The task is to calculate this integral using complex analysis: I know some methods, which involve calculating the integral around half of the circle, also it would be wise to add here $i cos(s^{\frac{1}{2}})$ and then take the real part of the result. But here comes my problem, this function isn't even, so I am unable to move the limit 0 to $-\infty$, which makes it problematic, because integrating over one fourth of the circle involves integrating along the imaginary line, which is not simpler than this. Any help appreciated.
$$\int_0^{\infty}\frac{\sin(s^{\frac{1}{2}})}{(1+s)^{2}} ds$$
Best Answer
Integrating by parts gives $$\int_0^{\infty}\frac{\sin(\sqrt s)}{(1+s)^{2}} ds=\int_0^\infty \frac{\cos (\sqrt s)}{1+s}\frac{ds}{2\sqrt s}$$ Now substitution $s=x^2$ and the integral become $$\int_0^\infty \frac{\cos x}{1+x^2}dx$$ Since the integrand is an even function and the real part of $e^{ix}$ is $\cos x$ we have the following equality $$\int_0^\infty \frac{\cos x}{1+x^2}dx=\frac12\int_{-\infty}^\infty \frac{e^{ix}}{(x+i)(x-i)}dx$$ Choosing the contour to be the upper semi-plane where only the pole $z=i$ exists for the function $$f(z)=\frac{e^{iz}}{(z+i)(z-i)}$$ Give the integral to be $$\frac12 \cdot 2\pi i \lim_{z\to i} \cdot (z-i)\cdot \text{Res} f(z)=\pi i \lim_{z\to i} \frac{e^{iz}}{z+i} =\pi i \frac{e^{-1}}{2i}=\frac{\pi}{2e}$$