$$\int_{0}^{\infty} e^{-x^2} dx$$

I couldn't find a comparison in $[0,\infty)$, but I found one $[1,\infty)$ not sure if I can do it like this:

Since I know that $f(x)$ is positive for all natural numbers the hypothesis is met.

For $$x \in [1,\infty), f(x) = \frac{1}{e^{x^2}} \leq \frac{1}{e^x} = g(x)$$

Consider $$\int_{1}^{\infty} g(x)dx = \lim_{A\to\infty} -e^{-x}\bigg|_{1}^{A} = \frac{1}{e}$$

Therefore $\int g(x)dx$ converges and by C.T. for $$\int_{1}^{\infty} f(x)dx$$ converges too.

Can I end here or do I gotta show $\int_{0}^{1} f(x)dx$ converges too?

  • 2
    $\begingroup$ Essentially you are done. Notice that $e^{-x^2}$ is continuous on $[0, 1]$, so it is absolutely integrable on $[0, 1]$ as well. $\endgroup$ – Sangchul Lee Apr 16 '17 at 22:41
  • $\begingroup$ en.wikipedia.org/wiki/Gaussian_integral $\endgroup$ – Jack Apr 16 '17 at 22:41

$f(x)=e^{-x^2}$ is positive and bounded by $1$ on the interval $[0,1]$, bounded by $e^{-x}$ on the interval $[1,+\infty)$. It follows that:

$$ \int_{0}^{+\infty}e^{-x^2}\,dx \leq \int_{0}^{1}1\,dx + \int_{1}^{+\infty}e^{-x}\,dx = 1+\frac{1}{e}. $$ Actually, we don't even need to split the integration range. We have $e^{-x^2}\leq \frac{1}{1+x^2}$ and $$ \int_{0}^{+\infty}e^{-x^2}\,dx \leq \int_{0}^{+\infty}\frac{dx}{1+x^2}=\frac{\pi}{2}.$$


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