Gamma Function

The gamma function, denoted by \(\Gamma(s)\), is defined by the formula

\[\Gamma (s)=\int_0^{\infty} t^{s-1} e^{-t}\, dt,\]

which is defined for all complex numbers except the nonpositive integers. It is frequently used in identities and proofs in analytic contexts.

The above integral is also known as Euler's integral of second kind. It serves as an extension of the factorial function which is defined only for the positive integers. In fact, it is the analytic continuation of the factorial and is defined as

\[\Gamma (n)=(n-1)! .\]

However, the gamma function is but one in a class of multiple functions which are also meromorphic with poles at the nonpositive integers.

The functional equation

\[\Gamma(s+1)=s\Gamma(s)\]

is true for all values of \(s \in \mathbb{C}\); this can be derived from an application of integration by parts.

Recalling the definition of the gamma function above, we can see that by applying integration by parts,

\[\Gamma(s+1) = \int_0^{\infty} t^{s} e^{-t}\, dt = -t^{s}e^{-t}\Big|_0^{\infty} + s \int_0^{\infty} t^{s-1} e^{-t}\, dt = s\Gamma(s).\]

Hence, the functional equation holds. \(_\square\)

Noting that \(\Gamma(1) = 1\), we can easily see \(\Gamma(s) = (s-1)!\) for all positive integers \(s\) by simple induction.

Evaluate

\[\frac{\Gamma\left(\frac{5}{2}\right)}{\Gamma\left(\frac{1}{2}\right)}.\]

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Using the functional equation for the gamma function, we obtain that

\[\frac{\Gamma\left(\frac{5}{2}\right)}{\Gamma\left(\frac{1}{2}\right)} = \frac{\dfrac{3}{2}\Gamma\left(\frac{3}{2}\right)}{\Gamma\left(\frac{1}{2}\right)} = \frac{\dfrac{3}{2}\frac{1}{2}\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)} =\frac{3}{4}. \ _\square\]

\[\Gamma(s)=\lim_{n\to \infty}\left(\dfrac{n^s}{s} \prod_{k=1}^n \dfrac{k}{s+k}\right)\]

Consider the integral

\[S=\int_0^n t^{s-1} \left(1-\dfrac{t}{n}\right)^n\, dt.\]

Using integration by parts,

\[S=\left.\dfrac{t^s}{s} \left(1-\dfrac{t}{n}\right)^n\right|_0^n+\dfrac{n}{ns} \int_0^n t^{s} \left(1-\dfrac{t}{n}\right)^{n-1}dt=\dfrac{n}{ns} \int_0^n t^{s} \left(1-\dfrac{t}{n}\right)^{n-1}dt.\]

We can deduce that by integrating by parts \(n-1\) times, we will get

\[S=\dfrac{n}{ns}\dfrac{n-1}{n(s+1)}\dfrac{n-2}{n(s+2)}\cdots \dfrac{1}{n(s+n-1)} \int_0^n t^{s+n-1} dt.\]

Evaluating the integral, we have

\[S=\dfrac{n^s}{s} \prod_{k=1}^n \dfrac{k}{s+k}. \]

Setting \(n\) toward infintiy,

\[\lim_{n\to \infty}\int_0^n t^{s-1} \left(1-\dfrac{t}{n}\right)^ndt=\lim_{n\to \infty}\left(\dfrac{n^s}{s} \prod_{k=1}^n \dfrac{k}{s+k}\right).\]

Notice that

\[\lim_{n\to \infty}\int_0^n t^{s-1} \left(1-\dfrac{t}{n}\right)^ndt=\int_0^\infty t^{s-1} e^{-t}dt=\Gamma(s).\]

Therefore,

\[\Gamma(s)=\lim_{n\to \infty}\left(\dfrac{n^s}{s} \prod_{k=1}^n \dfrac{k}{s+k}\right).\ _\square\]

\[\Gamma(s)=\dfrac{e^{-\gamma s}}{s} \prod_{k=1}^\infty e^{s/k} \left(1+\dfrac{s}{k}\right)^{-1} \]

Consider

\[\Gamma(s)=\lim_{n\to \infty}\left(\dfrac{n^s}{s} \prod_{k=1}^n \dfrac{k}{s+k}\right).\]

Then rewrite it as

\[\begin{align} \Gamma(s)&=\lim_{n\to \infty}\left(\dfrac{1}{s}\exp\big(sH_k-sH_k+s\ln(n)\big) \prod_{k=1}^n \dfrac{k}{s+k}\right)\\\\ &=\lim_{n\to \infty}\left(\dfrac{1}{s}\exp\left(\sum_{k=1}^n \dfrac{s}{k}-s\gamma\right)\prod_{k=1}^n \left(1+\dfrac{s}{k}\right)^{-1}\right) \\\\ &=\lim_{n\to \infty}\left(\dfrac{e^{-\gamma s}}{s} \prod_{k=1}^n e^{s/k} \left(1+\dfrac{s}{k}\right)^{-1}\right)\\\\ &=\dfrac{e^{-\gamma s}}{s} \prod_{k=1}^\infty e^{s/k} \left(1+\dfrac{s}{k}\right)^{-1}.\ _\square \end{align}\]

The gamma function has a very nice duplication formula. It can be stated as

\[ \sqrt { \pi } \ \Gamma (2z)={ 2 }^{ 2z-1 }\Gamma (z)\Gamma \left (z+\frac { 1 }{ 2 } \right ). \]

This is notoriously simple to prove.

We start with the representation of beta function and a relation between beta function and gamma function: \[\displaystyle \int _{ 0 }^{ \pi /2 }{ \sin ^{ 2m-1 }{ x } \cos ^{ 2n-1 }{ x }\, dx } = \dfrac{B(m,n)}{2} = \frac { \Gamma (m)\Gamma (n) }{2\Gamma (m+n) }.\] Put \( m=z\) and \( n=z\) to get \[\begin{align} \displaystyle I = \frac { \Gamma (z)\Gamma (z) }{ 2\Gamma (2z) }
&= \int _{ 0 }^{ \pi /2 }{ \sin ^{ 2z-1 }{ x } \cos ^{ 2z-1 }{ x }\, dx } \\ &= \frac { 1 }{ { 2 }^{ 2z-1 } } \int _{ 0 }^{ \pi /2 }{ \sin ^{ 2z-1 }{ 2x }\, dx }. \end{align} \] Change of variables \( 2x=t\) and \(2\,dx=dt \) gives \[ \displaystyle I = \frac { 1 }{ { 2 }^{ 2z } } \int _{ 0 }^{ \pi }{ \sin ^{ 2z-1 }{ t }\, dt }. \] Now, using the property \(\displaystyle \int _{ 0 }^{ 2a }{ f(x)\, dx } =2\int _{ 0 }^{ a }{ f(x)\, dx } \) if \(f(x)=f(2a-x),\) we have \[ \displaystyle I = \frac { 1 }{ { 2 }^{ 2z-1 } } \int _{ 0 }^{ \pi /2 }{ \sin ^{ 2z-1 }{ t }\, dt }. \] Again using beta function, we have \[ I = \dfrac { 1 }{ { 2 }^{ 2z } } \dfrac { \Gamma (z)\Gamma \left(\dfrac { 1 }{ 2 } \right) }{ \Gamma \left(z+\dfrac { 1 }{ 2 } \right) }.\] Since \( I = \dfrac { \Gamma (z)\Gamma (z) }{ 2\Gamma (2z) }, \) we have \[\dfrac { \Gamma (z)\Gamma (z) }{ \Gamma (2z) } =\dfrac { 1 }{ { 2 }^{ 2z-1 } } \dfrac { \Gamma (z)\Gamma \left(\dfrac { 1 }{ 2 } \right) }{ \Gamma \left(z+\dfrac { 1 }{ 2 } \right) }.\] Using \( \Gamma\left(\dfrac{1}{2}\right) = \sqrt{\pi} \) , we finally have \[ { 2 }^{ 2z-1 }\Gamma (z)\Gamma \left(z+\dfrac { 1 }{ 2 } \right)=\sqrt { \pi }\, \Gamma (2z).\] Hence proved. \(_\square\)

Euler's reflection formula is as follows:

\[\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin \pi z},\]

where \(z \in \mathbb{C}.\) \(_\square\)

Consider

\[\Gamma(s)=\lim_{n\to \infty}\left(\dfrac{n^s}{s} \prod_{k=1}^n \dfrac{k}{s+k}\right).\]

Replace \(s\) with \(-s\) to get

\[\Gamma(-s)=\lim_{n\to \infty}\left(\dfrac{n^{-s}}{-s} \prod_{k=1}^n \dfrac{k}{-s+k}\right).\]

Multiply both of these to get

\[\Gamma(s)\Gamma(-s)=\dfrac{-1}{s^2} \prod_{k=1}^\infty \left(1-\dfrac{s^2}{k^2}\right)^{-1}.\]

The product on the RHS is the famous Weierstrass product, so we have

\[\begin{align} \Gamma(s)\Gamma(-s)&=\dfrac{-1}{s^2} \dfrac{\pi s}{\sin(\pi s)}\\\\ \Gamma(s)\big(-s\Gamma(-s)\big)&=\dfrac{\pi}{\sin(\pi s)}\\\\ \Gamma(s)\Gamma(1-s)&=\dfrac{\pi}{\sin(\pi s)}.\ _\square \end{align}\]

Find \(\Gamma\left(\frac{1}{2}\right)^{4}\).

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Using Euler's reflection formula,

\[\Gamma\left(\frac{1}{2}\right)^{4} = \left[\Gamma\left(\frac{1}{2}\right)\Gamma\left(1-\frac{1}{2}\right)\right]^{2} = \left(\frac{\pi}{\sin \frac{\pi}{2}}\right)^{2} = \pi^{2}. \ _\square\]

There is also an Euler reflection formula for the digamma function

Main article: Beta Function

The gamma and beta functions satisfy the identity

\[B(x, y) = \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} = \int_0^1 t^{x-1} (1-t)^{y-1}\, dt= \int_0^{\pi/2}2 \sin^{2x-1} (t)\cos^{2y-1}(t)\, dt.\]

The above integral is indeed known as Euler's integral of first kind.

Main article: Digamma Function

The digamma function is defined as

\[\psi(s)=\dfrac{\Gamma'(s)}{\Gamma(s)}.\]

Using this on the Euler reflection formula and Legendre duplication formula, we have

\[\begin{align} \psi(1-s)-\psi(s)&=\pi\cot\big(\pi(s)\big)\\ 2\psi(2s)&=2\ln(2)+\psi(s)+\psi\left(s+\dfrac{1}{2}\right). \end{align}\]

Main article: Riemann Zeta Function

The zeta function and gamma functions are very closely related.

The gamma function turns up in the zeta functional equations:

\[\begin{align} \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right)\zeta(s)&=\pi^{-\frac{1-s}{2}} \Gamma\left(\frac{1-s}{2}\right)\zeta(1-s)\\\\ \zeta(s)&=2^s \pi^{s-1} \sin\left(\frac{\pi s}{2}\right) \Gamma(1-s)\zeta(1-s). \end{align}\]

It also has an integral closely related to it:

\[\Gamma(s) \zeta(s) =\int_0^\infty\dfrac{x^{s-1}}{e^x-1} dx. \]

Main article: Polylogarithm

An integral representation of gamma function is

\[\Gamma(s)\text{Li}_s(z)= \int_0^\infty\frac{x^{s-1}}{\frac{e^x}{z}-1} dx.\]

The Bohr-Mollerup theorem states that the gamma function \(\Gamma\) is the unique function on the interval \(x > 0\) such that \(\Gamma(1) = 1, \Gamma(s+1) = s\Gamma(s),\) and \(\ln \Gamma(s)\) is a convex function.

• Ramanujan's Master Theorem
• Integration by Parts
• Polylogarithm
• Riemann Zeta Function
• Digamma Function
• Beta Function