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 
$$\int_{0}^{t}\frac{e^{x}}{\sqrt{x}}dx> \frac{e^{t}}{\sqrt{t}},\qquad t\geq 1$$.
Any ideas?
 A: 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 $t=1$.
A: Let us start without considering $t\geq 1.$ 
Set $F(t)=e^{-t}\int_{0}^t\frac{e^x}{\sqrt{x}} dx.$ 
Then 
$
F'(t)=-F(t)+\frac{1}{\sqrt{t}}$ or equivalently $$F(t)=-F'(t)+\frac{1}{\sqrt{t}}.$$
From the other side, $$F(t)\leq \int_0^{t}\frac{dx}{\sqrt{x}}=2\sqrt{t}.$$
Putting together we get
 $$-F'(t)+\frac{1}{\sqrt{t}}\leq 2\sqrt{t}.$$ 
This leads to $$F'(t)\geq \frac{1-2t}{\sqrt{t}},$$
which is positive if $0\leq t \leq 1/2,$ but gives not an information for $t\geq 1.$
A: This is Dawson function
$$F(t^2)=e^{-t^2}\int_{0}^{t^2}\frac{e^{x}}{\sqrt{x}}dx=2e^{-t^2}\int_0^te^{x^2}\ dx$$
let say $$g(t)=e^{-t^2}\int_0^te^{x^2}\ dx\implies g'(t)=1-2te^{-t}\int_0^te^{x^2}\ dx<0$$
because
$$2te^{-t}\int_0^te^{x^2}\ dx=e^{-t}\int_0^t2te^{x^2}\ dx>e^{-t}\int_0^t2xe^{x^2}\ dx>e^{-t}(e^{t^2}-1)>1$$
Wolfram shows this is true only for $t>\sim1.2$.
A: Hint: The first derivative with respect to $t$ is given by
$$f'(t)-{{\rm e}^{-t}}\sqrt {\pi}{\it erfi} \left( \sqrt {t} \right) +{\frac 
{{{\rm e}^{-t}}{{\rm e}^{t}}}{\sqrt {t}}}
$$
