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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

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  • $\begingroup$ 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. $\endgroup$
    – davidlowryduda
    Apr 16 '15 at 22:52
  • $\begingroup$ you can see en.wikipedia.org/wiki/Error_function $\endgroup$
    – E.H.E
    Apr 16 '15 at 23:03
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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?

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

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

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

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  • $\begingroup$ nice one robjohn! $\endgroup$
    – Lucas
    Apr 16 '15 at 23:35
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    $\begingroup$ The integral is actually asymptotic to $\frac{e^{x^2}}{2x}$. That is $$\lim_{x\to\infty}2xe^{-x^2}\int_0^xe^{t^2}\,\mathrm{d}t=1$$ $\endgroup$
    – robjohn
    Apr 17 '15 at 0:00
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Or just use $e^{x^2} \ge 1$ on $[0,\infty)$ to see $\int_0^x e^{t^2}dt \ge x \to \infty.$

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  • $\begingroup$ here are several methods... to prove divergent, but I think your method is the easiest! $\endgroup$
    – Lucas
    Apr 16 '15 at 23:29
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It is vary simple: $$y(x)=\int_0^xe^{t^2}dt=\frac{1}{2} \sqrt{\pi } \text{erfi}(x)$$ So: $$\lim_{x \rightarrow+\infty}y(x)=+\infty$$

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