Gamma Function

The gamma function, denoted by \(\Gamma(z)\), is an extension of the factorial function to complex and real numbers, excluding non-positive integers where it is not defined. It was introduced by the Swiss mathematician Leonhard Euler and later generalized by Carl Friedrich Gauss.

The gamma function is defined by the following integral:

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

for complex numbers ( z ) with a positive real part. This integral converges for values of ( z ) with positive real parts, allowing the gamma function to be defined for a wide range of complex numbers.

Some key properties and facts about the gamma function include:

Factorial Relationship:

- For positive integers ( n ), \(\Gamma(n) = (n-1)!\). This property makes the gamma function a generalization of the factorial function.

Derivation:

When (z) is a positive integer, say (z = n), the integral becomes:

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

Apply integration by parts to the integral:

\[u = t^{n-1} \]

\[dv = e^{-t} \, dt\]

\[du = (n-1)t^{n-2} \, dt \]

\[v = -e^{-t}\]

The integration by parts formula is given by \(\int u \, dv = uv - \int v \, du\). Applying this formula:

\[\Gamma(n) = \left| -t^{n-1} e^{-t} \right|_0^\infty + (n-1) \int_0^\infty t^{n-2} e^{-t} \, dt\]

The first term evaluates to 0 at both limits.

\[\Gamma(n) = (n-1) \int_0^\infty t^{n-2} e^{-t} \, dt\]

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

\[\Gamma(n) = (n-1) \Gamma{(n-1)}\]

By repeatedly applying the recursion relation:

\[\Gamma(n) = (n-1)(n-2) \ldots 2 \cdot 1 \cdot \Gamma(1) = (n-1)! \]

Functional Equation:

- The gamma function satisfies the functional equation \(\Gamma(z+1) = z \Gamma(z)\), which extends the factorial relationship to non-integer values.

Reflection Formula:

The reflection formula for the gamma function is given by \(\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)}\).

Derivation I:

Introduce a Complex Variable:

Considering the integral over the contour (C) that consists of the real line from 0 to \(\infty\) and a semicircular arc in the upper half-plane.

\[\Gamma(z) = \int_C t^{z-1} e^{-t} \, dt\]

3. Introduce the Reflection Formula:

We use the reflection formula \(\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)}\) to express the product of two gamma functions.

\[\Gamma(z) \Gamma(1-z) = \int_C t^{z-1} e^{-t} \, dt\]

4. Expand (\(\sin(\pi z)\)):

The denominator (\sin(\pi z)) can be expanded using the exponential form of sine:

\[\sin(\pi z) = \frac{e^{i\pi z} - e^{-i\pi z}}{2i}\]

Substitute this into the expression:

\[ \Gamma(z) \Gamma(1-z) = \frac{\pi}{\frac{1}{2i} \left(e^{i\pi z} - e^{-i\pi z}\right)}\]

\[ \Gamma(z) \Gamma(1-z) = \frac{2i\pi}{e^{i\pi z} - e^{-i\pi z}}\]

6. Contour Integration:

Apply the residue theorem by considering the poles inside the contour. The poles of the integrand occur when \(e^{i\pi z} = 1\) (for (z) an integer). The residue at each pole is \(2\pi i\).

\[ \text{Res}_{z=n} \left[ \frac{2i\pi}{e^{i\pi z} - e^{-i\pi z}} \right] = 2\pi i\]

Summing the residues over all poles, we get the integral as the sum of the residues:

\[\Gamma(z) \Gamma(1-z) = 2\pi i \sum_{n=-\infty}^{\infty} \text{Res_{z=n} [ \frac{1}{e^{i\pi z} - e^{-i\pi z}}} ]\]

7. Evaluate the Sum:

The sum is equivalent to \(\frac{1}{2i} \sum_{n=-\infty}^{\infty} (1 - e^{2i\pi n})\). The terms with \(e^{2i\pi n}\) cancel each other, leaving (1).

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

This derivation provides insight into the connection between the gamma function, the sine function, and complex analysis.

Derivation II:

The reflection formula can be derived using the polar transformation. The idea is to consider a double integral involving the product of two gamma functions, perform a polar transformation, and then simplify the resulting expression.

\[ \Gamma (z) \Gamma (w) = \int_0^\infty t^{z-1} e^{-t}\, dt\int_0^\infty u^{w-1} e^{-u} \, du \]

\[ \Gamma (z) \Gamma (w) = \int_0^\infty \int_0^\infty t^{z-1} u^{w-1} e^{-(t+u)} \, du \, dt \]

Considering \(w=1-z\):

\[ \Gamma(z) \Gamma(1-z) = \int_0^\infty \int_0^\infty t^{z-1} u^{-z} e^{-(t+u)} \, dt \, du \]

\[ \Gamma(z) \Gamma(1-z) = \int_0^\infty \int_0^\infty {(\frac tu)}^{z} t^{-1} e^{-(t+u)} \, dt \, du \]

Performing the polar transformation (noting that t,u>0)

\[x_{(r)}^2=u = r^2 \cos^2(\theta)\]

\[y_{(\theta)}^2=t = r^2 \sin^2(\theta)\]

\[\frac tu=(\frac yx)^2=[\tan (\theta)]^2\]

\[t+u=r^2\]

\[dt=2y\ dy=2r^2\sin (\theta)\cos(\theta)\ d\theta\]

\[du=2x\ dx=2r\cos^2(\theta)\ dr\]

\[dt \ du=4r^3\sin (\theta)\cos^3(\theta)\ dr\ d\theta\]

\(t=0\) so \(\theta =0\)and \(\lim_{t\rightarrow \infty}=\frac \pi 2\) (noting that t>0)

\(u=0\) so \(r=0\)and \(\lim_{u\rightarrow \infty}= \infty\)

\[ \Gamma(z) \Gamma(1-z) = \int_0^\infty \int_0^{\pi/2} {[\frac {\sin ^2 (\theta)}{\cos ^2 (\theta )}]}^{z} [r^2 \sin ^2 (\theta)]^{-1} e^{-r^2} \, 4r^3\sin (\theta)\cos^3(\theta)\ dr\ d\theta\]

\[ \Gamma(z) \Gamma(1-z) = 4\int_0^\infty \int_0^{\pi/2} {[\frac {\sin (\theta)}{\cos (\theta )}]}^{2z} [ \sin (\theta)]^{-1} e^{-r^2} \, r\cos^3(\theta)\ dr\ d\theta\]

\[ \Gamma(z) \Gamma(1-z) = 4\int_0^\infty \int_0^{\pi/2} {[\frac {\sin (\theta)}{\cos (\theta )}]}^{2z-1} e^{-r^2} \, r\cos^4(\theta)\ dr\ d\theta\]

\[ \Gamma(z) \Gamma(1-z) = 4\int_0^\infty e^{-r^2} \, r\ dr\int_0^{\pi/2} {[\frac {\sin (\theta)}{\cos (\theta )}]}^{2z-1} \ \cos^4(\theta)d\theta\]

\[ I=2\int_0^\infty e^{-r^2} \, r\ dr=\int_0^\infty e^{-r^2}dr^2=\int_0^\infty e^{-g} \, dg\]

\[ I=\lim_{a \to \infty} \int_0^a e^{-g} \, dg=\lim_{a \to \infty} [-e^{-g}]_0^a=\lim_{a \to \infty} \left[-e^{-a} - (-e^0)\right]=\lim_{a \to \infty} \left[e^{-a} + 1\right] = 1\]

\[ \Gamma(z) \Gamma(1-z) = 2\int_0^{\pi/2} {[\frac {\sin (\theta)}{\cos (\theta )}]}^{2z-1} \ \cos^4(\theta)d\theta\]

\[\frac {\sin (\theta)}{\cos(\theta)}=\tan (\theta)\]

\[1+\tan^2 x=\frac {\cos^2 x}{\cos ^2x}+\frac {\sin^2 s}{\cos ^2 x}=\frac 1{\cos^2x}\]

\[\cos^2 (\theta)=\frac {1}{1+\tan^2(\theta)}\]

\[ \Gamma(z) \Gamma(1-z) = 2\int_0^{\pi/2} {[\tan (\theta)]}^{2z-1} \ \frac {1}{[1+\tan^2(\theta)]^2}d\theta\]

\[s=\tan (\theta)\]

\(\theta =0\)so \(s=0\)and \(\lim_{\theta \to \pi/2}=\infty\)

\[ds=[1+\tan^2 (\theta)]d\theta\]

\[\theta =\frac{ds}{1+\tan^2 (\theta)}=\frac {ds}{1+s^2}\]

\[ \Gamma(z) \Gamma(1-z) = 2\int_0^{\infty} {s}^{2z-1} \ \frac {1}{[1+s^2]^3}ds\]

Derivation III:

\[ \Gamma(z) \Gamma(1-z) = \int_0^\infty \int_0^{\pi/2} [r\cos(\theta)]^{z-1} [r\sin(\theta)]^{-z} e^{-r[\cos(\theta) + \sin (\theta)]} \, [\sin (\theta)\cos(\theta)(dr)^2+\cos^2(\theta)rd\theta dr -r \sin^2 (\theta) \ d\theta dr -r^2 \sin (\theta)\cos (\theta) (d\theta)^2]\]

\[ \Gamma(z) \Gamma(1-z) = \int_0^\infty \int_0^{\pi/2} [r\cos(\theta)]^{z-1} [r\sin(\theta)]^{-z} e^{-r} r \, d\theta \, dr\]

3. Simplify the Integral:

Simplify the expression by combining terms and using trigonometric identities:

[ \Gamma(z) \Gamma(1-z) = \int_0\infty \int_0^{\pi/2} r^{2z-2} e^{-r} \cos^{z-1}(\theta) \sin^{-z}(\theta) \, d\theta \, dr ]

4. Evaluate the (\theta) Integral:

The integral with respect to (\theta) can be evaluated using the beta function:

[ \Gamma(z) \Gamma(1-z) = \int_0\infty r^{2z-2} e^{-r} \cdot B\left(z-1, 1-z\right) \, dr ]

Using the definition of the beta function, we have:

[ B(x, y) = \int_0¹ t^{x-1}(1-t){y-1} \, dt ]

5. Beta Function Integration:

Substitute (u = \frac{r}{1+r}) to simplify the integral:

[ \Gamma(z) \Gamma(1-z) = \int_0¹ u^{z-1}(1-u){-z} \, du ]

6. Recognize the Result:

The resulting integral is the beta function evaluated at (B(z, 1-z)):

[ \Gamma(z) \Gamma(1-z) = B(z, 1-z) = \frac{\Gamma(z) \Gamma(1-z)}{\Gamma(z+1-z)} = \Gamma(z) \Gamma(1-z) ]

7. Solve for (\Gamma(z) \Gamma(1-z)):

Equate the result to the original expression (I):

[ \Gamma(z) \Gamma(1-z) = \int_0\infty \int_0\infty t^{z-1} u^{-z} e^{-(t+u)} \, dt \, du ]

8. Result:

The reflection formula is obtained:

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

Conclusion:

The reflection formula can indeed be derived using the polar transformation by introducing polar coordinates, simplifying the resulting expression, and recognizing the beta function, ultimately leading to the reflection formula involving the gamma function and the sine function.

Integral Representation:

- The integral representation of the gamma function allows for an alternative expression in terms of an integral.

Duplication Formula:

- The duplication formula states that \(\Gamma(z) \Gamma(z + \frac{1}{2}) = 2^{1-2z} \sqrt{\pi} \Gamma(2z)\).

Relation to Beta Function:

- The gamma function is related to the beta function through the identity

\[B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}\]

Analytic Continuation:

- The gamma function can be analytically continued to the entire complex plane, except for poles at non-positive integers.

Special Values:

\[ \Gamma(1) = 1\]

\[\Gamma(\frac{1}{2}) = \sqrt{\pi}\]

\[\Gamma(2) = 1\]

\(\Gamma(n) = (n-1)!\) for positive integers ( n ).

The gamma function is widely used in mathematics, physics, and engineering. It appears in many areas, including complex analysis, number theory, probability, and the solution of differential equations. The properties of the gamma function make it a powerful tool for expressing and manipulating mathematical expressions involving factorials and related functions.