So I am trying to prove the reflection formula for the gamma function by showing that $$\int_{0}^{\infty} \frac{v^{s-1}}{1+v}dv=\frac{\pi}{\sin(\pi s)}$$ for $0 < \Re(s) < 1$ , as these two statements are (almost) equivalent. I want to do this with elementary means if possible (I was hoping that it was possible to prove it without actually using complex integration, since the integrand is real, treating s "as if" it was simply real.)

My first attempt was this: assume that

$$\frac{d}{dv}\left \{ \frac{f(v)}{g(v)} \right \}= \frac{v^{s-1}}{1+v}$$ so that $$\frac{f'g-g'f}{g^2}=\frac{v^{s-1}}{1+v}$$ Thus, we have $g(v)=\sqrt{1+v}$ . Multiplying with the denominator yields: $$f'g-g'f=v^{s-1}$$ Or equivalently: $$\sqrt{1+v} f'(v)-\frac{f(v)}{2\sqrt{1+v}}=v^{s-1}$$ I thought about trying to solve this using Laplace transform, but got nowhere. The reason is that I don't know the Laplace transform of $v^{s-1}\sqrt{1+v}$
I also tried expressing $$\frac{v^{s-1}}{1+v}$$ as a Laurent series and using integration term by term, without success. Does anyone know how to prove the given identity (in a way as simple as possible) ?

Thanks, R :)

  • $\begingroup$ Correction: I ment $0<\Re(s)<1$. $\endgroup$ – AfterMath Feb 13 '19 at 13:12
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    $\begingroup$ Possible duplicate of Closed form for $ \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$ $\endgroup$ – mrtaurho Feb 13 '19 at 13:12
  • $\begingroup$ No, I don't think it's a duplicate, because I'm trying to prove the reflection formula for the gamma function using this integral, while in the link they are using the refection formula to prove this integral .... $\endgroup$ – AfterMath Feb 13 '19 at 13:19
  • $\begingroup$ @AfterMath You should mention that. In fact, only concerning the evaluation of the integral, it is indeed an duplicate. Proving Euler's Reflection Formula is something completely different. I would suggest to add this crucial detail within your post. $\endgroup$ – mrtaurho Feb 13 '19 at 13:20
  • $\begingroup$ It is done now :) $\endgroup$ – AfterMath Feb 13 '19 at 13:24

Your given integral is closely related to the Mellin transform and can be evaluated by using Ramanujan's Master Theorem.

Ramanujan's Master Theorem

Let $f(v)$ be an analytic function with a MacLaurin Expansion of the form $$f(v)=\sum_{k=0}^{\infty}\frac{\phi(k)}{k!}(-v)^k$$then the Mellin Transform of this function is given by $$\int_0^{\infty}v^{s-1}f(v)dv=\Gamma(s)\phi(-s)$$

In order to get there we can expand the fraction as a geometric series

\begin{align*} \int_0^\infty \frac{v^{s-1}}{1+v}\mathrm dv&=\int_0^\infty v^{s-1}\sum_{k=0}^\infty (-v)^k\mathrm dv\\ &=\int_0^\infty v^{s-1}\sum_{k=0}^\infty \frac{\Gamma(k+1)}{k!}(-v)^k\mathrm dv \end{align*}

Now we may use the aforementioned theorem with $s=s$ and $\phi(k)=\Gamma(k+1)$ to obtain

\begin{align*} \int_0^\infty v^{\nu-1}\sum_{k=0}^\infty \frac{\Gamma(k+1)}{k!}(-v)^k\mathrm dv&=\Gamma(s)\Gamma(1-s)\\ &=\frac\pi{\sin(\pi s)} \end{align*}

where we used Euler's Reflection Formula in order to perform the last step.

$$\therefore~\int_0^\infty \frac{v^{s-1}}{1+v}\mathrm dv~=~\frac\pi{\sin(\pi s)}$$

  • $\begingroup$ Maybe I'm missing something, but isn't the geometric series only valid for $0<v<1$? You would need to deal with $v\geq1$ separately, correct? $\endgroup$ – Clayton Feb 13 '19 at 13:59
  • $\begingroup$ @Clayton That is what I like about Ramanujan's Master Theorem: the radius of convergence does not play an important rule overall. Of course, the geometric series only converges for $|v|<1$ but since I only use this series expansion in order to recognize the hidden structure suitable for the RMT I do not have to worry about this. Another interesting situation is discussed here; all in all the same situation: a series that converges for $|x|<1$, the one of the Polylogarithm, but nevertheless I am allowed to use the RMT which does not care about this. $\endgroup$ – mrtaurho Feb 13 '19 at 14:16
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    $\begingroup$ Good old RMT! Nice work (+1) :) $\endgroup$ – clathratus Feb 13 '19 at 16:20
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    $\begingroup$ @Clayton I guess roughly speaking you can draw this conclusion even though I am not sure whether the same principle lies underneath. However, considering an actual proof of the RMT, via the Mellin Inversion formula or by umbra calculus as it is done here, you may notice that the radius of convergence does not is involved in the argumentation; at least not in the classical sense. But to be honest I am not sure whether I am qualified enough to answer your question properly. Perhaps ask a question here on MSE dealing with this issue? $\endgroup$ – mrtaurho Feb 13 '19 at 18:29
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    $\begingroup$ @mrtaurho: I asked the question here. Feel free to add anything you feel appropriate. $\endgroup$ – Clayton Feb 13 '19 at 19:06

Another method.

Recall the definition of the Beta function: $$\mathrm{B}(a,b)=\int_0^1t^{a-1}(1-t)^{b-1}\mathrm dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\mathrm{B}(b,a)$$ Then recall the Gamma reflection formula: $$\Gamma(s)\Gamma(1-s)=\frac\pi{\sin\pi s}$$ So with $a=s$ and $b=1-s$, we have $$\int_0^1t^{s-1}(1-t)^{-s}\mathrm dt=\int_0^1t^{-s}(1-t)^{s-1}\mathrm dt=\frac\pi{\sin\pi s}$$ Then use the substitution $x=\frac{1-t}{t}$ to see that $$\int_0^\infty \frac{x^{s-1}}{1+x}\mathrm dx=\int_0^1t^{-s}(1-t)^{s-1}\mathrm dt=\frac\pi{\sin\pi s}$$ As desired.

  • $\begingroup$ The latter formula is not elementary. $\endgroup$ – FDP Feb 13 '19 at 17:39
  • $\begingroup$ @FDP The Gamma reflection formula? Also: how do you define "elementary"? $\endgroup$ – clathratus Feb 13 '19 at 17:45
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    $\begingroup$ Vast question but, actually i know an "elementary" way to prove the formula demanded. I will post it if i have time. $\endgroup$ – FDP Feb 13 '19 at 18:00

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