Thursday 9 June 2016

Evaluation of definite integral using complex analysis

I want to evaluate the following indefinite integral
$$ \int_0^{\infty} x^{p - 1} \cos (ax) dx$$ where $0 < p < 1$ and $a > 0$. I was considering the function $f(z) = z^{p - 1} e^{iaz}$ and integrate it over the contour $\gamma = [-R, -\epsilon] - C_{\epsilon}^+ + [\epsilon, R] + C_R^+$. (Here $C_r^+$ denote the upper half circle centered at $0$ with radius $r$). However I only get
$$ \int_0^{\infty} x^{p - 1} e^{iax} dx - e^{i \pi p} \int_0^{\infty} x^{p - 1} e^{-iax} dx = 0$$ instead, which only gives some relation between $\int_0^{\infty} x^{p - 1} \sin (ax) dx$ and $\int_0^{\infty} x^{p - 1} \cos (ax) dx$. I know the result is somehow related to gamma function that is $$ \int_0^{\infty} x^{p - 1} \cos(x) dx =\frac{ \pi }{ 2 \Gamma(1 - p) \sin( (1 - p) \pi / 2) }$$ So would the regular method using complex analysis can still evaluate this integral?

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