Let $m,n,k$ be nonnegative integers. How might I go about evaluating the following integral?

$$ \int_{-\infty}^\infty \left( \frac{\mathrm{d}^m}{\mathrm{d}x^m} e^{-x^2} \right) \left( \frac{\mathrm{d}^n}{\mathrm{d}x^n} e^{-x^2} \right) x^k e^{x^2} \mathrm{d}x $$

  • $\begingroup$ Where did this problem come from? $\endgroup$ – science Mar 24 '15 at 4:40
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    $\begingroup$ Well, you can start by taking the $m$ and $n$-th derivative of $e^{-x^2}$. $\endgroup$ – MCT Mar 24 '15 at 4:44

HINT, not yet a complete answer.

Hermite polynomials: $$H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}$$

so integral becomes:

$$ I_{m,n,k}=(-1)^{m+n}\int_{-\infty}^\infty H_m(x) H_n(x) x^k e^{-x^2} \mathrm{d}x $$

Particular case $k=0$:

$$ I_{m,n,0}=(-1)^{m+n}\int_{-\infty}^\infty H_m(x) H_n(x) e^{-x^2} \mathrm{d}x = (-1)^{m+n}\sqrt\pi 2^nn!\delta_{mn}$$

EDIT: The general case

Remembering that:

$$x^k = \frac{k!}{2^k} \sum_{j=0}^{\lfloor \tfrac{k}{2} \rfloor} \frac{1}{j!(k-2j)! } ~H_{k-2j}(x)~$$

and replacing in the general integral, we find:

$$I_{m,n,k}=(-1)^{m+n}\int_{-\infty}^\infty H_m(x) H_n(x) e^{-x^2} \frac{k!}{2^k} \sum_{j=0}^{\lfloor \tfrac{k}{2} \rfloor} \frac{1}{j!(k-2j)! } ~H_{k-2j}(x)~ \mathrm{d}x =$$ $$=\sum_{j=0}^{\lfloor \tfrac{k}{2} \rfloor} \frac{1}{j!(k-2j)! } \frac{k!}{2^k} (-1)^{m+n}\int_{-\infty}^\infty H_m(x) H_n(x) H_{k-2j} e^{-x^2} (x)~ \mathrm{d}x =$$

Now we can make use of the following integral:

$$\int_{-\infty}^\infty H_m(x) H_n(x) H_l(x) e^{-x^2} dx = \frac{2^{\frac{m+n+l}{2}}l!m!n!\sqrt\pi}{\left(\frac{m+l-n}{2}\right)!\left(\frac{n+l-m}{2}\right)!\left(\frac{m+n-l}{2}\right)!}$$ when $\frac{m+n+l}{2}$ is integer and $m+n\ge l$ and $m+l \ge n$ and $l+n\ge m$ ; Zero otherwise.


$$I_{m,n,k}=\sum_{j=0}^{\lfloor \tfrac{k}{2} \rfloor} \frac{1}{j!(k-2j)! } \frac{k!}{2^k} (-1)^{m+n} \frac{2^{\frac{m+n+k-2j}{2}}(k-2j)!m!n!\sqrt\pi}{\left(\frac{m+k-2j-n}{2}\right)!\left(\frac{n+k-2j-m}{2}\right)!\left(\frac{m+n-k+2j}{2}\right)!}=$$

Simplifying ( we have employed the fact that if $m+n-k$ is an even integer also $\pm m \pm n \pm k$ and $\pm m \pm n \pm k \pm 2j$ are even integers) we can rewrite:

$$I_{m,n,k} = \begin{cases}2^{\frac{m+n-k}{2}}(-1)^{m+n}m!n!k! \sqrt\pi \sum_{j=0}^{\lfloor \tfrac{k}{2} \rfloor} \frac{2^{-j}[m+n+\ge k-2j][m+k-2j \ge n][k-2j+n\ge m] }{j!\left(\frac{m+k-2j-n}{2}\right)!\left(\frac{n+k-2j-m}{2}\right)!\left(\frac{m+n-k+2j}{2}\right)!} & \text{when $\frac{m+n+k}{2}$ is integer} \\ 0 & \text{ otherwise} \end{cases}$$

where the Iverson convention (see "Concrete Mathematics" Graham,Knuth,Patshnik, and http://en.wikipedia.org/wiki/Iverson_bracket ) has been employed:

$$[statement]=\begin{cases}1 & \text{when statement is true} \\ 0 & \text{otherwise}\end{cases}$$

ADDENDUM (Alternative solution) At Wolfram:


Equation (52) gives another solution of aboveseen integral. I think this alternative solution has a complexity comparable to that I have proposed.


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    $\begingroup$ Even if not complete, it is nice ! $+1$ $\endgroup$ – Claude Leibovici Mar 24 '15 at 7:23
  • $\begingroup$ Now, this is a splendid answer ! Congratulations for your work ! $\endgroup$ – Claude Leibovici Mar 25 '15 at 5:17
  • $\begingroup$ @ClaudeLeibovici: thank you! But I think the summation has to be simplified. I hope to find the time to perform such simplification. $\endgroup$ – giorgiomugnaini Mar 25 '15 at 6:51
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    $\begingroup$ Giorgio, you probably didn't know about this, but after your edit number #10, the site software automatically flagged your answer summoning a moderator to the scene. See this meta discussion as well as threads linked to that explaining why several edits in a short time span occasionally irritate other users. No harm, no foul. I just wanted to make sure that you know about. If you sense the need to fine tune your presentation a lot again, you may consider using the sandbox. A nice answer, by the way :-) $\endgroup$ – Jyrki Lahtonen Mar 28 '15 at 14:41
  • $\begingroup$ Ooops! I am sorry. Ok, I will use the sandbox.. $\endgroup$ – giorgiomugnaini Mar 28 '15 at 16:45

I do not see how a closed form could be found for the integral.

What you can notice is that each derivative term write as a polynomial $$\frac{{d}^m}{{d}x^m} \Big(e^{-x^2}\Big)=P_m(x)~~ e^{-x^2} $$ which makes $$\frac{{d}^m}{{d}x^m} \Big(e^{-x^2}\Big)~~\frac{{d}^n}{{d}x^n} \Big(e^{-x^2}\Big)~~x^k~~e^{x^2}=P_{m+n+k}(x) ~~e^{-x^2}$$ and, so, you are let with a sum of integrals $$I_p=\int_{-\infty}^\infty x^p ~e^{-x^2} \,dx=\frac{1}{2}~ \Big(1+(-1)^p\Big)~~ \Gamma \Big(\frac{p+1}{2}\Big)$$ You can notice that $$I_{2p-1}=0$$ $$I_{2p}=\Gamma\Big(p+\frac{1}{2}\Big)$$ which, more than likely, will make the entire integral as the product of a polynomial in $k$ by $\Gamma \Big(k+\frac{1}{2}\Big)$.

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    $\begingroup$ Hermite polynomials: $$H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}$$ so integral becomes: $$ (-1)^{m+n}\int_{-\infty}^\infty H_m(x) H_n(x) x^k e^{-x^2} \mathrm{d}x $$ $\endgroup$ – giorgiomugnaini Mar 24 '15 at 6:53
  • $\begingroup$ @giorgiomugnaini. Thank you for this nice answer ! May I suggest you replace this comment by an answer ? $\endgroup$ – Claude Leibovici Mar 24 '15 at 7:01
  • $\begingroup$ But it is not yet a complete answer! $\endgroup$ – giorgiomugnaini Mar 24 '15 at 7:03
  • $\begingroup$ May be, but it is beautiful ! $\endgroup$ – Claude Leibovici Mar 24 '15 at 7:03
  • $\begingroup$ Ok, I followed your suggestion, but I don't know when I will work again to improve the answer... $\endgroup$ – giorgiomugnaini Mar 24 '15 at 7:21

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