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


  • $\begingroup$ After some scaling, substitution it is enough to do: $$\int_{\mathbb {R}^+}\frac{x^{a-1}\sin(-bx+a\pi/2)}{x^2+1}\,dx$$ $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.