# How we can solve that $\lim _{_{x\rightarrow \infty }}\int _0^x\:e^{t^2}dt$?

How we can solve that $\lim _{_{x\rightarrow \infty }}\int _0^x\:e^{t^2}dt$ ?

P.S: This is my method as I thought: $\int _0^x\:\:e^{t^2}dt>\int _1^x\:e^tdt=e^x-e$ which is divergent, so all your answers, helped me to think otherwise, maybe my method help something else :D

• Ideally, you have the question in both the title (if it fits; if not, then good indication of the question) and the body of the post. More ideally, you motivate the question as well in the body of the post. Apr 16 '15 at 22:52
• you can see en.wikipedia.org/wiki/Error_function Apr 16 '15 at 23:03

Since $$e^{x^2}=1+x^2+\frac{x^4}{2!}+\dotsb$$ we have that, for $x\ge0$, $e^{x^2}\ge1+x^2$. So $$\int_{0}^{x}e^{t^2}\,dt\ge\int_{0}^x(1+t^2)\,dt=x+\frac{x^3}{3}$$ Can you finish?

• hahahaha, nice try and intresting method! "can you help to finish?" Apr 16 '15 at 23:24
• What happens as $x \to \infty$? Apr 16 '15 at 23:27
• marty? was a joke because he think I am so fool... :( Apr 16 '15 at 23:27
• @Lucas Surely I'm not thinking you're a fool. It's usual here to close an answer/hint with that phrase. Apr 16 '15 at 23:33

This function diverges extremely fast. Notably, $e^{t^2}$ is monotone increasing with limit $\infty$ as $t \to \infty$. Thus your integral diverges (and it get very, very large very, very quickly).

$$e^{t^2}> e^t\text{ from 1 to \infty, and the part from 0 to 1 is finite.}$$ and the integral$\int_1^\infty e^t$ diverges. Therefore, by the comparison test, it diverges too.

\begin{align} \int_0^xe^{t^2}\,\mathrm{d}t &\ge\int_0^x\frac tx\,e^{t^2}\,\mathrm{d}t\\ &=\frac1{2x}\left(e^{x^2}-1\right) \end{align} As $x\to\infty$, the function on the right goes to $\infty$ extremely fast.

• nice one robjohn! Apr 16 '15 at 23:35
• The integral is actually asymptotic to $\frac{e^{x^2}}{2x}$. That is $$\lim_{x\to\infty}2xe^{-x^2}\int_0^xe^{t^2}\,\mathrm{d}t=1$$
– robjohn
Apr 17 '15 at 0:00

Or just use $e^{x^2} \ge 1$ on $[0,\infty)$ to see $\int_0^x e^{t^2}dt \ge x \to \infty.$

• here are several methods... to prove divergent, but I think your method is the easiest! Apr 16 '15 at 23:29

It is vary simple: $$y(x)=\int_0^xe^{t^2}dt=\frac{1}{2} \sqrt{\pi } \text{erfi}(x)$$ So: $$\lim_{x \rightarrow+\infty}y(x)=+\infty$$