Integral bounded below A function $f:\mathbb{R} \rightarrow \mathbb{R}$ is defined by
\begin{equation*}
f(x) := \int_{1}^{x^2} e^{t^2} \; dt.
\end{equation*}
Is $f$ bounded below?
$f$ is bounded below if there is an $m\in \mathbb{R}$ with $f(x)\geq m$, $\forall x\in \mathbb{R}$.
The answer is yes it is bounded below by $f(0) = -\frac{\sqrt{\pi}}{2}$.
My question is how do you get this value and how do you use the above definition to show this? Thanks.
 A: Clearly $f(x)$ is even. That is to say $f(x)=f(-x)$. Then we have
$$f'(x)=2xe^{x^4}\gt0\qquad\forall x\gt0$$
Hence $f(x)$ has a global minimum at $x=0$. This is because $f(x)$ would appear 'bowl-like' (it increases away from $f(0)$ when $x$ tends in either direction from $x=0$) if sketched on a plane. Hence $f(x)\ge f(0)$ for all $x\in\mathbb{R}$.
A: $$f(x) = \frac{\sqrt{\pi}}{2} \text{erfi}(x^2) - \frac{\sqrt{\pi}}{2} \text{erfi}(1)$$
where $$ \text{erfi}(y) = \frac{2}{\sqrt{\pi}} \int_0^y \exp(t^2)\; dt$$
Since $\text{erfi}(x^2) \ge 0$ the minimum value 
$$ -\frac{\sqrt{\pi}}{2} \text{erfi(1)} = - \int_0^1 \exp(t^2)\; dt$$
occurs at $x=0$.  $\text{erfi}$ is a "well-known" special function, but not an elementary one.  You could expand the exponential in a Maclaurin series and integrate term-by-term, so
$$ f(0) = -1 - \frac{1}{3} - \frac{1}{10} - \frac{1}{42} - \ldots = - \sum_{k=0}^\infty \frac{1}{(2k+1)k!}$$
But it's certainly not $-\sqrt{\pi}/2$.  Numerically, $f(0) \approx -1.462651746$ while $-\sqrt{\pi}/2 \approx -0.8862269255$.
