Is the Integral Finite for $n$ Fixed?

[evinda]

Let $\lambda$ be a positive number and $n$ a natural number. I want to show that

$$\int_{-\infty}^{+\infty} x^{2n} e^{-2 \lambda x^2} dx<+\infty.$$

There is the following hint: $e^{\lambda x^2} \geq \frac{1}{n!}(\lambda x^2)^n$, thus $x^{2n} e^{-\lambda x^2}\leq \frac{n!}{\lambda^n}$.I have done the following:We have that $$e^{2 \lambda x^2} \geq \frac{(2 \lambda x^2)^n}{n!}.$$

Then

$$e^{-2 \lambda x^2} \leq \frac{n!}{(2 \lambda x^2)^n} \Rightarrow x^{2n} e^{-2 \lambda x^2} \leq \frac{n!}{2^n \lambda^n}$$But the integral $\int_{-\infty}^{+\infty} \frac{n!}{2^n \lambda^n} dx$ isn't finite. Is it?

So have I done something wrong? Or can we not use the hint? (Thinking)

 

[GJA]

Hi evinda,

But the integral $\int_{-\infty}^{+\infty} \frac{n!}{2^n \lambda^n} dx$ isn't finite. Is it?

This improper integral does not exist, because, by the definition of $\displaystyle\int_{-\infty}^{\infty}\frac{n!}{2^n \lambda^n}\,dx,$ the two separate integrals $\displaystyle\int_{c}^{\infty}\frac{n!}{2^n \lambda^n}\,dx$ and $\displaystyle\int_{-\infty}^{c}\frac{n!}{2^n \lambda^n}\,dx$ must both exist/converge (Note: The value of $c$ can be fixed arbitrarily, though typically it is taken to be zero).

There is the following hint: $e^{\lambda x^2} \geq \frac{1}{n!}(\lambda x^2)^n$, thus $x^{2n} e^{-\lambda x^2}\leq \frac{n!}{\lambda^n}$.I have done the following:We have that $$e^{2 \lambda x^2} \geq \frac{(2 \lambda x^2)^n}{n!}.$$

Then

$$e^{-2 \lambda x^2} \leq \frac{n!}{(2 \lambda x^2)^n} \Rightarrow x^{2n} e^{-2 \lambda x^2} \leq \frac{n!}{2^n \lambda^n}$$

So have I done something wrong? Or can we not use the hint? (Thinking)

You didn't do anything wrong, your work is reasonably thought out. It is commonplace when working in analysis to derive bounds that are not of the best utility for the problem at hand. I would suggest going back to the inequality $x^{2n} e^{-\lambda x^2}\leq \frac{n!}{\lambda^n}$ and multiplying both sides by $e^{-\lambda x^{2}}$ and see what you can do from there.

 

[evinda]

Ok, so then we have that $x^{2n} e^{-2 \lambda x^2} \leq \frac{n!}{\lambda^n} e^{-\lambda x^2}$.

Thus, $\int_{-\infty}^{+\infty} x^{2n} e^{-2 \lambda x^2} dx \leq \frac{n!}{\lambda^n} \int_{-\infty}^{+\infty} e^{-\lambda x^2} dx=\sqrt{\frac{\pi}{\lambda}} \frac{n!}{\lambda^n}$.

And do we check what happens when $n \to +\infty$ at the last quantity that we got? Or is it finite since it doesn't depend on $x$ ? (Thinking)

 

[GJA]

Ok, so then we have that $x^{2n} e^{-2 \lambda x^2} \leq \frac{n!}{\lambda^n} e^{-\lambda x^2}$.

Thus, $\int_{-\infty}^{+\infty} x^{2n} e^{-2 \lambda x^2} dx \leq \frac{n!}{\lambda^n} \int_{-\infty}^{+\infty} e^{-\lambda x^2} dx=\sqrt{\frac{\pi}{\lambda}} \frac{n!}{\lambda^n}$.

This is correct, well done!

And do we check what happens when $n \to +\infty$ at the last quantity that we got? Or is it finite since it doesn't depend on $x$ ? (Thinking)

Good question. We were asked to show that for $n$ fixed the integral is finite, which you have done. If you were to take the limit as $n\rightarrow \infty$, however, you would find that the bounds for the sequence of integral values corresponding to $n=1, 2, 3,\ldots$ goes to infinity (because factorials grow more rapidly than exponentials). Because we are dealing with upper bounds, we cannot deduce anything about the limit
$$\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty}x^{2n} e^{-2 \lambda x^2} dx$$
from this information alone.

 

[evinda]

This is correct, well done!
Good question. We were asked to show that for $n$ fixed the integral is finite, which you have done. If you were to take the limit as $n\rightarrow \infty$, however, you would find that the bounds for the sequence of integral values corresponding to $n=1, 2, 3,\ldots$ goes to infinity (because factorials grow more rapidly than exponentials). Because we are dealing with upper bounds, we cannot deduce anything about the limit
$$\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty}x^{2n} e^{-2 \lambda x^2} dx$$
from this information alone.

I see... thanks a lot! (Smile)