IMC 2015 · Day 2 · P7

We prove that $$\lim_{A \to +\infty} \frac{1}{A} \int_1^A A^{\frac{1}{x}}\,dx = 1.$$ For $A > 1$ the integrand is greater than 1, so $$\frac{1}{A} \int_1^A A^{\frac{1}{x}}\,dx > \frac{1}{A} \int_1^A 1\,dx = \frac{1}{A} (A - 1) = 1 - \frac{1}{A}.$$ In order to find a tight upper bound, fix two real numbers, $\delta > 0$ and $K > 0$, and split the interval into three parts at the points $1 + \delta$ and $K \log A$. Notice that for sufficiently large $A$ (i.e., for $A > A_0(\delta, K)$ with some $A_0(\delta, K) > 1$) we have $1 + \delta < K \log A < A$.) For $A > 1$ the integrand is decreasing, so we can estimate it by its value at the starting points of the intervals: $$\begin{align*} \frac{1}{A} \int_1^A A^{\frac{1}{x}}\,dx &= \frac{1}{A} \left( \int_1^{1+\delta} + \int_{1+\delta}^{K \log A} + \int_{K \log A}^{A} \right) < \\$$

&= \frac{1}{A} \Bigl( \delta \cdot A + (K \log A - 1 - \delta) A^{\frac{1}{1+\delta}} + (A - K \log A) A^{\frac{1}{K \log A}} \Bigr) < \\ &< \frac{1}{A} \Bigl( \delta A + K A^{\frac{1}{1+\delta}} \log A + A \cdot A^{\frac{1}{K \log A}} \Bigr) = \delta + K A^{-\frac{\delta}{1+\delta}} \log A + e^{\frac{1}{K}}. \end{align*} Hence, for $A > A_0(\delta, K)$ we have $$1 - \frac{1}{A} < \frac{1}{A} \int_1^A A^{\frac{1}{x}}\,dx < \delta + K A^{-\frac{\delta}{1+\delta}} \log A + e^{\frac{1}{K}}.$$ Taking the limit $A \to \infty$ we obtain $$1 \le \liminf_{A \to \infty} \frac{1}{A} \int_1^A A^{\frac{1}{x}}\,dx \le \limsup_{A \to \infty} \frac{1}{A} \int_1^A A^{\frac{1}{x}}\,dx \le \delta + e^{\frac{1}{K}}.$$ Now from $\delta \to +0$ and $K \to \infty$ we get $$1 \le \liminf_{A \to \infty} \frac{1}{A} \int_1^A A^{\frac{1}{x}}\,dx \le \limsup_{A \to \infty} \frac{1}{A} \int_1^A A^{\frac{1}{x}}\,dx \le 1,$$ so $\liminf\limits_{A \to \infty} \frac{1}{A} \int_1^A A^{\frac{1}{x}}\,dx = \limsup\limits_{A \to \infty} \frac{1}{A} \int_1^A A^{\frac{1}{x}}\,dx = 1$ and therefore $$\lim_{A \to +\infty} \frac{1}{A} \int_1^A A^{\frac{1}{x}}\,dx = 1.$$

We will employ l'Hospital's rule.

Let $f(A, x) = A^{\frac{1}{x}}$, $g(A, x) = \frac{1}{x} A^{\frac{1}{x}}$, $F(A) = \int_1^A f(A, x)\,dx$ and $G(A) = \int_1^A g(A, x)\,dx$. Since $\frac{\partial}{\partial A} f$ and $\frac{\partial}{\partial A} g$ are continuous, the parametric integrals $F(A)$ and $G(A)$ are differentiable with respect to $A$, and $$F'(A) = f(A, A) + \int_1^A \frac{\partial}{\partial A} f(A, x)\,dx = A^{\frac{1}{A}} + \int_1^A \frac{1}{x} A^{\frac{1}{x} - 1}\,dx = A^{\frac{1}{A}} + \frac{G(A)}{A},$$ and $$\begin{align*} G'(A) &= g(A, A) + \int_1^A \frac{\partial}{\partial A} g(A, x)\,dx = \frac{A^{\frac{1}{A}}}{A} + \int_1^A \frac{1}{x^2} A^{\frac{1}{x} - 1}\,dx = \\ &= \frac{A^{\frac{1}{A}}}{A} + \left[ \frac{-1}{\log A} A^{\frac{1}{x} - 1} \right]_1^A = \frac{A^{\frac{1}{A}}}{A} - \frac{A^{\frac{1}{A}}}{A \log A} + \frac{1}{\log A}. \end{align*}$$ Since $\lim\limits_{A \to \infty} A^{\frac{1}{A}} = 1$, we can see that $\lim\limits_{A \to \infty} G'(A) = 0$. Aplying l'Hospital's rule to $\lim\limits_{A \to \infty} \frac{G(A)}{A}$ we get $$\lim_{A \to \infty} \frac{G(A)}{A} = \lim_{A \to \infty} \frac{G'(A)}{1} = 0,$$ so $$\lim_{A \to \infty} F'(A) = \lim_{A \to \infty} \left( A^{\frac{1}{A}} + \frac{G(A)}{A} \right) = 1 + 0 = 1.$$ Now applying l'Hospital's rule to $\lim\limits_{A \to \infty} \frac{F(A)}{A}$ we get $$\lim_{A \to +\infty} \frac{1}{A} \int_1^A A^{\frac{1}{x}}\,dx = \lim_{A \to \infty} \frac{F(A)}{A} = \lim_{A \to \infty} \frac{F'(A)}{1} = 1.$$