Contour for $\int_0^\infty \arctan(z) e^{-z^2}\,dz$ or some variant I'm trying to practice my contour integration skills and got interested in the following integral:
$$\int_0^\infty \arctan(z) e^{-z^2}\,dz$$
I know that the usual way to calculate integrals on $[0,\infty)$ is to use the keyhole-contour, but the problem in this case is that $\arctan(z)$ has branch points at $z=\pm i$, with branch cuts usually chosen on $[i,i\infty)$ and $[-i,-i\infty)$, which I think means that the keyhole contour doesn't work in this case. I have tried a rectangle contour made up of $C:[0,R]\cup [R+i/2] \cup [R+i/2,i/2] \cup [i/2,0]$ but that didn't work out. Because my function is holomorphic on $\mathbb{C}\setminus [i,i\infty) \cap [-i,-i\infty)$ there also wouldn't be any residues to calculate, which doesn't necessarily have to be a problem as Cauchy's theorem could be used to try and compute the integral. 
If I'm not mistaken, I would then get somerhing like $\int_0^\infty f(x)-f(x+i/2)\,dx=0$. One difficulty I ran into is simplify $f(z+i/2)$ (where $f(z)=\arctan(x) e^{-z^2}$) into some other form such as $\alpha f(z)+\beta g(z)$ for some $g(z)$ whose integral can be calculated on $[0,\infty)$ and $\alpha, \beta\in\mathbb{C}$. This would let me then solve for $\int_0^\infty f(x)\,dx$. 
I also thought that, if that makes it easier, we could extend the range of integration to $\mathbb{R}$, as long as we found an odd function $q(a,x)$, such that $q(0,x)=1$ and compute
$$\lim_{a\to 0}  \frac{1}{2} \int_{-\infty}^\infty  q(a, x) \arctan(x)e^{-x^2}\,dx$$
We could either use Cauchy's/the Residue theorem depending on whether $q(z)$ has poles or not. I have not been able to use this approach. 
Any ideas?
 A: Applying of the Residue theorem is a very hard task, because the function $e^{-x^2}$ is not bounded when $z\to i\cdot \infty.$
On the other hand, parametric method can be used.
Let us consider the integral
$$\begin{align}
&I(p) = \int\limits_0^\infty\arctan pz\, e^{-z^2}\,\mathrm dz,\\[4pt]
&I'(p) = \int\limits_0^\infty\dfrac{ze^{-z^2}}{p^2z^2+1}\,\mathrm dz
 = \dfrac12 p^{-2}\int\limits_0^\infty\dfrac{e^{-z^2}}{z^2+p^{-2}}\,\mathrm dz^2.\\[4pt]
\end{align}$$
Using known definite integral
$$\int_0^\infty\dfrac{e^{-ay}}{b+y}\,\mathrm dy=e^{ab}\mathrm {E}_1(ab),$$
one can get
$$I'(p)=\dfrac12 p^{-2}e^{p^{-2}}\mathrm {E}_1(-p^2),$$
$$I(p) = \dfrac12 \int\limits_0^p p^{-2}e^{p^{-2}}\mathrm {E}_1(-p^2) dp 
= \left|p=\frac1t,\, t=\frac1p,\,\mathrm dt=-\frac{\mathrm dp}{p^2}\right|
= \dfrac12\int\limits_{p^{-1}}^\infty e^{t^2}\mathrm {E}_1(t^2)\, dt,$$
$$\boxed{I(1) = \dfrac12\int\limits_{1}^\infty e^{t^2}\mathrm {E}_1(t^2)\, dt}.$$
Numeric calculations give the same result $I(1)\approx0.40978$ for the both expressions, but this result can not presented via elementary functions.
