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Evaluate: $$\int_{0}^{\pi/2}(\cos x)^{a+b}\cos (b-a)x\, dx,0<a<1,a+b>1$$

This integral looks very simple, but I can't solve it. Any help is appreciated.

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  • $\begingroup$ Hint: Integrate $${z^p}{(\frac{{z + 1/z}}{{2i}})^q}$$ around the vertical semicircle in the 1st & 4th quadrant enables you to conclude $$\int_0^{\pi /2} (\cos x)^q \cos(px) dx = \frac{{{2^{ - q - 1}}\Gamma (1 + q)}}{{\Gamma (1 + \frac{{p + q}}{2})\Gamma (1 - \frac{{p - q}}{2})}} $$ $\endgroup$
    – pisco
    Jan 20, 2018 at 8:13
  • $\begingroup$ @pisco125 Thank you for your hint, Sorry. I didn't learn contour integral. Can you help me? $\endgroup$
    – JamesJ
    Jan 20, 2018 at 8:16
  • $\begingroup$ @pisco125 Great hint!,thank you very much,The result is beautiful! $\endgroup$
    – JamesJ
    Jan 20, 2018 at 8:20

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