Computation of the limit of $\frac{1}{A} \int_{1}^{A} A^{1/x} \,dx$ as $A\to \infty$ Compute the limit (if there is one) of 
$$ \frac{1}{A} \int_{1}^{A} A^{1/x}\, dx $$
as $A \to \infty$. 
--
I actually managed to solve it, and although I know the policy of MathSE, I found it so nice I wanted to share this with you guys (this is also the purpose of this forum to share interesting problems) : enjoy ! For those interested, many more similar questions may be found here https://www.ens.fr/sites/default/files/2019_mathsulm_sujets-1.pdf
-- 
As mentioned by @zhw in the comments, the question has already been raised at least three times in this forum:
How to compute: $ \lim\limits_{a\to \infty}\frac1a\int_1^a a^{\frac1x} dx$
Computing $\lim_{A\to\infty} \frac{1}{A} \int\limits_1^A \! A^{\frac{1}{x}} \, \mathrm{d}x.$
Problem 7 IMC 2015 - Integral and Limit
Beware the accepted answer to the first of these questions is formally wrong (no domination). Among these solutions, the quicker ones are the one that perform a change of variable, $A^{1/x}= A^{y}$ or $A^{1/x}=e^z$, and the one below. One may also dispense from the use of L'Hopital rule.
-- 
To be complete, let me add this question seems to be due to Jan Sustek, University of Ostrava, and that the solution sheet for the contest they come from (ICM 2015) is to be found here
 A: Use L'Hospital's rule and Leibniz integral rule to get
$$
\lim_{A \to \infty} \frac{1}{A} \int_{1}^{A} A^{1/x} dx = \lim_{A \to \infty} \big(A^{1/A} + \frac{1}{A}\int_{1}^A \frac{A^{1/x}}{x} dx \big) = 1 + \lim_{A \to \infty} \frac{1}{A}\int_{1}^A \frac{A^{1/x}}{x} dx
$$
In order to compute obtained integral use the same technique once again:
$$
\lim_{A \to \infty} \frac{1}{A}\int_{1}^A \frac{A^{1/x}}{x} dx = \lim_{A \to \infty}  \big( \frac{A^{1/A}}{A} + \frac{1}{A}\int_{1}^A \frac{A^{1/x}}{x^2} dx\big) = \lim_{A \to \infty} \frac{1}{A}\int_{1}^A \frac{A^{1/x}}{x^2} dx
$$
Now the last integral can be evaluated directly with substitution $t = 1/x$. Once it's calculated, we establish
$$
\lim_{A \to \infty} \frac{1}{A}\int_{1}^A \frac{A^{1/x}}{x^2} dx = 0,
$$
and the answer to the problem is
$$
\lim_{A \to \infty} \frac{1}{A} \int_{1}^{A} A^{1/x} dx = 1.
$$
A: I think that we can obtain more than the limit itself.
$$\int A^{\frac{1}{x}}  dx=x A^{\frac{1}{x}}-\log (A) \text{Ei}\left(\frac{\log (A)}{x}\right)$$
$$\int_1^A A^{\frac{1}{x}}  dx=A^{\frac{1}{A}+1}+\log (A) \text{Ei}(\log (A))-\log (A) \text{Ei}\left(\frac{\log(A)}{A}\right)-A$$
and use 
$$\text{Ei}(t)=\gamma+\log (t) +t+\frac{t^2}{4}+O\left(t^3\right)$$
$$\text{Ei}(t)=e^t
   \left(\frac{1}{t}+\frac{1}{t^2}+O\left(\frac{1}{t^3}\right)\right)$$ 
which make
$$\int_1^A A^{\frac{1}{x}}  dx\sim A^{\frac{1}{A}+1}-\frac{\log ^3(A)}{4 A^2}-\frac{\log ^2(A)}{A}+\frac{A}{\log
   (A)}-\log (A) \left(\log \left(\frac{\log (A)}{A}\right)+\gamma \right)+\cdots$$
$$\frac 1 A \int_1^A A^{\frac{1}{x}}  dx\sim A^{\frac{1}{A}}-\frac{\log ^3(A)}{4 A^3}-\frac{\log ^2(A)}{A^2}+\frac{1}{\log
   (A)}-\frac{\log (A)}A \left(\log \left(\frac{\log (A)}{A}\right)+\gamma \right)+\cdots$$
