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How can I solve this problem? $$ \int_{-\infty}^{+\infty}\frac1{1+x^2}\left(\frac{\mathrm d^n}{\mathrm dx^n}e^{-x^2}\right)\mathrm dx $$

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    From the Rodrigues formula for the [Hermite polynomials](http://mathworld.wolfram.com/HermitePolynomial.html), we have $$\dfrac{\mathrm d^n}{\mathrm dx^n}\exp(-x^2)=(-1)^n \exp(-x^2) H_n(x)$$ If $n$ is odd, $H_n(x)$ is odd, so the integral is zero for odd $n$.2011-12-22
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    With some *Mathematica* assistance, I obtained $$\int_{-\infty}^\infty \frac{\exp(-x^2)}{1+x^2}H_{2n}(x)\mathrm dx=e\sqrt{\pi}(-1)^n \frac{(2n)!}{n!}\sum_{k=0}^n (-1)^k \binom{n}{k} E_{k+\frac12}(1)$$ where $E_p(z)$ is the [generalized exponential integral](http://dlmf.nist.gov/8.19.E2). The generalized exponential integral satisfies a [recursion relation](http://dlmf.nist.gov/8.19.E12), so further simplification is very much possible.2011-12-22
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    My goal of solving this integral is to find $a_{n}$ for fourier-hermit series expansion of $f(x)=\frac{1}{1+x^2}=\displaystyle\sum_{n=0}^{\infty}a_{n}H_{n}(x)$, where $a_{n}$ is $$\frac{1}{2^nn!\sqrt{\pi}}\int_{-\infty}^{+\infty}e^{-x^{2}}f(x)H_{n}(x)dx$$.2011-12-22
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    You should have said that earlier, you know. I hope that your $a_n$ are zero for odd $n$ is obvious to you.2011-12-22
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    yes,It must be found $H_{2n}(x)$.2011-12-22
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    could you please help me to find $a_{n}$ for even n.2011-12-22
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    You've seen my second comment, no?2011-12-22
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    Yes,thank you for your good comments!2011-12-22

3 Answers 3