Inequality between $\int_x^{2x}e^{-t^2}dt$ and $xe^{-x^2}$

Question

Here's the problem I'm facing. The function $f$ is defined as follows, over $\mathbb{R}_+ $: $$ \forall x \in \mathbb{R}_+, f(x) = \int_x^{2x}{e^{-t^2}\mathrm{d}t} $$ And we must figure the following inequality (in order to study the limit of $f$ in positive infinity) $$ f(x) \leq xe^{-x^2} $$ A preliminary question asked to study variations of $f$ on its domain, and I successfully managed to differentiate $f$: $$ f'(x) = e^{-x^2}(2e^{-3x^2}-1) $$ And conclude that f is increasing on $\left[0;\sqrt{\dfrac{\ln(2)}{3}}\right]$ and then decreasing. So I tried to study the difference $xe^{-x^2} - f(x)$ And found out it's decreasing at least from 1 to infinity, but I didn't see what I could do with that. Then I looked for an inequality that I could integrate like so: I tried to find a function $g$ satisfying: $$ xe^{-x^2} = \int_x^{2x}g'(t)\mathrm{d}t \iff xe^{-x^2} = g(2x)-g(x) $$ So if $g'(x) \geq e^{-x^2}$ hence $f(x) \leq xe^{-x^2}$. But sadly I couldn't find such a function of the forms I could think to. But then I couldn't think of anything else except some unfruitful relations ships between $(xe^{-x^2})'$ and $f'(x)$. Any help is welcomed, thanks to you for your time.

Answer 1

Differentiation is reasonable but unfortunately doesn’t work out here. The difference between the two is not monotone.

Rather, we can bound the integrand above by $e^{-x^2}$, and the size of the integration domain is $2x-x=x$. Thus the integral is bounded above by $x\cdot e^{-x^2}$.

Answer 2

Hint: if $h$ is a positive decreasing function, then $\int_0^5 h \leq 5 h(0)$.