Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am looking for the asymptotics of the following integral

$\int_{\mathbb{R}} H_m^2(x) {\rm e}^{-2 \alpha^2 x^2} {\rm d} x = 2^{m-1/2} \alpha^{-2m -1} (1-2\alpha^2)^m \ \Gamma(m+1/2) ~ _2F_1\left(-m,m,1/2-m,\frac{\alpha^2}{2\alpha^2-1}\right)$

where $H_m$ is the $m^{\rm th}$ Hermite polynomial (orthogonal under the weight ${\rm e}^{-x^2}$), and $_2F_1$ is the hypergeometric function.

I found this formula from p. 803 of "Table of Integrals, Series, and Products" by Gradshteyn-Ryzhik. However, I have idea about the asymptotics of the $_2F_1$ term. Can anyone enlighten me on the asymptotics of

$_2F_1\left(-m,m,1/2-m,\beta\right)$

when $m$ is large? In fact I tried mathematica and it seems $_2F_1\left(-m,m,1/2-m,\beta\right) \sim |4 \beta|^m$. Any reference on this issue?

Now given the above asymptotics is true, observe that the norm of $H_m$ under the weight ${\rm e}^{-2 \alpha^2 x^2}$ has the same exponent for all $alpha$, including the original weight ($\alpha^2 = 1/2$). Is this a common phenomenon for orthogonal polynomials?

share|improve this question
    
Note that if m is an integer, the hypergeometric function degenerates to a polynomial in $(2-\alpha^{-2})^{-1}$ of degree m. –  J. M. Aug 15 '10 at 13:11
add comment

1 Answer

I suppose I should be more precise than what I wrote in my comment: as already mentioned, whenever one of the "numerator parameters" of a hypergeometric function is a negative integer -m, the series terminates (the Pochhammer symbols in the terms of degree higher than m vanish), i.e. your hypergeometric function becomes a polynomial.

Using this, the degree m term of the polynomial is

$\frac{(-m)_m (m)_m}{\left(\frac1{2}-m\right)_m}\frac{\beta^m}{m!}$

or, using the properties of the Pochhammer symbols and the factorial:

$2^{2m-1}\beta^m$

Replacing $\beta$ with $(2-\alpha^{-2})^{-1}$ and multiplying by the factors in front of the hypergeometric expression nets you

$\frac{(-1)^m 2^{3m-\frac{3}{2}}\Gamma\left(\frac1{2}+m\right)}{\alpha}$

What complicates things, however, are the factors in front of the hypergeometric function, which when expanded is a polynomial in odd powers of the variable $\alpha^{-1}$.

It's a bit late here, so I suppose I'll let someone else finish the analysis...

share|improve this answer
    
Now that I have gotten to look at my copy of G&R, I suppose you should have mentioned that your integral of interest was a specialization of equation 7.374, #5. Note that there is a restriction in there that $\alpha^2\neq\frac1{2}$; the correct formula for $\alpha^2=\frac1{2}$ is given by 7.374, #1, second case. The kicker: after trying out random values of the general integral in Mathematica (keeping also in mind the restriction that the sum of the degrees of the two Hermite polynomials in the integrand should be even); I seem to be unable to reproduce the identity. I better investigate… –  J. M. Aug 15 '10 at 22:29
    
thanks. i think the special case $\alpha^2 = 1/2$ can be deduced from the formula via taking limit. –  mr.gondolier Aug 16 '10 at 4:53
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.