Contour integral of $\int_{-\infty} ^{\infty} \frac{x\sin(ax)}{x^2+b^2}\,dx$ for $a,b<0$. Bounding $\int_{Arc}\frac{z\sin(az)}{z^2+b^2}\,z$ step

Question

Well I wish to use the Contour method to calculate $$\int_{-\infty} ^{\infty} \frac{x\sin(ax)}{x^2+b^2}\,dx$$ . Well I started with the standard method - $$\lim_{y\rightarrow\infty}\int_{z\in[-y,y]}\frac{z\sin(az)}{z^2+b^2} = \lim_{y\rightarrow\infty}\int_{z\in \gamma_y}\frac{z\sin(az)}{z^2+b^2} - \int_{C_y}\frac{z\sin(az)}{z^2+b^2}$$ when $C_y$ is the upper half arc of a circle with radius $y$ and $\gamma_y = C_y \cup[-y,y]$ in order to include $-i\beta$ a singularity point and use Cauchy's theorem to estimate $$\int_{z\in \gamma_y}\frac{z\sin(az)}{z^2+b^2} = \int_{z\in \gamma_y}\frac{\frac{z\sin(az)}{(z-ib)}}{z-(-ib)} = 2\pi i\cdot I(\gamma_y,-ib)\cdot (\frac{z\sin(az)}{z-ib}|_{-ib} )$$ with is independent of $y$. However I don't succeed to bound the integral $$|\int_{C_y}\frac{z\sin(az)}{z^2+b^2}\,dz|$$ in order to show that the limit: $$\lim_{y\rightarrow \infty }\int_{C_y}\frac{z\sin(az)}{z^2+b^2}\,dz =0$$ , a bound for $\frac{|z|}{|z^2+b^2|}$ when $z\in C_y$ is $$\frac{|y|}{||z^2|-|b^2||} =\frac{|y|}{|y^2-b^2|}$$ which indeed tends to zero when $y$ approaches infinity. But I don't see how to bound $|\sin(az)|$ when $a<0$ $z\in C_y$.

Answer 1

One of the points on your arc is $z=iy$, and at this point you have $$|\sin(az)|=|\sin(aiy)|=|i\sinh(ay)|=|\sinh(ay)|\ .$$ Since $|\sinh(ay)|\to\infty$ as $y\to\infty$, this expression is unbounded, and your attempt is doomed.

That is why the standard way to evaluate your integral is to consider $$\int_C\frac{ze^{iaz}}{x^2+b^2}\,dz\qquad\hbox{and not}\qquad \int_C\frac{z\sin(az)}{x^2+b^2}\,dz\ .$$