So, I've been trying to prove the following integral related to the gamma function, and I'm really banging my head against the wall over this:

$$\int_0^\infty{\cos{t^p}}dt=\frac{1}{p}\cos({\frac{\pi}{2p}})\Gamma({\frac{1}{p}})$$ Note: $p>1$

Does anyone have any ideas or resources which could point me in the right direction for proving this? I have tried complex contour integration using a semi-circle, messing around with Hankel's representation, and a few other things. However, I can't come up with anything that looks remotely similar to this. Any insight would be GREATLY appreciated!

Edit: I have also tried ideas in this post: Calculating $\int_{0}^{\infty} x^{a-1} \cos(x) \ \mathrm dx = \Gamma(a) \cos (\pi a/2)$ . For some reason, I am still not ending up with the desired expression.

  • $\begingroup$ Since $e^{ix}=\cos x+i\sin x$, and $\displaystyle\int_0^\infty e^{-x^n}dx=\Gamma\Big(1+\tfrac1n\Big)=\tfrac1n\cdot\Gamma\Big(\tfrac1n\Big)$, it should come as no surprise that $\displaystyle\int_0^\infty\cos(x^n)dx = \cos\frac\pi{2n}\cdot\Gamma\Big(1+\tfrac1n\Big)$ and $\displaystyle\int_0^\infty\sin(x^n)dx = \sin\frac\pi{2n}\cdot\Gamma\Big(1+\tfrac1n\Big)$. Also, a similar question has been asked just recently. $\endgroup$ – Lucian Apr 5 '14 at 3:27
  • $\begingroup$ @Lucian I don't know; that is definitely not obvious to me. Could you please flesh out the details a bit? $\endgroup$ – Incognito Apr 5 '14 at 3:48
  • $\begingroup$ Let $t^n=ix^n$. What are the real and imaginary parts of $\sqrt[n]i$ ? $\endgroup$ – Lucian Apr 5 '14 at 4:02
  • $\begingroup$ @Lucian Well, yeah; that would give us the $\cos({\frac{pi}{2n}})$ term, but that substitution messes up the bounds, right? We'd have to integrate from 0 to $i\infty$, correct? $\endgroup$ – Incognito Apr 5 '14 at 4:30
  • $\begingroup$ Not necessarily. See Riemann sphere. Besides: a little bit of handwaving never hurt anyone. ;-) $\endgroup$ – Lucian Apr 5 '14 at 5:33

If you want to derive the result from scratch using contour integration (and justify the substitution that Lucian made), consider $f(z) = e^{iz^{p}}$ and integrate around a sector/wedge of radius $R$ that makes an angle of $\frac{\pi}{2p}$ with the positive real axis.

I'm just going to consider the case where $p$ is an integer greater than $1$.


$$\int_{0}^{\infty} f(x) \ dx + \lim_{R \to \infty} \int_{0}^{\pi/(2p)} f(Re^{it}) iRe^{it}\ dt + \int^{\infty}_{0} f(te^{i \pi / (2p)}) e^{i \pi/(2p)} \ dt =0 $$


$$\Big| \int_{0}^{\pi/(2p)} f(Re^{it}) iRe^{it}\ dt \Big| =\int_{0}^{\pi/(2p)} |e^{iR^{p}e^{ipt}} | R \ dt = \int_{0}^{\pi/(2p)} e^{-R^{p} \sin pt} R \ dt$$

$$ \le \int_{0}^{\pi /(2p)} e^{-R^{p} \frac{2}{\pi} pt} R \ dt \ \ (\text{Jordan's inequality})$$

$$ = R \frac{\pi}{2} \frac{1}{pR^{p}} \Big(1-e^{-R^{p}} \Big) \to 0 \ \text{as} \ R \to \infty \ \text{since p >1}$$


$$\int_{\infty}^{0} f(te^{i \pi / (2p)}) e^{i \pi/(2p)} \ dt = - \int_{0}^{\infty} e^{it^{p}e^{i \pi/2}}e^{i \pi/(2p)} \ dt = -e^{i \pi/(2p)}\int_{0}^{\infty} e^{-t^{p}} \ dt $$

So we have

$$ \int_{0}^{\infty} e^{ix^{p}} \ dx = e^{i \pi /(2p)} \int_{0}^{\infty} e^{-t^{p}} \ dt$$


$$ \int_{0}^{\infty} \Big( \cos(x^{p}) + i \sin(x^{p}) \Big) \ dx = \left[ \cos \left(\frac{\pi}{2p} \right) + i\sin \left( \frac{\pi}{2p} \right) \right]\frac{1}{p}\Gamma\left( \frac{1}{p} \right)$$

If you want to extend the result to all real values of $p$ greater than $1$, indent the contour around the origin and show that it's contribution is vanishing small as the radius of the indentation goes to $0$. The computation is very similar to the one showing that the integral around the big wedge vanishes.

  • $\begingroup$ Very good response; thank you! Do we not need a branch cut for p>1 though? What if p = 3/2 for example? $\endgroup$ – Incognito Apr 5 '14 at 16:01
  • $\begingroup$ Yes. If p is not an integer, there is a branch point at the origin. So we need a branch cut somewhere. Have it running down the negative imaginary axis (or along the negative real axis) and then indent the contour around the origin. Then what is left, as I mentioned above, is to show that the contribution from the indentation goes to $0$ as the radius of the indentation goes to $0$. Basically just replace $R$ with $r$ in the estimation I did above, and then let $r$ go to $0$. $\endgroup$ – Random Variable Apr 5 '14 at 16:18
  • $\begingroup$ Thank you for the insight, sir! $\endgroup$ – Incognito Apr 5 '14 at 16:46

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.