Computing $\lim_{A\to\infty} \frac{1}{A} \int\limits_1^A \! A^{\frac{1}{x}} \, \mathrm{d}x.$ On this year's IMC there was this problem:
Compute
$$
\lim_{A\to\infty} \frac{1}{A} \int\limits_1^A \! A^{\frac{1}{x}} \, \mathrm{d}x.
$$
In addition to the two official solutions, I am curious as to if there exist other solutions. 
I heard that someone wrote that this is actually a probability density function (just for this one line, that person got 6 out of 10 points), so it would be great to see an answer involving this claim.
Also, I thought of turning this integral (using substitutions) to something I can evaluate using the Gamma function. Could someone hint a substitution which could lead to that?
EDIT: The official solutions can be found here, Problem 7.
 A: Squeezing seems to provide a quick solution. Assume $A>e^2$. We have:
$$ A^{\frac{1}{x}}=\exp\left(\frac{\log A}{x}\right)\geq 1+\frac{\log A}{x}\tag{1}$$
hence:
$$ \int_{1}^{A}A^{\frac{1}{x}}\,dx \geq A+\left(\log^2(A)-1\right).\tag{2} $$
On the other hand:
$$\begin{eqnarray*}\int_{1}^{A}A^{\frac{1}{x}}\,dx = \int_{\frac{1}{A}}^{1}\frac{A^x}{x^2}\,dx &=&(A-1)-\left.\frac{A^x-1}{x}\right|_{\frac{1}{A}}^{1}+\log(A)\int_{\frac{1}{A}}^{1}\frac{A^x}{x}\,dx\\&=&A\left(A^{\frac{1}{A}}-1\right)+\log^2(A)+\log(A)\int_{\frac{1}{A}}^{1}\frac{A^x-1}{x}\,dx \end{eqnarray*}$$
and $\frac{A^x-1}{x}$ is a convex function on $(0,1]$, hence it is not difficult to find a tight upper bound for the LHS of $(2)$ through the Hermite-Hadamard inequality or Jensen's inequality. Another chance is given by proving, then exploiting,
$$ \forall x\in(0,1),\qquad \frac{A^{1-x}-1}{1-x}\leq (A-1)\exp\left[\log\left(\frac{\log A}{A-1}\right)\,x\right]\tag{3} $$
so by $(2)$ and $(3)$ it follows that the wanted limit is just $\color{red}{1}$.
A: $$\begin{aligned}\lim_{A\to+\infty}\frac1A\int_1^AA^{\frac1x}\,dx&\stackrel{\ln A=xt}{\!=\!=\!=}\lim_{A\to+\infty}\frac{\int_{1}^{\ln A}t^{-2}e^{t}\,dt-\int_{1}^{\frac{\ln A}{A}}t^{-2}e^{t}\,dt}{\frac{A}{\ln A}} \notag \\&\stackrel{(*)}{=}\lim_{A\to+\infty}\frac{d\left(\int_{1}^{\ln A}t^{-2}e^{t}\,dt-\int_{1}^{\frac{\ln A}{A}}t^{-2}e^{t}\,dt\right)/dA}{d\left(\frac{A}{\ln A}\right)/dA}\notag \\ &=\lim_{A\to+\infty}\frac{1+A^{1/A}(\ln A-1)}{\ln A-1}=1\end{aligned}$$
$(*)$ using strong DLH. 
A: Here is an alternative:
Upper bound: First, do the change of variables $y=x/A$. Then
$$
\frac{1}{A}\int_1^A A^{1/x}\,dx=\int_{1/A}^1 (A^{1/A})^{1/y}\,dy.
$$
Now, using the generalization of Bernoulli's inequality,
$$
(A^{1/A})^{1/y}=\bigl(1+(A^{1/A}-1)\bigr)^{1/y}\leq 1+\frac{1}{y}\bigl(A^{1/A}-1\bigr).
$$
Integrating, we get
$$
\int_{1/A}^1 (A^{1/A})^{1/y}\,dy\leq(1-1/A)+\log A\bigl(A^{1/A}-1\bigr)
$$
If we write
$$
0\leq \log A\bigl(A^{1/A}-1\bigr)=\log^2A\int_0^{1/A}A^x\,dx\leq\log^2(A)A^{1/A}\frac{1}{A},
$$
we see that the last term tends to zero as $A\to+\infty$.
Lower bound: Here, I'd like to use the Hermite-Hadamard inequality, also mentioned by Jack in his answer.
Since $x\mapsto A^{1/x}$ is convex,
$$
\frac{1}{A-1}\int_1^A A^{1/x}\,dx\geq A^{2/(1+A)},
$$
so
$$
\frac{1}{A}\int_1^A A^{1/x}\,dx\geq \frac{A-1}{A}A^{2/(1+A)},
$$
The limit of the right-hand side is clearly $1$.
Conclusion: It follows by the upper and lower bounds, and the squeeze theorem that
$$
\frac{1}{A}\int_1^A A^{1/x}\,dx \to 1,\quad\text{as}\quad A\to+\infty.
$$
A: $$
\begin{align}
\frac1a\int_1^aa^{1/x}\,\mathrm{d}x
&=\frac1a\int_1^a\sum_{k=0}^\infty\frac1{k!}\left(\frac{\log(a)}x\right)^k\,\mathrm{d}x\\
&=\frac1a\left(a-1+\log(a)^2+\sum_{k=2}^\infty\frac{\log(a)^k}{(k-1)k!}\left(1-\frac1{a^{k-1}}\right)\right)
\end{align}
$$
Since $\frac1{k-1}\le\frac3{k+1}$ for $k\ge2$, we have
$$
\begin{align}
\frac1a\sum_{k=2}^\infty\frac{\log(a)^k}{(k-1)k!}\left(1-\frac1{a^{k-1}}\right)
&\le\frac3{a\log(a)}\sum_{k=2}^\infty\frac{\log(a)^{k+1}}{(k+1)!}\\
&\le\frac3{\log(a)}
\end{align}
$$
Thus,
$$
\frac1a\int_1^aa^{1/x}\,\mathrm{d}x=1+O\!\left(\frac1{\log(a)}\right)
$$
Therefore,
$$
\lim_{a\to\infty}\frac1a\int_1^aa^{1/x}\,\mathrm{d}x=1
$$
