Integration with residue theorem

Question

I am trying to calculate the following integral: $$ \int\limits_{-\infty}^{\infty}\frac{x \arctan(x-a)}{(x-b)^2+y^2}d x $$ where $a$, $b$ are positive constants and also y is a positive variable. (Actually the integral comes from Poisson's formula for the half-space). I have calculated a similar integral without $x$ factor by assuming $h(z)=\ln(1+I(z-a))$, $R(z)=\frac{1}{(z-b)^2+y^2}$ and $f(z)=h(z)R(z)$. Then calculation of $$ \int_C f(z)dz $$ from residue theorem gives the correct results for the considered integral, as can be found in Yu. V. Sidorov, M. V. Fedoryuk, M. I. Shabunin, Lectures on the Theory of Functions of a Complex Variable. In the book it is stated that in order to have zero contribution from the semicircle of the integration counter for $$ \int_C z^{\alpha-1} R(z)dz $$ the function $R(z)$ must have $\displaystyle\lim_{z\to 0}|z|^\alpha R(z)=0$ and $\displaystyle\lim_{z\to \infty}|z|^\alpha R(z)=0$. However, for the above interval $\displaystyle\lim_{z\to \infty}|z|^2 R(z)\neq 0$, indicating there must be a contribution from the semicircle that I could not evaluate. Can anyone suggest how to calculate this integral or to obtain the contribution from the semicircle?