Ytukyg

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 $[i,iinfty)$ and $[-i,-iinfty)$, 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 $mathbb{C}setminus [i,iinfty) cap [-i,-iinfty)$ 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 $alpha, betainmathbb{C}$. 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

$$lim_{ato 0} frac{1}{2} int_{-infty}^infty q(a, x) arctan(x)e^{-x^2},dx$$
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?

Applying of the Residue theorem is a very hard task, because the function $e^{-x^2}$ is not bounded when $zto icdot infty.$

On the other hand, parametric method can be used.

Let us consider the integral
$$begin{align}
&I(p) = intlimits_0^inftyarctan pz, e^{-z^2},mathrm dz,\[4pt]
&I'(p) = intlimits_0^inftydfrac{ze^{-z^2}}{p^2z^2+1},mathrm dz
= dfrac12 p^{-2}intlimits_0^inftydfrac{e^{-z^2}}{z^2+p^{-2}},mathrm dz^2.\[4pt]
end{align}$$
Using known definite integral
$$int_0^inftydfrac{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 intlimits_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|
= dfrac12intlimits_{p^{-1}}^infty e^{t^2}mathrm {E}_1(t^2), dt,$$
$$boxed{I(1) = dfrac12intlimits_{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.