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?

  • $\begingroup$ The value of the integral is $$G_{3, 2}^{1, 3} \left( 1 \middle| {1, 1, \frac 3 2 \atop 1, \frac 1 2} \right),$$ $G$ being the Meijer G-function. It's unlikely to have any significantly simpler closed form, it doesn't even seem to be expressible as a sum of generalized hypergeometric functions. $\endgroup$ – Maxim Jan 3 at 1:46
  • $\begingroup$ Are we allowed to use any other method to solve this integral? 'cause the Cauchy theorem doesn't seem to be applicable here......... $\endgroup$ – Mostafa Ayaz Jan 4 at 15:13

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.


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