# Evaluate $\int_0^\infty x^{a-1}\frac{\sin(\frac{1}{2}a \pi-bx)}{x^2+r^2}\, r\,dx$

Evaluate $$\int_0^\infty x^{a-1}\frac{\sin(\frac{1}{2}a \pi-bx)}{x^2+r^2}\, r\,dx$$

with $0<a<2$, $b>0$, $r>0$, using methods of complex analysis.

I can't find a proper contour because of the sine term (when $z=Re^{i\theta}$ it doesn't go to zero in the bottom complex plane as $R$ go to infinity).

The answer is $\frac{1}{2}\pi r^{a-1}e^{-br}$

Thanks

• After some scaling, substitution it is enough to do: $$\int_{\mathbb {R}^+}\frac{x^{a-1}\sin(-bx+a\pi/2)}{x^2+1}\,dx$$ – Shashi Feb 6 '18 at 17:13

Taking principal logarithm, integrate $$f(z) = \frac{{{z^{a - 1}}}}{{{z^2} + {r^2}}}{e^{ibz}}$$ around semicircle contour in the upper-half plane, the integral along the circle vanishes (Jordan's lemma). Hence $$\int_{ - \infty }^\infty {\frac{{{x^{a - 1}}}}{{{x^2} + {r^2}}}{e^{ibx}}dx} = \int_0^\infty {\frac{1}{{{x^2} + {r^2}}}\left[ {{x^{a - 1}}{e^{ibx}} + {{( - x)}^{a - 1}}{e^{ - ibx}}} \right]dx} = 2\pi i{(ir)^{a - 1}}\frac{{{e^{ib(ir)}}}}{{2ir}}$$
Some simplification then gives $$\int_0^\infty {\frac{{{x^{a - 1}}}}{{{x^2} + {r^2}}}\left[ {{e^{ - (a - 1)\frac{{\pi i}}{2}}}{e^{ibx}} + {e^{(a - 1)\frac{{\pi i}}{2}}}{e^{ - ibx}}} \right]dx} = \pi {r^{a - 1}}\frac{{{e^{ - br}}}}{r}$$ so $$\int_0^\infty {\frac{{{x^{a - 1}}}}{{{x^2} + {r^2}}}\cos \left[ {(a - 1)\frac{\pi }{2} - bx} \right]dx} = \pi {r^{a - 1}}\frac{{{e^{ - br}}}}{{2r}}$$ as desired.