Show that $e^{-t}\int_{0}^{t}\frac{e^{x}}{\sqrt{x}}dx$ is strictly decreasing for $t\geq1$

Question

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?

Answer 1

Hint: The first derivative with respect to $t$ is given by $$f'(t)-{{\rm e}^{-t}}\sqrt {\pi}{\it erfi} \left( \sqrt {t} \right) +{\frac {{{\rm e}^{-t}}{{\rm e}^{t}}}{\sqrt {t}}} $$

Answer 2

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 $t>\sim1.2$.

Answer 3

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 $t\geq 1.$

Answer 4

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 $t=1$.