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?

  • 2
    $\begingroup$ Terminology correction: this is not an indefinite integral, it is an improper integral. $\endgroup$ – Santiago Canez Apr 28 '14 at 23:52
  • $\begingroup$ Hint: $~\displaystyle\Gamma(p)=\int_0^\infty x^{p-1}e^{-x}dx\quad=>\quad\int_0^\infty x^{p-1}\cos\big(ax\big)dx=\Re\int_0^\infty x^{p-1}e^{-iax}dx.~$ Let $t=iax$. $\endgroup$ – Lucian Apr 29 '14 at 6:57
  • $\begingroup$ You might want to check the $\sin{(\pi (1-p)/2)}$ factor in your expression; I got $\sin{(\pi p/2)}$ instead. WA agrees. Also note that your expression remains finite as $p \to 0$. $\endgroup$ – Ron Gordon Apr 29 '14 at 17:55

Use a contour in the first quadrant, that is, consider

$$\oint_C dz \, z^{p-1} \, e^{i a z}$$

where $C$ is the contour consisting of $[\epsilon,R]$, the quarter-circle of radius $R$ in the first quadrant centered at the origin, $[i R,i \epsilon]$, and a quarter circle of radius $\epsilon$ about the origin.

The integral then becomes

$$\int_{\epsilon}^R dx \, x^{p-1} \, e^{i a x} + i R^p \int_0^{\pi/2} d\theta \, e^{i p \theta} \, e^{i a R e^{i \theta}} \\ + i^p \int_R^{\epsilon} dy \, y^{p-1} \, e^{-a y} + i \epsilon^p \int_{\pi/2}^0 d\phi \, e^{i p \phi} \, e^{i a \epsilon e^{i \phi}}$$

The fourth integral vanishes as $\epsilon \to 0$ as $0 \lt p \lt 1$. Also, the second integral vanishes as $R \to \infty$, as its magnitude is bounded by

$$R^p \int_0^{\pi/2} d\theta \, e^{-a R \sin{\theta}} \le R^p \int_0^{\pi/2} d\theta \, e^{-2 a R \theta/\pi} \le \frac{\pi}{2 a R^{1-p}}$$

and, again, $0\lt p \lt1$.

The contour integral is zero by Cauchy's theorem as there are no poles inside the contour. Therefore, in the above limits, we have

$$\int_0^{\infty} dx \, x^{p-1} \, e^{i a x} = i^p \int_0^{\infty} dy \, y^{p-1} \, e^{-a y} = e^{i p \pi/2} \frac{\Gamma(p)}{a^p}$$

Use the relationship

$$\Gamma(p) \Gamma(1-p) = \frac{\pi}{\sin{\pi p}}$$

and we have, taking real parts of the above,

$$\int_0^{\infty} dx \, x^{p-1} \, \cos{(a x)} = \frac{\pi \cos{(\pi p/2)}}{a^p \sin{(\pi p)} \Gamma(1-p)} = \frac{\pi}{2 a^p \sin{(\pi p/2)} \Gamma(1-p)}$$

  • $\begingroup$ Who downvoted this? Ridiculous. $\endgroup$ – Bennett Gardiner Apr 30 '14 at 1:15

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