Bounding an intriguing integral from above I would like to show that the following function of the positive integers k and N does not grow to infinity with k, at least for some values of N.
$$
F(k,N) = \int_1^{k^2} e^{\log (N/k) \sqrt x } e^{ x/k } \textrm{d}x
$$
In other words, my goal is to find $N \in \mathbb{N}_{>0}$ and $M < \infty$ such that $f(k,N) < M$ for every $k$.  How could I upper bound the integral efficiently? In the ideal case such a finite upper bound on $F(k,N)$ (uniform in $k$) will not grow faster than linearly with $N$...
 A: This is more of a supplement to Gary's answer, with the goal of avoiding the use of special functions.
There are two distinct behaviours depending on the size of $\log(N/k)$. Setting $m = \log(N/k)$, we have an integral of the form $$ \int_1^{k^2} e^{\frac{x +mk \sqrt{x}}{k}}.$$ Now notice that $(x +mk\sqrt{x}) < 0 \iff \sqrt{x} < -mk \iff x < m^2 k^2.$ So, in particular, if $m <- 1,$ then the integrand is upper bounded by $$ \int_1^{k^2} e^{(m+1)\sqrt{x} }\mathrm{d}x \le \int_0^\infty e^{(m+1)\sqrt{x}} \mathrm{d}x = \int_0^\infty \frac{2t}{(m+1)^2} e^{-t} \mathrm{d}t = \frac{2}{(m+1)^2},$$ where I've used that $\int _0^\infty te^{-t} = 1.$

If, on the other hand, $m > -1,$ then notice that the integrand is maximised near $k^2$, and grows quite large around there. This sets up the calculation $$ \int_1^{k^2} e^{\frac{x + mk\sqrt{x}}{k}} = \int_{1/k}^1 2k^2 t e^{k(t^2 +mt)} \le \int_0^1 2k^2 t e^{k(t^2 + mt)} \le e^{k(1+m)} \int_0^12k^2t = k^2 e^{k(1+m)}.$$
I'll note that for fixed $N/k$, the growth rate $k^2 e^{k(1+m)}$ is correct up to a factor of $k$. For a lower bound (for $k\ge 4$), observe that the slope of $t^2 + mt$ over $(0,1)$ is at most $2 + m$, and thus in $(1-1/k, 1),$ $t^2 + mt \ge 1 - (2+m)/k,$ giving $$ \int_{1/k}^1 2k^2 t e^{k(t^2 + mt)} \ge \int_{1 - 1/k}^1 2k^2 te^{k(t^2 + mt)} \ge \frac{2}{4e^{2}} ke^{-m} e^{k(1+m)} = \frac{k}{2e^2 (N/k)} e^{k(1+m)}$$
This quite clearly shows that $f(k,N)$ has a phase transition around the line $N/k = 1/e$ as $k$ grows large.
A: Let $m=\log(N/k)$ and assume $k \gg N$. Then
\begin{align*}
0 & < \int_1^{k^2 } {\exp \left( {x/k + \sqrt x \log (N/k)} \right)dx} 
\\ &
 = 2k^2 \int_{1/k}^1 {\exp \left( {k\left[ {t^2  + tm} \right]} \right)tdt} 
\\ &
 \le 2k^2 \int_0^1 {\exp \left( {k\left[ {t^2  + tm} \right]} \right)tdt} 
\\ &
 = \frac{{\sqrt \pi  }}{2}\left| m \right|e^{ - km^2 /4} k^{3/2} \left( {\operatorname{erfi}\left( {\frac{{\left| m \right|}}{2}\sqrt k } \right) - \operatorname{erfi}\left( {\frac{{\left| m \right| - 2}}{2}\sqrt k } \right)} \right) - k + ke^{ - k(\left| m \right| - 1)} 
\\ &
 < \frac{{\sqrt \pi  }}{2}\left| m \right|e^{ - km^2 /4} k^{3/2} \operatorname{erfi}\left( {\sqrt k \frac{{\left| m \right|}}{2}} \right) - k + ke^{ - k(\left| m \right| - 1)} 
\\ &
 \sim \frac{2}{{m^2 }} + ke^{ - k(\left| m \right| - 1)}  \sim \frac{2}{{m^2 }}
\end{align*}
for large $k$. Here I used the following two-term asymptotics of the imaginary error function:
$$
\operatorname{erfi}(z) \sim \frac{{e^{z^2 } }}{{\sqrt \pi  }}\left( {\frac{1}{z} + \frac{1}{{2z^3 }}} \right),\quad z\to +\infty.
$$
For small values of $k$ the integral is clearly bounded and for large values of $k$ it even tends to zero.
A: From an algebraic point of view.
We have the antiderivative
$$I=\int e^{a \sqrt{x}+\frac{x}{k}}\,dx$$
$$I=k e^{a+\frac{1}{k}} \left(a \sqrt{k} F\left(\frac{a k+2}{2 \sqrt{k}}\right)-1\right)-k
   e^{(a+1) k} \left(a \sqrt{k} F\left(\frac{(a+2) \sqrt{k}}{2} \right)-1\right)$$ where appears Dawson integral function.
Then, the analytical expression of $f(n,k)$is
$$\color{blue}{f(n,k)=e^{\frac{1}{k}} n \left(\sqrt{k} \log \left(\frac{n}{k}\right) F\left(\frac{k \log
   \left(\frac{n}{k}\right)+2}{2 \sqrt{k}}\right)-1\right)-}$$ $$\color{blue}{e^k k
   \left(\frac{n}{k}\right)^k \left(\sqrt{k} \log \left(\frac{n}{k}\right)
   F\left(\frac{1}{2} \sqrt{k} \left(\log
   \left(\frac{n}{k}\right)+2\right)\right)-1\right)}$$
Considering $k$ as a continuous variable, the maximum of $f(n,k)$ is attained at $k\sim n+\frac32$ which makes that we can compute exactly the maximum value of the function. The only problem is that the formulae are quite nasty.
What is pleasant is that
$$f(n,n)=\left(e^n-e^{\frac{1}{n}}\right) n$$ and, numerically,
$$\frac{f\left(n,n+\frac{3}{2}\right)}{f(n,n)} \sim 1+\frac 1{n}$$
If $k$ is an integer, it has been verified that the maximum value of the function happens for $k=n+1$ as soon as $n \geq 3$.
So, the maximum value of the function is $f(n,n+1)$ and a rather good approximation is, for $n \geq 10$,
$$f(n,n+1)\sim e^{n+1}\, n^{n+1} \,(n+1)^{-n} $$
