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).