An exponential improper integral Evaluate : 
$$\int_{0}^{\infty }{\frac{{{\text{e}}^{-{{x}^{2}}}}}{{{\pi }^{2}}+{{\left( \gamma +x \right)}^{2}}}\text{d}x}$$
 A: Before I begin and draw the reader in, I will say that this exposition outlines a general approach to integrals like these.  Unfortunately, I was unable to evaluate the integral analytically, but I was able to break it into pieces, some of which did have a closed form.
A good approach with integrals like these is to use Parseval's theorem (or Plancherel's Theorem, depending on your preference).  That is, I see two functions in the integrand whose Fourier transform I recognize.  For the record, I define the FT of a function $f$ as
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: f(x) e^{i k x}$$
Parseval's theorem then states that, for functions $f$ and $g$, both of whom have FT's $\hat{f}$ and $\hat{g}$, respectively:
$$\int_{-\infty}^{\infty} dx \: f(x) g^*(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: \hat{f}(k) \hat{g}^*(k) $$
In this case, $f(x) = e^{-x^2} \theta(x)$ and $g(x) = [\pi^2 + (\gamma + x)^2]^{-1}$, where $\theta(x)$ is the Heaviside step function.  The FT's of these functions are
$$\hat{f}(k) = \frac{\sqrt{\pi}}{2} e^{-\frac{k^2}{4}} - i F \left ( \frac{k}{2} \right ) $$
where $F$ is Dawson's integral
$$F(z) = e^{-z^2} \int_0^z dt \: e^{t^2}$$
and
$$\hat{g}(k) = e^{-i \gamma k} e^{-\pi |k|}$$
In multiplying the FT's together, we get a real and imaginary part.  The imaginary part is an odd fucntion of $k$ and its integral over the real line is zero.  The integral we want is then
$$\frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: \left [  \frac{\sqrt{\pi}}{2} e^{-\frac{k^2}{4}} \cos{\gamma k} + F \left ( \frac{k}{2} \right ) \sin{\gamma k} \right ] e^{-\pi |k|} $$
The first part of the integral may be evaluated in terms of Dawson's integral:
$$\frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: \frac{\sqrt{\pi}}{2} e^{-\frac{k^2}{4}} e^{-\pi |k|} \cos{\gamma k} = \frac{1}{2} e^{\pi^2 - \gamma^2} \cos{2 \pi \gamma} - \frac{1}{\sqrt{\pi}} \Im{F(\gamma + i \pi)}$$
The second part may be manipulated into the following form:
$$\begin{align}\frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: F \left ( \frac{k}{2} \right ) e^{-\pi |k|} \sin{\gamma k} &= \frac{1}{4 \pi} \int_0^1 du \: \int_{-\infty}^{\infty} dk \: k \, e^{-\frac{(1-u^2) k^2}{4}} e^{-\pi |k|} \sin{\gamma k}\\ &= -\frac{1}{4 \pi} \frac{\partial}{\partial \gamma}  \int_0^1 du \: \int_{-\infty}^{\infty} dk \: e^{-\frac{(1-u^2) k^2}{4}} e^{-\pi |k|} \cos{\gamma k}\\  \end{align}$$
Although the inner integral looks very much the the integral from the first piece above, I have not been able to evaluate it yet.  I will leave the evaluation here for now.
