Show that $e^{-t^2} \in \mathcal{L}^1$

Question

I have some homework where I need to determine if $u(t)=1/e^{t^2} \in \mathcal{L}^1$ on $(\mathbb{R},\mathcal{B}(\mathbb{R},\lambda))$

I have checked a theorem that says if it is Riemann integrable then it is also Lebesgue integrable. And I also found a theorem saying that if $|u|\leq w$ for some $w \in \mathcal{L}^1$ then $u \in \mathcal{L}^1$

However, my big problem is that we are integrating on the entire real line. I see that $u(t)\rightarrow 0$ for $|x|\rightarrow\infty$.

I was thinking of making $w$ to some simple function but this gives me the same problem, on how to show the simple function is in $\mathcal{L}^1$ or maybe somehow use a convergent series as w but a series is not based on real numbers but natural numbers

But how do I use these things to show that $u(t) \in \mathcal{L}^1$?

Answer 1

From multivariable calculus, one computes integral of $e^{-t^2}$ directly to be $\sqrt\pi$.

As far as you just want to show the integral $\int_{\infty}^\infty e^{-t^2} dt$ exists, by the Integral test, it suffices to show $\sum_{n=1}^{\infty} e^{-n^2}$ converges. The series is shown to be convergent by estimating term-wise to $1/n^2$.

Answer 2

$e^{t^{2}} \geq 1+t^{2}$ so $e^{-t^{2}} \leq \frac 1 {1+t^{2}}$ There are many ways of showing that $\frac 1 {1+t^{2}}$ is integrable; the simplest is to split the integral over the line into integral over $|t|<1$ and integral over $|t|\geq 1$. [ In the latter $\frac 1 {1+t^{2}} \leq \frac 1 {t^{2}}]$

Answer 3

Consider $$I=\int_{\mathbb R^2}e^{-(x^2+y^2)}dxdy.$$Using Fubini-Tonelli: $$I=\left(\int_{\mathbb R}e^{-x^2}dx\right)\left(\int_{\mathbb R}e^{-y^2}dy\right)=\alpha^2.$$ See @Nick answer to understand why the integral is finite.

Then using polar coordinates $$I=2\pi\int_0^{\infty}\rho e^{-\rho^2}d\rho=\pi.$$

So $\alpha=\left(\int_{\mathbb R}e^{-x^2}dx\right)=\sqrt I=\sqrt \pi.$