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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.

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  • $\begingroup$ 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
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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 $1$.

Then

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

But

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

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

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

And

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

So we have

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

Or

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

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.

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  • $\begingroup$ 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

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