# The Hankel Integral Representation for $\Gamma(z)$

I am trying to understand some details hidden in the proof of the Hankel integral representation for the gamma function: $$\frac{1}{\Gamma(z)} = -\frac{1}{2\pi i} \int_{\mathcal{H}} (-t)^{-z} e^{-t} dt$$ for all $$z \in \mathbb{C}$$. Here $$\mathcal{H}$$ denotes the Hankel contour: $$\mathcal{H} = [i + \infty,i] + \mathcal{H}_{sc} + [-i,-i + \infty]$$, where $$\mathcal{H}_{sc}$$ joins $$i$$ with $$-i$$ along a positively oriented semicircle centered at $$0$$.

A typical approach to the proof, as far as I understand it, goes as follows:

• Cut the plane along the positive real axis and choose a fixed branch of the multifunction $$(-t)^{-z}$$ by taking its principal branch for negative real $$t$$, and by continuing this branch analytically to the cut plane.
• Let $$\varepsilon\mathcal{H}$$ denote $$\mathcal{H}$$ scaled by $$\varepsilon$$, i.e., after applying the transformation $$z \mapsto \varepsilon z$$.
• The integral along $$\varepsilon\mathcal{H}$$ is then said to be the same as the one along $$\mathcal{H}$$ by Cauchy's theorem. This is a first step that I find unclear: I understand that the integrand is analytic in $$\mathbb{C} \setminus [0,\infty)$$; however I do not know about any deformation theorem for improper contours. Could someone describe a rigorous argument that is used here?
• Assume $$z < 0$$ and take $$\varepsilon \to 0$$. The integral can then be decomposed into three integrals, two of which can be manipulated to obtain an integral much alike the usual integral representation of $$\Gamma(1-z)$$ for $$\mathrm{Re}(1-z) > 0$$, while the remaining one can be shown to be negligible when $$\varepsilon \to 0$$, thanks to the assumption $$z < 0$$. The Hankel representation is then proved for $$z < 0$$.
• Finally, the result is extended to the whole complex plain via analytic continuation. This is a second step that I find unclear, as it can only be performed if one knows that $$I(z) = \int_{\mathcal{H}} (-t)^{-z} e^{-t} dt$$ is an analytic function of $$z$$. This property is usually qualified as obvious. Nevertheless, I have no idea about why it is obvious.

Could someone explain the rigorous arguments needed to perform the two critical steps mentioned above?

I would also be very grateful for pointers to literature that treats the Hankel representation rigorously (the treatements that I have found seem more-or-less sketchy to me).

• Cauchy integral theorem implies $\int_{\mathcal{H}} (-t)^{-z} e^{-t} dt= \int_{\varepsilon\mathcal{H}} (-t)^{-z} e^{-t} dt$ and for $\Re(z) < 0$, $\lim_{\varepsilon \to 0} \int_{\varepsilon\mathcal{H}} (-t)^{-z} e^{-t} dt$ converges to a well-defined integral which is $\int_\infty^0 e^{-i\pi (-z)} u^{-z}e^{-u}du+ \int_0^\infty e^{i\pi (-z)} u^{-z}e^{-u}du=-2i \sin(\pi z) \Gamma(1-z)$. I renamed the $t$ in $u$ to make clear that the function of $t$ is continued analytically along the contour whereas for the latter it is a function of the real variable $u$. – reuns Feb 9 at 14:40
• @reuns Thanks for a comment. However, I seem to "understand" this. My questions were: (1) why does the Cauchy integral theorem imply the equality of both integrals? (The contour is improper here, so I do not find this straightforward.) (2) The reasoning leads to the representation valid for $\mathrm{Re}\, z > 0$; to prove the formula for all $z$, you need the fact that $I(z)$ is analytic. My question was why this is true. – Jára Cimrman Feb 9 at 14:52
• $\int_{\mathcal{H}}- \int_{\varepsilon\mathcal{H}} (-t)^{-z} e^{-t} dt$ is the integral of a function over a close contour enclosing a region where it is analytic Thus it is $=0$. The only difference with the standard Cauchy integral theorem is that the boundary of that region has infinite length, this can be easily circumvented by the exponential decay at $+\infty$. – reuns Feb 9 at 15:05
• $\int_{\mathcal{H}} (-t)^{-z} e^{-t} dt+2i\sin(\pi z) \Gamma(1-z)$ is entire in $z$, thus that it is $=0$ for $\Re(z) < 0$ implies it is $=0$ for all $z$. – reuns Feb 9 at 15:09

I had the same question as other people here, here or here.

1. (Your first point was not very clear to me as I use the word "branch" to mean the branch cut. At the end, it seems that you do cut out the positive reals and use the principal determination of the logarithm: $$(-t)^{-z}=e^{-z \log(-t)} = e^{-z \big( \ln \lvert t\rvert + i \mathrm{Arg}(-t)\big)}$$ with $$\mathrm{Arg}(-t) \in ]-\pi,\pi[$$)
2. For the deformation of an open contour $$\mathcal{C}_1$$ into $$\mathcal{C}_2$$, we can use the same idea as for closed contours but we'll have to add the contribution of the segments relating the end points. Hopefully under some limit, these contributions will vanish. For example, let us consider the mentionned integral but with at first a "finite Hankel contour": $$[i+R,i]\cup \mathcal{H}_{sc} \cup [-i, -i+R]$$ and its rescaled version. The extra contributions I mentionned are $$\int_{\epsilon i+R}^{i+R} (−w)^{−𝑧}𝑒^{−w}\, dw + \int_{-i+R}^{-\epsilon i+R} (−w)^{−𝑧}𝑒^{−w}\, dw = \int_{[-1,-\epsilon]\cup [\epsilon,1]}\hspace{-10mm} (-R-is)^{-z} e^{-R -is}\times i\, ds$$ In modulus, this is bounded by $$2(1-\epsilon) \sup_{s\in[\cdots]\cup[\cdots]} \left(e^{-\mathrm{Re}(z) \ln \lvert-R-is\rvert + -\mathrm{Im}(z) \mathrm{Arg}(-R-is) }\right) e^{-R} \underset{R\to \infty}{\longrightarrow} 0$$
3. (What I personally found unclear is that this is (at this step) not a result of the residue theorem as we do not close the contour. Then in some proof of the Reflection formula (French) we do use the residue theorem, and also a Hankel contour which is this time closed by a big circle whose contribution vanishes)
4. To prove analyticity I would simply use the dominated convergence theorem for holomorphic functions (this seems not to be given any name in english litterature. In France it is taught in this readily useable form, as e.g. in thm at the top of p.2. Basically, one needs only dominate the family (indexed by $$z$$) of functions of $$t$$ and not the derivatives w.r.t. $$z$$ by sthg integrable)

All this is done (with different conventions) in "Advanced Complex Analysis - A Comprehensive Course in Analysis Part 2B" (AMS 2015), Barry Simon, Thm 14.7.1 p.153, and also in "Complex Analysis 1" (UTX 2009), Rolf Busam, Eberhard Freitag, Exercise 17 p.209, Correction p.477

second question

Why is $$I(z) = \int_{\mathcal{H}} (-t)^{-z} e^{-t} dt$$ an analytic function of $$z$$?

For a fixed $$t \in \mathcal H$$, the function $$z \mapsto (-t)^{-z} e^{-t}$$ is analytic.

Let $$\sigma$$ be a closed curve in $$\mathbb C$$. Then for a fixed $$t \in \mathcal H$$, $$\oint_\sigma (-t)^{-z} e^{-t}\;dz = 0$$ Interchange integrals: $$\oint_\sigma I(z)\;dz = \oint_{\mathcal H}\oint_\sigma (-t)^{-z} e^{-t}\;dz\;dt = 0$$ By Morera's Theorem, $$I(z)$$ is analytic.

• Thank you for the answer, looks nice. However, I have one more question: what precisely justifies the interchange of integrals? I know that continuity of the integrand is sufficient in case both contours are proper. However, I am not sure about the improper Hankel contour. Possibly some combination of the proper-contour-case with uniform convergence? Or some standard theorem I do not know about? – Jára Cimrman Feb 9 at 15:17