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I am trying to compute the following integral for $k$ and $k'$ integers:

$$I(k,k')=\int_{0}^{+\infty} \mathrm{d}x\; e^{-x^2} H_{2k+1}(x) H_{2k'}(x)$$ where $H_{k}$ are Hermite functions : $H_k(x)=(-1)^k e^{x^2}\frac{d^k}{dx^k}e^{-x^2}$.

This is not the usual integral we find where we would use simple orthogonality, the interval $[0,+\infty[$ makes things more complicated.

I haven't make lots of progress apart from reformulating the problem in terms of generalized Laguerre polynomials :

$$I(k,k')=C\int_{0}^{+\infty} \mathrm{d}x\; e^{-x} L_{k}^{(1/2)}(x) L_{k'}^{(1/2)}(x)$$ where C is a proportional factor between Hermite and Laguerre polynomials.

I could not either find a closed form for this integral.

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  • $\begingroup$ What is the parity of the Hermite functions for n=2k and n=2k'+1? How can that impact the range of your integral? $\endgroup$ – AlphaNumeric Jan 19 '17 at 17:46
  • $\begingroup$ Hermite functions have the parity of their exponent, so $H_{2k+1}$ is odd and $H_{2k'}$ is even. An integral over $\mathbb{R}$ would lead 0 but because we restrict ourselves to the positive axis, this is more complicated. $\endgroup$ – Alexandre Krajenbrink Jan 19 '17 at 17:51
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    $\begingroup$ The $e^{-x^2}$ factor cancels that from $H_k\propto e^{x^2}$ so integration by parts looks attractive if you substitute the expression for $H_{2k'}$ into the integral. $\endgroup$ – Winther Jan 19 '17 at 17:51
  • $\begingroup$ Putting $k=0$, we obtain $I(0,k')=(-)^{k'}2(2k')!/k'!$ $\endgroup$ – tired Jan 19 '17 at 18:02
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Here is the result $$I=\frac{(-1)^{k+k'} 2^{2 k+2 k'+1} \Gamma \left(k+\frac{3}{2}\right) \Gamma \left(k'+\frac{1}{2}\right)}{\pi (2 k-2 k'+1)}= \frac{(-2)^{k+k'} (2k+1)!!\,(2k'-1)!!}{2k-2k'+1}.$$

The derivation goes along the following lines. First, we use these expressions to write the Hermite polynomials as explicit power series. Interchanging the sums with the integral and employing $$\int_0^\infty x^{n} e^{-x^2} = \frac{1}{2}\Gamma\left(\frac{n+1}{2} \right) ,$$ we immediately obtain the representation as the double sum $$I=\sum_{l=0}^k\sum_{l'=0}^{k'}\frac{ (-1)^{k+k'-l-l'} 2^{2 l+2 l'} (2 k+1)! (2 k')! (l+l')! }{(2 l+1)! (2 l')! (k-l)! (k'-l')!}.$$

This final sum (as many sums involving products of factorials) can be evaluated explicitly. In this case, Mathematica produces the result quoted above.

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