Finding an Upper Bound for e^(-x^2) for Easy Integration

[Ted123]

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

 

[LCKurtz]

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

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

 

[Ted123]

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

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

 

[LCKurtz]

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

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

 

[Ted123]

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

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

 

[LCKurtz]

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

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

 

[Ted123]

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

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

 

[HallsofIvy]

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

 

[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.)

 

[D H]

$\int_{-n}^n 1 \; dx =0$

No, it isn't.

 

[Ted123]

No, it isn't.

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.

 

[Whovian]

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

 

[Ted123]

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

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.

 

[D H]

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

But doesn't that contradict the comparison test?

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.

 

[Ted123]

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.

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?

 

[D H]

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

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 $\sqrt a \le a$.

 

[Mentallic]

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

 

[Ted123]

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

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

 

[D H]

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.

 

[Ted123]

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.

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

 

[Mentallic]

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

Not only that, but if $0\leq x < 1$ then it still holds that $e^{-x^2} > xe^{-x^2}$
But it is clear that since $e^{-x^2}$ is symmetric about the y-axis as f(x)=f(-x), and the function is finite over $x\in [0,1]$ then all you need to do is show that $xe^{-x^2}$ is convergent for $x\in [1,\infty)$.