I want to know if the residue method is applicable to calculate the integral (which kind of contour?) $$ \int_0^{+ \infty} \frac{\sin x }{1+x^2} dx. $$ We know that it is easy to calculate this integral if we replace $ \sin $ by $ \cos $ because of this function is even and then we can pass to the integral over $ ]- \infty, + \infty[. $
Thank you in advance.
Elaborating after @Sangchul Lee 's comment, consider first $$\frac{1}{x^2+1}=\frac{1}{(x+i)(x-i)}=\frac i 2\left(\frac 1{x+i} -\frac 1{x-i}\right)$$ and so $$\int \frac{e^{ix}}{x^2+1} dx=\frac i 2\left(\int \frac {e^{ix}}{x+i}dx-\int \frac {e^{ix}}{x-i}dx\right)$$ For the first one, let $x+i=y$ $$\int \frac {e^{ix}}{x+i}dx=e\int \frac {e^{iy}}{y}dy=e \,\text{Ei}(i y)$$ Do the same for the second one and go back to $x$ $$\int \frac{e^{ix}}{x^2+1} dx=\frac i 2\left(e \,\text{Ei}(i x-1)-\frac{\text{Ei}(i x+1)}{e}\right)$$ $$\int_0^\infty \frac{e^{ix}}{x^2+1} dx=\frac{\pi }{2 e}+\frac i {2e}\left( \text{Ei}(1)-e^2 \text{Ei}(-1)\right)$$ and, unfortunately for you, $$\int_0^\infty \frac{\sin(x)}{x^2+1} dx=\frac{e\operatorname{Ei}(-1)+e^{-1}\operatorname{Ei}(1)}{2}\approx 0.646761$$