$\int_{-\infty}^{+\infty}\frac1{1+x^2}\left(\frac{\mathrm d^n}{\mathrm dx^n}e^{-x^2}\right)\mathrm dx$ Evaluate How can I solve this problem?
$$
\int_{-\infty}^{+\infty}\frac1{1+x^2}\left(\frac{\mathrm d^n}{\mathrm dx^n}e^{-x^2}\right)\mathrm dx
$$
 A: 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 $a_{2n}$.
A: 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 
$$B(t) = \frac{-1}{\sqrt{4t+1}} e^{-1 + 1/(4t+1)}$$
A: $$
\int_{-\infty}^{+\infty}\frac1{1+x^2}\left(\frac{\mathrm d^n}{\mathrm dx^n}e^{-x^2}\right)\mathrm dx = \,U\left(1+\frac{n}{2},\frac{3}{2},1\right)\cos\left(\frac{\pi\,n}{2}\right)\,n!\,\sqrt{\pi},
$$
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.
