As part of analysing an algorithm I stumbled upon the integral $\displaystyle \int_1^X \frac{\exp(B/\sqrt x)}{\sqrt x} dx$ as an approximation for the corresponding sum. Wolfram alpha gives the antiderivative in terms of non-elementary functions. This seems more trouble than it's worth since I'm only interested in a readable bound for the integral.
One very poor bound comes from rounding the exponential up to $e^B$ and get $\displaystyle e^B \int_1^X \frac{dx}{\sqrt x} = 2e^B(\sqrt X - 1) = O(e^B \sqrt X) $.
One better bound comes from breaking the intgral into two parts like this:
$$\displaystyle \int_1^X \frac{\exp(B/\sqrt x)}{\sqrt x} dx= \int_1^{B^2}\frac{\exp(B/\sqrt x)}{\sqrt x} dx + \int_{B^2}^X \frac{\exp(B/\sqrt x)}{\sqrt x} dx$$
For the first part do the same rounding up to get $$ \int_1^{B^2}\frac{\exp(B/\sqrt x)}{\sqrt x}\le e^B \int_1^{B^2}\frac{dx}{\sqrt x} \le 2e^B \sqrt {B^2} = 2B e^B.$$ For the second part the denominator is small and we get
$$ \int_{B^2}^X \frac{\exp(B/\sqrt x)}{\sqrt x} dx \le \int_{B^2}^X \frac{e}{\sqrt x} dx = 2 e (\sqrt X - \sqrt {B^2}) \le 2e \sqrt X.$$
Putting it back together we can bound the original integral by $2Be^B + 2e \sqrt X$. We still have the exponential term but not it is a constant and not multiplied by the $\sqrt X$.
I wonder can we do any better than this? Can we replace the dependence on $B$ with something weaker?
I do not know if this will answer your question.
Effectively, after one integration by parts, we end with $$I=\int_1^X\frac{e^{\frac{B}{\sqrt{x}}}}{\sqrt{x}}\, dx=-2 \left(B \,\text{Ei}\left(\frac{B}{\sqrt{X}}\right)-B\, \text{Ei}(B)-\sqrt{X} \, e^{\frac{B}{\sqrt{X}}}+e^B\right)$$ What is doable is to expand as a series around $B=0$; this would give $$I=2\left( \sqrt{X}-1\right)+B\, \log (X)+ \left(1-\frac{1}{\sqrt{X}}\right)B^2+O\left(B^3\right)$$