Goodwin Staton integral $G(x) = \int_0^\infty \frac{e^{-t^2}}{t+x}dt$ and its symmetry The Goodwin Staton integral 
$$G(x) = \int_0^\infty \frac{e^{-t^2}}{t+x}dt$$ 
is said on Wikipedia to have the symmetry 
$$G(x) = -G(-x)$$
I'm not convinced by this symmetry... indeed if we consider $G(-x)$ and we choose $k = -t$ this integral becomes
$$G(-x) = \int_0^{-\infty} \frac{e^{-k^2}}{-k-x}(-dk)$$ 
or
$$G(-x) = -\int_{-\infty}^{0} \frac{e^{-k^2}}{k+x}dk$$ 
which does not seem to be equal to $-G(x)$ to me...
Any suggestions ?
EDIT : actually the symmetry of this integral is part from my problem. My final goal is to compute this integral :
$$PV. \left( \int_{0}^{\infty} \frac{e^{-a^2(k-q)^2} \; k\; dk}{k_0^2-k^2} -\int_{0}^{\infty} \frac{e^{-a^2(k+q)^2} \; k\; dk}{k_0^2-k^2}\right) $$ 
If i'm not wrong, with $k\to -k$ in second integral we get that
$$PV. \left( \int_{-\infty}^{\infty} \frac{e^{-a^2(k-q)^2} \; k\;dk}{k_0^2-k^2} \right) $$ 
Then by taking $k\to k+k_0$, we finally have
$$ - PV. \left( \int_{-\infty}^{\infty} \frac{e^{-a^2(k-q)^2} \;dk}{k+2k_0} + k_0 \int_{-\infty}^{\infty} \frac{e^{-a^2(k-q)^2} \;dk}{k(k+2k_0)} \right) $$ 
which I do not know how to handle. These seem to be closely related to these Goodwin-Staton integrals / Dawson functions... But. Meh

I think Wikipedia is wrong. The integral does not converge for $x<0, $ try e.g. int(e^(-t^2)/(t-2),t=0..infinity) in Wolfram Alpha. It can be interpreted as a Cauchy principle value (see Nico Temme's answer http://mathforum.org/kb/message.jspa?messageID=7389647). I use 
$$G(-x) = - \frac{1}{2} e^{-x^2}\left(\pi\; \mathrm{erfi}(x) + \mathrm{Ei}(x^2)\right), \quad x>0$$
or
$$G(-x) = \sqrt{\pi} F(-x) - \frac{1}{2} e^{-x^2}\mathrm{Ei}(x^2), \quad x>0.$$
i.e. http://dlmf.nist.gov/7.5.E13 extended to negative $x$
(with the Dawson integral $F$ and exponential integral $\mathrm{Ei}$). The Dawson integral is odd, and therefore $G(x)$ is not odd.
