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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?

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  • $\begingroup$ 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. $\endgroup$ May 30, 2016 at 13:10
  • $\begingroup$ @gammatester Can you add this as an answer? :P $\endgroup$ May 30, 2016 at 13:14
  • $\begingroup$ I could. But it will be just a quote from Lebedev. Is this OK? $\endgroup$ May 30, 2016 at 13:16
  • $\begingroup$ @gammatester As long as you link the source, it should be. :P I suppose that I'd have to cite the source, though. :\ $\endgroup$ May 30, 2016 at 13:17
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    $\begingroup$ @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. $\endgroup$
    – pyon
    Apr 10, 2020 at 16:02

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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.

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  • $\begingroup$ 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/ $\endgroup$ May 30, 2016 at 13:42
  • $\begingroup$ 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. $\endgroup$
    – D.R.
    Mar 2, 2017 at 20:40
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    $\begingroup$ 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? $\endgroup$
    – user170231
    May 2, 2017 at 18:39
  • $\begingroup$ @gammatester How do you find the general formula of the last Integral? I cannot find where "Titchmarsh" is. $\endgroup$
    – Ma Joad
    Jan 29, 2019 at 13:45
  • $\begingroup$ @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: highvoltagemath.wixsite.com/highvoltagemath/…. I hope this helps, otherwise tell me and I will try it with the integrals in question. $\endgroup$ Apr 15, 2020 at 15:58

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