1 + 2/\sqrt{\ln A}$. For large enough $A$, we have $1/x \le 1 - 1/\sqrt{\ln A}$. This means for $x \ge 1 + 2/\sqrt{\ln A}$, $$\exp\big(\tfrac{\ln A}{x}\big) \le A \exp\big(-\tfrac{\ln A}{\sqrt{\ln A}}\big) = A \exp(-\sqrt{\ln A}).$$

This is considerably smaller than $A$, so we are free to use this bound for a sizable range of $x$. We'd like to go beyond $x = \ln A$ which is where $A^{1/x}$ actually gets close to $1$. Happily, this goal is in reach as (for all large $A$) $\sqrt{\ln A} > 3 \ln \ln A$, so $$ \exp(-\sqrt{\ln A}) < \exp(-3 \ln \ln A) = 1/(\ln A)^3.$$

That means we can set the next cutoff at $x = (\ln A)^2$, and we still get $$\int_{1 + 2/\sqrt{\ln A}}^{(\ln A)^2} A^{1/x} - 1 \, dx < \int_0^{(\ln A)^2} A \exp(-\sqrt{\ln A})\, dx < A \frac{(\ln A)^2}{(\ln A)^3} = o(A).$$

3) Now the last piece from $x = (\ln A)^2$ to $A$. We chose the previous cutoff to make $A^{1/x}$ very close to one, indeed for $x \ge (\ln A)^2$ we have

$$A^{1/x} \le \exp(1/\ln A) = 1 + o(1),$$ since $\exp(1/\ln A) \to 1$ as $A \to \infty$ (we could be more precise using Taylor expansion and say that the RHS is $1 + O(1/\ln A)$). Therefore the contribution from this third interval is

$$\int_{(\log A)^2}^A (A^{1/x} - 1)\, dx \le \int_0^A (\exp(1/\ln A) - 1)\, dx. = \int_0^A o(1)\, dx = o(A).$$

Combining 1), 2), 3) we have the desired upper bound. Although the calculations are a little messy the argument is elementary and doesn't require integrating infinite series term-by-term.