TITLE: Show that $e^{-t}\int_{0}^{t}\frac{e^{x}}{\sqrt{x}}dx$ is strictly decreasing for $t\geq1$
QUESTION [6 upvotes]: I want to prove that $e^{-t}\int_{0}^{t}\frac{e^{x}}{\sqrt{x}}dx$ is decreasing
on $[1,\infty[$.
First of all numerical experiments verify this.
I am trying the first derivative test, but stuck with showing that
the sign of the derivative is negative, which is equivalent to showing that 
$$\int_{0}^{t}\frac{e^{x}}{\sqrt{x}}dx> \frac{e^{t}}{\sqrt{t}},\qquad t\geq 1$$.
Any ideas?

REPLY [2 votes]: I think I solved it. 
Let $f(t):=e^{-t}\int_{0}^{t}\frac{e^{\xi}}{\sqrt{\xi}}$.
We need to show that 
$f^{\prime}(t)=\frac{\frac{e^t}{\sqrt{t}}-\int_{0}^{t}\frac{e^{\xi}}{\sqrt{\xi}}}{e^{t}}<0$ for $t>1$.
The latter follows if we show that the function $g$ where
$$g(t):=\int_{0}^{t}\frac{e^{\xi}}{\sqrt{\xi}}-\frac{e^t}{\sqrt{t}}$$
is positive for $t>1$.
Now
$$g(1):=\int_{0}^{1}\frac{e^{\xi}}{\sqrt{\xi}}-e$$
It turns out that $g(1)=\sqrt{\pi}\,\mathrm{erfi}(1)-e>2.9-e>0$
(Can we do the latter estimate analytically ?) 
Finally, 
$$g^{\prime}(t):=\frac{e^{t}}{\sqrt{t}}-\frac{e^t}{\sqrt{t}}+
\frac{e^t}{2t^{\frac{3}{2}}}>0$$ 
So, $g$ is increasing and positive at $t=1$.