I have the following conjecture: \begin{equation} \text{Re}\left[(1+\text{i}y)\arctan\left(\frac{t}{1+\text{i}y}\right)\right] \ge \arctan(t), \qquad \forall y,t\ge0. \end{equation}
Which seems to be true numerically. Can anyone offer some advice on how to approach proving (or disproving) this?
It originates from a question involving the (complex) Hilbert transform of a symmetric non-increasing probability distribution: \begin{equation} h(y) = (1+\text{i}y)\int_{-\infty}^\infty \frac{1}{1 + \text{i}(y-t)}\text{d}G(t) \end{equation} and attempting to show $\text{Re}[h(y)]$ takes its minimum at $y=0$.
I start at the beginning, with a symmetric non-increasing density $\varphi$ on $\mathbb R$. The layer cake representation reduces the matter to $\varphi=\chi_{[-a,a]}$. Therefore, the goal is to show that $$\int_{-a}^a \operatorname{Re}\frac{1+iy}{1+i(y-t)}\,dt > \int_{-a}^a \operatorname{Re}\frac{1 }{1-i t}\,dt \tag1$$ for all $y>0$ and $a>0$. Let's find this real part: $$ \operatorname{Re}\frac{1+iy}{1+i(y-t)} = \frac{1+y^2-ty}{1+(y-t)^2} \tag2$$ and add the value at $-t$ so that we integrate over $[0,a]$ only: $$ \frac{1+y^2-ty}{1+(y-t)^2}+ \frac{1+y^2+ty}{1+(y+t)^2} = 2\,\frac{1+2y^2+t^2+y^4-y^2t^2}{(1+(y-t)^2)(1+(y+t)^2)} \tag3$$ Looks bad, but we have to keep going. The goal is to prove that
$$ \int_0^a 2\,\frac{1+2y^2+t^2+y^4-y^2t^2}{(1+(y-t)^2)(1+(y+t)^2)}\,dt > \int_0^a \frac{2}{1+t^2}\,dt \tag4$$ Take the difference of two sides, and it simplifies! $$ \int_0^a \frac{2 y^2 t^2 (3+y^2-t^2)}{(1+t^2)(1+(y-t)^2)(1+(y+t)^2)}\,dt > 0 \tag5$$ Sweet. The denominator is always positive. The numerator changes sign only once, from plus to minus. Therefore, as a function of $a$, the integral in (5) increases from $0$ (at $a=0$) to some positive value, and then decreases to its limit as $a\to\infty$. It turns out that the limit as $a\to\infty$ is $0$: $$ \int_0^\infty \frac{2 y^2 t^2 (3+y^2-t^2)}{(1+t^2)(1+(y-t)^2)(1+(y+t)^2)}\,dt = 0 \tag6$$ which completes the proof.
Well, I ought to prove (6) instead of trusting my computer. Let
$$ R(z) = \frac{2 y^2 z^2 (3+y^2-z^2)}{(1+z^2)(1+(y-z)^2)(1+(y+z)^2)},\quad z\in\mathbb C \tag7$$ This is an even rational function which is $O(|z|^{-2})$ at infinity. Hence, $\int_0^\infty R(z)\,dz$ is $\pi i$ times the sum of residues of $R$ in the upper halfplane. For computing the residues, it makes sense to go back to where $R$ came from: $$ R(z) = \frac{1+y^2-zy}{1+(y-z)^2}+ \frac{1+y^2+zy}{1+(y+z)^2} -\frac{2}{1+z^2} \tag8$$ From (8) we easily get $$ \begin{split} \operatorname{res}\limits_{z=i} R(z) &= -\frac{2}{2i} = i \\ \operatorname{res}\limits_{z=i-y} R(z) &= \frac{1+y^2+(i-y)y}{2(i-y)} =\frac{1+iy}{2(i-y)} =-\frac{i}{2}\\ \operatorname{res}\limits_{z=i+y} R(z) &= \frac{1+y^2-(i+y)y}{2(i+y)} = \frac{1-iy}{2(i+y)}= -\frac{i}{2} \end{split} \tag{9}$$ which indeed sum to $0$.