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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, 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 $n$. –  J. M. Dec 22 '11 at 13:58
    
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. The generalized exponential integral satisfies a recursion relation, so further simplification is very much possible. –  J. M. Dec 22 '11 at 14:31
    
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 $$\frac{1}{2^nn!\sqrt{\pi}}\int_{-\infty}^{+\infty}e^{-x^{2}}f(x)H_{n}(x)dx$$. –  gex Dec 22 '11 at 14:36
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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. –  J. M. Dec 22 '11 at 14:37
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You've seen my second comment, no? –  J. M. Dec 22 '11 at 15:09
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3 Answers 3

Write $f(x) \leftrightarrow F(\omega)$ to mean that $F(\omega)$ is the Fourier transform of $f(x)$, where $$F(\omega)) = \int_{-\infty}^\infty f(x) \exp(-i\omega x)\mathrm dx.$$ Then we have $$\begin{align*} \frac{1}{1+x^2} &\leftrightarrow \pi \exp(-\vert \omega \vert )\\ \exp(-x^2) &\leftrightarrow \sqrt{\pi}\exp(- \omega^2/4),\\ \frac{\mathrm d^n}{\mathrm dx^n}\exp(-x^2) &\leftrightarrow (i\omega)^n\sqrt{\pi}\exp(-\omega^2/4) \end{align*} $$ The inner-product version of Parseval's theorem is $\displaystyle \int_{-\infty}^\infty f^*(x) g(x)\mathrm dx = \frac{1}{2\pi}\int_{-\infty}^\infty F^*(\omega) G(\omega)\mathrm d\omega$ and so we have $$\begin{align*} \int_{-\infty}^\infty \frac{1}{1+x^2} \frac{\mathrm d^n}{\mathrm dx^n}\exp(-x^2) \mathrm dx &= \frac{1}{2\pi}\int_{-\infty}^\infty \pi\exp(-\vert \omega \vert) (i\omega)^n \sqrt{\pi} \exp(-\omega^2/4) \mathrm d\omega \end{align*}$$ where the integrand on the right is an odd function of $\omega$ when $n$ is odd and so the integral is $0$. Otherwise, convert to an integral on $\mathbb R^+$ to get $$\begin{align*} a_{2n} &= \frac{\sqrt{\pi}}{2}\int_{-\infty}^\infty \exp(-\vert \omega \vert) (i\omega)^{2n} \exp(-\omega^2/4) \mathrm d\omega\\ &= (-1)^n\sqrt{\pi}\int_0^\infty \omega^{2n}\exp(-\omega) \exp(-\omega^2/4) \mathrm d\omega\\ &= (-1)^n\sqrt{\pi}b_{2n} \end{align*}$$ where $\displaystyle b_n = \int_0^\infty \omega^{n}\exp(-\omega) \exp(-\omega^2/4) \mathrm d\omega$ can be integrated by parts to get $$\begin{align*} b_{n} &= \int_0^\infty \omega^{n-1}\exp(-\omega) \cdot \omega \exp(-\omega^2/4) \mathrm d\omega\\ &= \left[ -2\omega^{n-1}\exp(-\omega) \exp(-\omega^2/4)\biggr|_0^\infty \right . \\ &\quad \quad\quad \quad\quad \left . + \int_0^\infty 2[(n-1)\omega^{n-2}\exp(-\omega) - \omega^{n-1}\exp(-\omega)]\exp(-\omega^2/4) \mathrm d\omega\right ]\\ &= -2b_{n-1} + 2(n-1)b_{n-2} \end{align*}$$ or, with a slight change in indices $$b_{n+1} = -2b_n +2nb_{n-1}.$$ Note that $b_0 = e\sqrt{\pi}\text{erfc}(1)$ as noted here while $b_1 = 2(1-b_0)$. Thus, the recurrence relation can be used to find $b_{2n}$ and hence $a_{2n}$.

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It appears that your integral for even $n = 2 k$ has the exponential generating function (egf) $$F(t) = \frac{\pi (1 - \text{erf}(1/\sqrt{4t+1}))}{\sqrt{4t+1}} e^{1/(4t+1)}$$ That is $$\int_{-\infty}^{\infty} \frac{1}{1+x^2} \frac{d^{2k}}{dx^{2k}} e^{-x^2} \ dx = F^{(k)}(0) $$

Each of these values will be of the form $a_k \sqrt{\pi} + b_k \pi e (\text{erf}(1) - 1)$ where $a_k$ and $b_k$ are integers. The egf for $a_k$ is $$A(t) = \frac{\sqrt{\pi} (\text{erf}(1) - \text{erf}(1/\sqrt{4t+1}))}{\sqrt{4t+1}} e^{1/(4t+1)}$$ and the egf for $b_k$ is $$B(t) = \frac{-1}{\sqrt{4t+1}} e^{-1 + 1/(4t+1)}$$

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$$ \int_{-\infty}^{+\infty}\frac1{1+x^2}\left(\frac{\mathrm d^n}{\mathrm dx^n}e^{-x^2}\right)\mathrm dx = \,U\left(1+\frac{n}{2},\frac{3}{2},1\right)\cos\left(\frac{\pi\,n}{2}\right)\,n!\,\sqrt{\pi}, $$ where $U(a,b,z)$ is the confluent hypergeometric function of the second kind.

The proof is quite long. I started with applying the Fourier transform to $e^{-x^2}$ so that the derivative $\frac{\mathrm d^n}{\mathrm dx^n}$turns into a multiplication by $(-i\, t)^n$. The inverse transform gives a general form for n-th derivative in terms of hypergeometric functions.

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