11

In my recent question about the Fransén-Robinson constant, an answer was given using the Gamma reflection formula. However, as an AP Calculus student, I didn't quite understand how the reflection formula worked. After two days of research, I have only found explanations for the Gamma reflection formula in terms of Weierstrass products, which I don't begin to understand.

Is there a proof for the Gamma reflection formula by which I can understand, or at least begin to understand, how this formula works?

Community
  • 1
Addison Crump
  • 347
  • Lebedev, Special functions, section 1.2 uses a double-integral approach:

    $$\Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}, d t \qquad (\Re z > 0)$$

    $$\Gamma(z)\Gamma(1-z) = \int_{0}^{\infty}\int_{0}^{\infty}s^{-z}t^{z-1}e^{-(s+t)},ds, dt \qquad (0<\Re z <1)$$

    The double-integral can be evaluated as $\frac{\pi}{\sin \pi z}$. The he uses continuation to extend the formula to all $z\in \mathbb{C}$ without the negative integers.

     – gammatester May 30 '16 at 13:10
  • @gammatester Can you add this as an answer? :P – Addison Crump May 30 '16 at 13:14
  • I could. But it will be just a quote from Lebedev. Is this OK? – gammatester May 30 '16 at 13:16
  • @gammatester As long as you link the source, it should be. :P I suppose that I'd have to cite the source, though. :\ – Addison Crump May 30 '16 at 13:17
  • I'm not sure if you know how to do contour integration, but here's a link for a proof using that https://highvoltagemath.wixsite.com/highvoltagemath/euler-s-reflection-formula-take-1 – Highvoltagemath Jul 24 '19 at 00:52
  • 1
    @Highvoltagemath: Wouldn't it be conceptually simpler to first rewrite $$\int_0^\infty \frac {v^{z-1}} {1+v} , dv \int_{-\infty}^\infty \frac {e^{zw}} {1 + e^w} , dw$$ and only then pass to the complex plane? The map $w \mapsto v = e^w$ is the universal cover of the punctured plane $\mathbb C - { 0 }$. In the universal cover, there are no branch cuts to talk about. – isekaijin Apr 10 '20 at 16:02
  • @pyron: There is a link included in the page I sent with my comment that brings you to this page https://highvoltagemath.wixsite.com/highvoltagemath/euler-s-reflection-formula-take-2. It's goes over the derivation of Euler's reflection formula without branch cuts.Thanks for bringing it up though! – Highvoltagemath Apr 15 '20 at 15:51
  • https://math.stackexchange.com/q/714482/321264 – StubbornAtom May 01 '21 at 11:44

1 Answers1

14

Note: This is a description from N.N. Lebedev, Special Functions and Their Applications, Dover, New York, 1972, it is not my work but it can be used as starting point.

Lebedev uses in his section 1.2 (Some Relations Satisfied by the Gamma Function) a double-integral approach. From the well-known integral formula

$$\Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}\, d t \qquad (\Re z > 0)$$

temporarily assume $ 0 < \Re z < 1,\,$ use the formula for $1-z\,$ and get

$$\Gamma(z)\Gamma(1-z) = \int_{0}^{\infty}\int_{0}^{\infty}s^{-z}t^{z-1}e^{-(s+t)}\,ds\, dt \qquad (0<\Re z <1)$$

With the new variables $u = s + t, v = t/s$ this gives $$\Gamma(z)\Gamma(1-z) = \int_{0}^{\infty}\int_{0}^{\infty}\frac{v^{z-1}}{1+v}e^{-u}\,du\, dv = \int_{0}^{\infty}\frac{v^{z-1}}{1+v} dv = \frac{\pi}{\sin \pi z}$$

For the last step he refers to Titchmarsh. Then he uses continuation to extend the formula to all $z\in \mathbb{C}$ without the integers.

gammatester
  • 19,147
  • 1
    While multivariable calc isn't in the AP Calculus syllabus, this is something that I personally understand and that my fellow students will understand. \o/ – Addison Crump May 30 '16 at 13:42
  • How do you do the variable change between the dsdt and the dudv? I can figure out the inside of the integral, but can't understand how you got from dsdt to dudv. – D.R. Mar 02 '17 at 20:40
  • 4
    Maybe this isn't relevant anymore, but with $u=s+t$ and $v=\dfrac ts$, and thus $s=\dfrac{uv}{v+1}$ and $t=\dfrac u{v+1}$, the Jacobian is (in absolute value) $\dfrac u{(v+1)^2}$, so the integral becomes $$\int_0^\infty\int_0^\infty t^{-1}\left(\frac ts\right)^ze^{-(s+t)},\mathrm ds,\mathrm dt=\int_0^\infty\int_0^\infty\frac{v+1}uv^{z-1}e^{-v}\frac u{(v+1)^2},\mathrm du,\mathrm dv$$ $$=\int_0^\infty\int_0^\infty\frac{v^{z-1}}{v+1}e^{-u},\mathrm du,\mathrm dv$$ But how do we arrive at the limits of integration? – user170231 May 02 '17 at 18:39
  • @gammatester How do you find the general formula of the last Integral? I cannot find where "Titchmarsh" is. – Ma Joad Jan 29 '19 at 13:45
  • @user170231: You can find the second set of limits by looking at the how the new variables are bounded. I have a good example of this through the Gaussian integral here: https://highvoltagemath.wixsite.com/highvoltagemath/the-gaussian-integral-redo. I hope this helps, otherwise tell me and I will try it with the integrals in question. – Highvoltagemath Apr 15 '20 at 15:58
  • @highvoltagemath It's been a minute since I posted that comment, but I think my question boiled down to, why is $v\in[0,\infty)$? Both $s,t$ are in $[0,\infty)$, so naturally $u=s+t$ is too. But I wasn't clear on why $v=\frac ts$ also belongs to this interval. Right now, looking at $s,t$ in terms of $u,v$, it's pretty clear why. – user170231 Apr 15 '20 at 16:32