What is the norm of the following function?

Question

I need to evaluate the norm of following function: $$f(t) = ate^{-bt^2}$$ I tried integration by part, but I cannot find the answer. "a" and "b" are constant real numbers.

Answer 1

For the sake of simplification, I will take $a=1$.

How to compute easily

$$\|f\|^2=\int_{-\infty}^{\infty} t^2e^{-2bt^2}dt=2\int_{0}^\infty t^2e^{-2bt^2}dt \ ? \tag{1}$$ We are here in a case where we can identify an integral we have to compute with a part of an integral involving a pdf (probability density function), a trick I have used more than once. Here, consider the Maxwell distribution (https://mathworld.wolfram.com/MaxwellDistribution.html) whose density has of course an integral equal to $1$:

$$\int_0^{\infty}\sqrt{\frac{2}{\pi}}\dfrac{x^2}{\sigma^3}e^{-\tfrac{x^2}{2\sigma^2}}dx=1\tag{2}$$

Otherwise said :

$$\int_0^{\infty}x^2e^{-\tfrac{x^2}{2\sigma^2}}dx=\sigma^3\sqrt{\frac{\pi}{2}} \tag{3}$$

It remains to identify the integrals in (1) and (2) by setting $\sigma = \frac{1}{2 \sqrt{b}}$, finaly giving

$$\|f\|^2=\dfrac{1}{4b}\sqrt{\frac{\pi}{2b}}$$

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Edit : there is another method, very similar to the previous one.

Consider formula $$\displaystyle \int_{-\infty}^{+\infty}x^2f(x) dx \ = \ 1\tag{4}$$

with :

$$\displaystyle f(x)=\frac {1}{\sigma \sqrt {2\pi }}e^{-\frac {1}{2}\left({\frac {x}{\sigma }}\right)^{2}}\tag{5}$$

expressing the second centered moment of the normal distribution $N(0,\sigma)$ (https://en.wikipedia.org/wiki/Normal_distribution#Moments) [all even centered moments of this distribution are equal to $1$].

Operating in the same way as before, we get, from the previous expressions,

$$2\int_{0}^{+\infty}x^2e^{-\frac {1}{2}\left({\frac {x}{\sigma }}\right)^{2}}=\sigma \sqrt {2\pi }$$

out of which, by identification with (1), gives the same result as (3).

Answer 2

$$||f||^2=\int_{-\infty}^\infty a^2t^2e^{-2bt^2}dt=a^2\int_{-\infty}^\infty t^2e^{-2bt^2}dt$$ Use the first function $g=t$, $g'=1$. The second function is $h'=te^{-2bt^2}$. Then $$h=\int te^{-2bt^2}dt$$ You solve this by doing the substitution $y^2=2bt^2$. Then $2ydy=4bt dt$. At the end you should get an answer proportional to $$\int_{-\infty}^\infty e^{-2bt^2}dt=\sqrt{\frac{\pi}{2b}}$$