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 $$\int_0^{\pi /2} (\cos x)^q \cos(px) dx = \frac{{{2^{ - q - 1}}\Gamma (1 + q)}}{{\Gamma (1 + \frac{{p + q}}{2})\Gamma (1 - \frac{{p - q}}{2})}} $$ $\endgroup$– piscoJan 20, 2018 at 8:13
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$\begingroup$ @pisco125 Thank you for your hint, Sorry. I didn't learn contour integral. Can you help me? $\endgroup$– JamesJJan 20, 2018 at 8:16
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$\begingroup$ @pisco125 Great hint!,thank you very much,The result is beautiful! $\endgroup$– JamesJJan 20, 2018 at 8:20
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