International Mathematics Competition for University Students

July 27 - Aug 2 2015, Blagoevgrad, Bulgaria

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Problem 7

7. Compute $$\lim_{A\to+\infty}\frac1A\int_1^A A^{\frac1x}\dx\,.$$

Proposed by Jan Ĺ ustek, University of Ostrava

Solution 1. We prove that $$\lim_{A\to+\infty}\frac1A\int_1^A A^{\frac1x}\dx = 1 \,.$$

For $A>1$ the integrand is greater than~$1$, so $$\frac1A \int_1^A A^{\frac1x}\dx > \frac1A \int_1^A 1\dx = \frac1A(A-1) = 1-\frac1A.$$

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: $$\frac1A \int_1^A A^{\frac1x}\dx = \frac1A\left( \int_1^{1+\delta} + \int_{1+\delta}^{K\log A} + \int_{K\log A}^A \right) <$$ $$= \frac1A\left( \delta\cdot A + (K\log A-1-\delta) A^{\frac1{1+\delta}} + (A-K\log A) A^{\frac1{K\log A}} \right) <$$ $$< \frac1A\left( \delta A + KA^{\frac1{1+\delta}}\log A + A\cdot A^{\frac1{K\log A}} \right) = \delta+A^{-\frac{\delta}{1+\delta}}\log A + e^{\frac1K}.$$

Hence, for $A>A_0(\delta,K)$ we have $$1-\frac1A < \frac1A \int_1^A A^{\frac1x}\dx < \delta+A^{-\frac{\delta}{1+\delta}}\log A + e^{\frac1K}.$$ Taking the limit $A\to\infty$ we obtain $$1 \le \liminf_{A\to\infty} \frac1A \int_1^A A^{\frac1x}\dx \le \limsup_{A\to\infty} \frac1A \int_1^A A^{\frac1x}\dx \le \delta + e^{\frac1K}.$$ Now from $\delta\to1+0$ and $K\to\infty$ we get $$1 \le \liminf_{A\to\infty} \frac1A \int_1^A A^{\frac1x}\dx \le \limsup_{A\to\infty} \frac1A \int_1^A A^{\frac1x}\dx \le 1,$$ so $\liminf\limits_{A\to\infty} \frac1A \int_1^A A^{\frac1x}\dx =\limsup\limits_{A\to\infty} \frac1A \int_1^A A^{\frac1x}\dx=1$ and therefore $$\lim_{A\to+\infty}\frac1A\int_1^A A^{\frac1x}\dx = 1 \,.$$

Solution 2. We will employ l'Hospital's rule.

Let $f(A,x)=A^{\frac1x}$, $g(A,x)=\frac1x A^{\frac1x}$, $F(A)=\int_1^A f(A,x)\dx$ 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 $$F'(A) = f(A,A) + \int_1^A \frac{\partial}{\partial A}f(A,x)\dx = A^{\frac1A} + \int_1^A \frac1x A^{\frac1x-1} \dx = A^{\frac1A} + \frac1A G(A),$$ and $$G'(A) = g(A,A) + \int_1^A \frac{\partial}{\partial A}g(A,x)\dx = \frac{A^{\frac1A}}{A} + \int_1^A \frac1{x^2} A^{\frac1x-1} \dx =$$ $$= A^{\frac1A} + \left[ \frac{-1}{\log A} A^{\frac1x-1} \right]_1^A = \frac{A^{\frac1A}}{A} - \frac{A^{\frac1A}}{A\log A} + \frac1{\log A} .$$

Since $\lim_{A\to\infty} A^{\frac1A}=1$, we can see that $\lim_{A\to\infty} G'(A)=0$. 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 $$\lim_{A\to\infty} F'(A) = \lim_{A\to\infty} \left( A^{\frac1A} + \frac{G(A)}{A} \right) = 1+0 = 1.$$

Now applying l'Hospital's rule to $\lim\limits_{A\to\infty}\frac{F(A)}{A}$ we get $$\lim_{A\to+\infty}\frac1A\int_1^A A^{\frac1x}\dx = \lim_{A\to\infty} \frac{F(A)}{A} = \lim_{A\to\infty} \frac{F'(A)}{1} = 1 .$$