Monotony of a function

Let $F:(0, +\infty) \rightarrow \mathbb{R}$ be defined as:

$$ F(x)= \int_{1/x^2}^{ \int_{0}^{x}\exp (s^2)\, {\rm d}s}e^{t^2}\, {\rm d}t $$

What is the monotony of the function?

Solution 



1st solution:

We are proving that the function is strictly increasing.

Note that $g(t)=e^{t^2}>0$ in $(0, +\infty)$. In addition $\displaystyle \varphi(x)=\int_0^x \exp(s^2)\, {\rm d} s$ is increasing as a function because the derivative is positive for every $x \in (0, +\infty)$, while $t(x)= 1/x^2$ is decreasing (pretty straight forward).

Let $x_1 , \; x_2 \in (0, +\infty)$ such that $x_1<x_2$. It is a simple observation to see that:

$$ F(x_1)=\int_{t(x_1)}^{\varphi(x_1)}g(t)\, {\rm d}t< \int_{t(x_2)}^{\varphi(x_2)}g(t)\, {\rm d}t= F(x_2)$$

which proves our claim and the exercise comes to an end!

2nd solution:
Differentiating the function we have that:
$$F'(x)=\left [ \int_{1}^{{\Large \int_{0}^{x}} e^{s^2}\, {\rm d}s}e^{t^2}\, {\rm d}t -\int_{1}^{1/x^2} e^{t^2}\, {\rm d}t   \right ]' = e^{\left ( {\Large \int_{0}^{x}}e^{s^2}\, {\rm d}s \right )^2}e^{x^2}+ \frac{2e^{1/x^4}}{x^3}>0 , \; \forall x >0$$

which proves our claim.