# Upper bound for e^(-x^2)

1. Apr 15, 2012

### Ted123

Can anyone suggest an upper bound for $e^{-x^2}$ that can be integrated easily?

2. Apr 15, 2012

### LCKurtz

On what interval? Just guessing maybe $[0,\infty)$? This is really easy. What have you thought of so far?

3. Apr 15, 2012

### Ted123

I've just realised it's less than 1 for all x.

4. Apr 15, 2012

### LCKurtz

You haven't answered the question about what interval, or told us what problem you are trying to solve.

5. Apr 15, 2012

### Ted123

It doesn't matter now; I know how to do it!

6. Apr 15, 2012

### LCKurtz

Not if you were trying a comparison test on $[0,\infty)$.

7. Apr 16, 2012

### Ted123

It's actually on a finite interval $[-n,n]$, but yes the integral of 1 would diverge on $[0,\infty)$

8. Apr 16, 2012

### HallsofIvy

Staff Emeritus
Are you asserting that "1 times infinity" is a finite number?

9. Apr 16, 2012

### Ted123

All I need to show is that $\int_{-n}^n e^{-x^2} dx$ converges as $n\to\infty$. This is the last part of a bigger question where I'm trying to use the Monotone Convergence Theorem to show that $e^{-x^2}$ is Lebesgue integrable over $\mathbb{R}$.

Since $0 \leqslant e^{-x^2} \leqslant 1$ and $\int_{-n}^n 1 \; dx =0 \to 0$ as $n\to\infty$ it follows from the comparison test that $\int_{-n}^n e^{-x^2} dx$ converges as $n\to\infty$.

(I'm not asked to find $\int_{\mathbb{R}} e^{-x^2}$, just to prove it is Lebesgue integrable using the MCT.)

10. Apr 16, 2012

### D H

Staff Emeritus
No, it isn't.

11. Apr 16, 2012

### Ted123

Oh yeah, it equals $2n$ which diverges as $n\to\infty$.

But doesn't that contradict the comparison test?

I'm trying to show that $\int_{-n}^n e^{-x^2}dx$ converges, but the comparison test implies it diverges.

12. Apr 16, 2012

### Whovian

Yep, it diverges, assuming you mean as $\displaystyle\lim_{n\to\infty}$. Why do you think it converges?

13. Apr 16, 2012

### Ted123

Well, OK. Let's start from scratch.

I'm trying to prove $f(x) = e^{-x^2} \in L^1(\mathbb{R})$ ; that is, that $f$ is Lebesgue integrable over $\mathbb{R}$.

Let $f(x) = e^{-x^2} \chi_{(-\infty , \infty)}(x)$ and $f_n = f \chi_{[-n,n]}$.

Since $f\geqslant 0$, $(f_n)$ is an increasing sequence of functions which converges everywhere to $f$.

We want to show that $f$ is integrable so it suffices to show $\int f_n$ converges as $n\to\infty$. It then follows from the Monotone Convergence Theorem that $f\in L^1 (\mathbb{R})$ (and $\int f = \lim_{n\to\infty} \int f_n$).

We use the given hint ("don't try to find $\int f$ ; integrate a simpler upper bound instead").

We see that $f_n$ is bounded by... The integral of this upper bound converges, so by comparison test $\int f_n$ converges.

14. Apr 16, 2012

### D H

Staff Emeritus
Of course not. Given functions $f(x)$ and $g(x)$ such that $0\le f(x) \le g(x)$ for all x and such that $\lim_{s\to\infty}\int_{-s}^s g(x) dx$ diverges says nothing about the convergence or divergence of $\lim_{s\to\infty}\int_{-s}^s f(x) dx$.

There are plenty of positive definite functions $f(x)$ that are bounded from above by $g(x)=1$ whose integral over the real number line does converge. $f(x)=\exp(-x^2)$ is one such function.

15. Apr 16, 2012

### Ted123

Oh yes, it's the other way around for divergence isn't it: if $\int f(x) dx$ diverges then so does $\int g(x) dx$.

OK, so I need an upper bound whose integral converges then. Any suggestions?

16. Apr 16, 2012

### D H

Staff Emeritus
No. That would be solving your homework for you.

Huge hint: if $0\le a \le 1$ then $\sqrt a \ge a$, and if $a\ge1$ then $\sqrt a \le a$.

Last edited: Apr 16, 2012
17. Apr 16, 2012

### Mentallic

What about something with $$xe^{-x^2}$$

18. Apr 16, 2012

### Ted123

But if $x<0$ then $e^{-x^2} > xe^{-x^2}$

19. Apr 16, 2012

### D H

Staff Emeritus
Do you care? exp(-x^2) is symmetric, so if the integral from 0 to infinity exists, then so does the integral from -infinity to infinity.

20. Apr 16, 2012

### Ted123

Bingo! If only I'd drawn a graph first...