Editing Gamma function (section)

=== Integral representations ===
There are many formulas, besides the Euler integral of the second kind, that express the gamma function as an integral. For instance, when the real part of {{mvar|z}} is positive,<ref>{{Cite book |last1=Gradshteyn |first1=I. S.|last2=Ryzhik |first2=I. M. |title=Table of Integrals, Series, and Products |edition=Seventh |publisher=Academic Press |year=2007 |isbn=978-0-12-373637-6|page=893}}</ref>
<math display="block">\Gamma (z)=\int_{-\infty}^\infty e^{zt-e^t}\, dt</math>
and<ref>Whittaker and Watson, 12.2 example 1.</ref>
<math display="block">\Gamma(z) = \int_0^1 \left(\log \frac{1}{t}\right)^{z-1}\,dt,</math>
<math display="block">\Gamma(z) = 2c^z\int_{0}^{\infty}t^{2z-1}e^{-ct^{2}}\,dt \,,\; c>0</math>
where the three integrals respectively follow from the substitutions <math>t=e^{-x}</math>, <math>t=-\log x</math> <ref>{{Cite journal |last=Detlef |first=Gronau |title=Why is the gamma function so as it is? |url=https://imsc.uni-graz.at/gronau/TMCS_1_2003.pdf |journal=Imsc.uni-graz.at}}</ref> and <math>t=cx^2</math><ref>{{cite journal |last1=Pascal Sebah |first1=Xavier Gourdon |title=Introduction to the Gamma Function |journal=Numbers Computation |url=https://www.csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf |access-date=30 January 2023 |archive-date=30 January 2023 |archive-url=https://web.archive.org/web/20230130155521/https://www.csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf |url-status=dead }}</ref> in Euler's second integral. The last integral in particular makes clear the connection between the gamma function at half integer arguments and the [[Gaussian integral]]: if <math>z=1/2,\; c=1</math> we get
<math display="block">
\Gamma(1/2)=2\int_{0}^{\infty}e^{-t^{2}}\,dt=\sqrt{\pi} \;.
</math>

Binet's first integral formula for the gamma function states that, when the real part of {{mvar|z}} is positive, then:<ref>Whittaker and Watson, 12.31.</ref>
<math display="block">\operatorname{log\Gamma}(z) = \left(z - \frac{1}{2}\right)\log z - z + \frac{1}{2}\log (2\pi) + \int_0^\infty \left(\frac{1}{2} - \frac{1}{t} + \frac{1}{e^t - 1}\right)\frac{e^{-tz}}{t}\,dt.</math>
The integral on the right-hand side may be interpreted as a [[Laplace transform]].  That is,
<math display="block">\log\left(\Gamma(z)\left(\frac{e}{z}\right)^z\sqrt{\frac{z}{2\pi}}\right) = \mathcal{L}\left(\frac{1}{2t} - \frac{1}{t^2} + \frac{1}{t(e^t - 1)}\right)(z).</math>

Binet's second integral formula states that, again when the real part of {{mvar|z}} is positive, then:<ref>Whittaker and Watson, 12.32.</ref>
<math display="block">\operatorname{log\Gamma}(z) = \left(z - \frac{1}{2}\right)\log z - z + \frac{1}{2}\log(2\pi) + 2\int_0^\infty \frac{\arctan(t/z)}{e^{2\pi t} - 1}\,dt.</math>

Let {{math|''C''}} be a [[Hankel contour]], meaning a path that begins and ends at the point {{math|∞}} on the [[Riemann sphere]], whose unit tangent vector converges to {{math|−1}} at the start of the path and to {{math|1}} at the end, which has [[winding number]] 1 around {{math|0}}, and which does not cross {{closed-open|0, ∞}}.  Fix a branch of <math>\log(-t)</math> by taking a branch cut along {{closed-open|0, ∞}} and by taking <math>\log(-t)</math> to be real when {{math|t}} is on the negative real axis.  Assume {{mvar|z}} is not an integer.  Then Hankel's formula for the gamma function is:<ref>Whittaker and Watson, 12.22.</ref>
<math display="block">\Gamma(z) = -\frac{1}{2i\sin \pi z}\int_C (-t)^{z-1}e^{-t}\,dt,</math>
where <math>(-t)^{z-1}</math> is interpreted as <math>\exp((z-1)\log(-t))</math>.  The reflection formula leads to the closely related expression 
<math display="block">\frac{1}{\Gamma(z)} = \frac{i}{2\pi}\int_C (-t)^{-z}e^{-t}\,dt,</math>
again valid whenever {{math|''z''}} is not an integer.