Show that $e^{-t}\int_{0}^{t}\frac{e^{x}}{\sqrt{x}}dx$ is strictly decreasing for $t\geq1$ 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 
$$\int_{0}^{t}\frac{e^{x}}{\sqrt{x}}dx&gt; \frac{e^{t}}{\sqrt{t}},\qquad t\geq 1$$.

Any ideas?
I think I solved it. 

Let $f(t):=e^{-t}\int_{0}^{t}\frac{e^{\xi}}{\sqrt{\xi}}$.

We need to show that 
$f^{\prime}(t)=\frac{\frac{e^t}{\sqrt{t}}-\int_{0}^{t}\frac{e^{\xi}}{\sqrt{\xi}}}{e^{t}}&lt;0$ for $t&gt;1$.
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 $t&gt;1$.

Now
$$g(1):=\int_{0}^{1}\frac{e^{\xi}}{\sqrt{\xi}}-e$$
It turns out that $g(1)=\sqrt{\pi}\,\mathrm{erfi}(1)-e&gt;2.9-e&gt;0$
(Can we do the latter estimate analytically ?) 

Finally, 

$$g^{\prime}(t):=\frac{e^{t}}{\sqrt{t}}-\frac{e^t}{\sqrt{t}}+
\frac{e^t}{2t^{\frac{3}{2}}}&gt;0$$ 
So, $g$ is increasing and positive at $t=1$.
Hint: The first derivative with respect to $t$ is given by
$$f'(t)-{{\rm e}^{-t}}\sqrt {\pi}{\it erfi} \left( \sqrt {t} \right) +{\frac 
{{{\rm e}^{-t}}{{\rm e}^{t}}}{\sqrt {t}}}
$$