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To show that $\int_0^\infty e^{x^2} \, dx$ does not converge, I can use polar coordinates, and say

$$\int_0^\infty \int_0^\infty e^{x^2+ y^2} \, dx \, dy = \int_0^\infty\int_0^{\pi/2} e^{r^2}r d\theta \, dr$$ $$ = \frac{\pi}{2}\int_0^\infty e^{r^2}r \, dr$$ $$= \lim_{r \to \infty} \frac{\pi}{2}\bigg(\frac{e^{r^2}}{2} - \frac{1}{2} \bigg) = \infty$$

So $\int_0^\infty e^{x^2} \, dx$ does not converge.

What are some other ways to show it does not converge?

Michael Hardy
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user80456
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    $\int_{n}^{n+1}e^{x^2},\mathrm dx\ge e^{n^2}\not\to 0$. – Hagen von Eitzen May 31 '13 at 21:17
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    $f(x)=e^{x^2}$ is bounded below by $1$ on $[0,to)$. – Brian M. Scott May 31 '13 at 21:18
  • Your procedure is a good way to find the important integral $\int_0^\infty e^{-x^2},dx$. It is much more complicated than necessary to prove non-existence of your integral: the function $e^{x^2}$ blows up. – André Nicolas May 31 '13 at 22:09
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    You're really swatting a fly with a nuclear weapon. Think about what the graph of $y=e^{x^2}$ looks like and you have the answer instantly. That's not a logically rigorous argument, but the step from there to logical rigor is very short. – Michael Hardy Jun 01 '13 at 02:16

6 Answers6

18

$e^{x^2}\ge1$, so $\int_0^M e^{x^2}\,dx\ge M$.

Harald Hanche-Olsen
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Well, $e^{z^{2}}\geq 1$ for $z\geq 0$ and integral $\int_{0}^{\infty}1dz=+\infty$.

Neph
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Observe that $$ \int_{0}^{\infty} e^{x^2} dx \ge \int_{1}^{\infty} e^{x^{2}}dx\ge \int_{1}^{\infty} e^{x}dx=\infty $$ as desired :)

Prism
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The easiest way to observe to observe this to me looks like to observe that $1 \leq e^{x^{2}}$ for all $x$. Hence $$\int_{0}^{\infty} e^{x^{2}}dx \geq \int_{0}^{\infty} 1 dx = \infty$$

Alex Wertheim
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2

In general if a function $f$ has a non zero limit $\ell$ (possibly infinite) at $+\infty$: i.e $$\lim_{x\to\infty}f(x)=\ell\neq0$$ then $$\int_a^\infty f(x)dx$$ is divergent.

1

$e^x \ge x$ for all $ x \in [0,\infty)$. Hence $\int_{0}^{\infty} e^{x^2} dx \ge \int_{0}^{\infty} x^{2}dx=\infty$

user148928
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