Difference between revisions of "Gamma"

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<strong>Proposition:</strong> The following formula holds:
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$$\Gamma(x)=2\displaystyle\int_0^{\infty} e^{-t^2}t^{2x-1}dt.$$
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<strong>Proof:</strong> █
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Revision as of 05:48, 16 May 2016

The gamma function is the function defined by the integral (initially for positive values of $x$) by the formula $$\Gamma(x)=\displaystyle\int_0^{\infty} \xi^{x-1}e^{-\xi} d\xi.$$ The analytic continuation of $\Gamma$ leads to a meromorphic function with poles at the negative integers.

Properties

Theorem

The following formula holds: $$\Gamma(1)=1,$$ where $\Gamma$ denotes the gamma function.

Proof

Compute using the fundamental theorem of calculus, $$\begin{array}{ll} \Gamma(1) &= \displaystyle\int_0^{\infty} \xi^{0} e^{-\xi} \mathrm{d}\xi \\ &= \displaystyle\int_0^{\infty} e^{-\xi} \mathrm{d}\xi \\ &= \left[ -e^{-\xi} \right.\Bigg|_{0}^{\infty} \\ &= 1, \end{array}$$ as was to be shown. █

References

Theorem

The following formula holds: $$\Gamma(z+1)=z\Gamma(z),$$ where $\Gamma$ denotes gamma.

Proof

Use integration by parts to compute $$\begin{array}{ll} \Gamma(z+1) &= \displaystyle\int_0^{\infty} \xi^z e^{-\xi} \mathrm{d}\xi \\ &= -\xi^z e^{-\xi}\Bigg|_0^{\infty}- \displaystyle\int_0^{\infty} z \xi^{z-1} e^{-\xi} \mathrm{d}\xi \\ &= z\Gamma(z), \end{array}$$ as was to be shown. █

References

Theorem

If $n \in \{0,1,2,\ldots\}$, then $$\Gamma(n+1)=n!,$$ where $\Gamma$ denotes the gamma function and $n!$ denotes the factorial of $n$.

Proof

References

Proposition: The following formula holds: $$\Gamma(x)=2\displaystyle\int_0^{\infty} e^{-t^2}t^{2x-1}dt.$$

Proof:

Proposition: The following formula holds: $$\displaystyle\int_0^{\frac{\pi}{2}} \cos^{2x-1}(\theta)\sin^{2y-1}(\theta) d\theta = \dfrac{\Gamma(x)\Gamma(y)}{2\Gamma(x+y)}.$$

Proof:

Theorem: The following formula holds for $\mathrm{Re}(z)>0$: $$\Gamma(z)=\displaystyle\int_0^1 \log \left( \dfrac{1}{t} \right)^{z-1} dt,$$ where $\Gamma$ denotes the gamma function and $\log$ denotes the logarithm.

Proof:

Proposition: The following formula holds: $$\Gamma(z)=\lim_{k \rightarrow \infty} \dfrac{k!k^z}{z(z+1)\ldots(z+k)}.$$

Proof:

Gamma function Weierstrass product

Theorem

The following formula holds: $$\Gamma(s)\zeta(s,a) = \displaystyle\int_0^{\infty} \dfrac{x^{s-1}e^{-ax}}{1-e^{-x}} \mathrm{d}x,$$ where $\Gamma$ denotes the gamma function and $\zeta$ denotes the Hurwitz zeta function.

Proof

References

Proposition: $\Gamma \left( \dfrac{1}{2} \right) = \sqrt{\pi}$.

Proof:

Corollary: $\displaystyle\int_0^{\infty} e^{-t^2} dt = \dfrac{1}{2}\sqrt{\pi}$.

Theorem (Convexity): The gamma function is logarithmically convex.

Proof:

Theorem (Legendre Duplication Formula): $$\Gamma(2x)=\dfrac{2^{2x-1}}{\sqrt{\pi}} \Gamma(x)\Gamma \left( x +\dfrac{1}{2} \right).$$

Proof:

Proposition: If $z=0,-1,-2,\ldots$ then $\Gamma(z)=\infty$.

Proof:

  1. REDIRECT Gamma(z)Gamma(1-z)=pi/sin(pi z)

Theorem

The gamma function is the unique function $f$ such that $f(1)=1$, $f(x+1)=xf(x)$ for $x>0$, and $f$ is logarithmically convex.

Proof

References

Theorem: The following formula holds: $$\displaystyle\lim_{t \rightarrow \infty} \dfrac{\Gamma(t+\alpha)}{\Gamma(t)t^{\alpha}}=\displaystyle\lim_{t \rightarrow \infty} \dfrac{\Gamma(t)t^{\alpha}}{\Gamma(t+\alpha)}=1.$$

Proof: proof goes here █

Videos

Gamma Function (playlist)
The Gamma Function: intro (5)
Gamma Integral Function - Introduction
Gamma function
Mod-04 Lec-09 Analytic continuation and the gamma function (Part I)
gamma function - Part 1
Beta Function, Gamma Function and their Properties
What's the Gamma Function?
euler gamma function
Thermodynamics 19 a : Gamma Function 1/2
The Gamma Function: why 0!=1 (5)
Gamma Function Of One-Half: Part 1
Gamma Function Of One-Half: Part 2
Gamma function at 1/2
Contour Integral Definition of the Gamma Function

See Also

Reciprocal gamma

References

The Gamma Function by Emil Artin
The sine product formula and the gamma function
Leonhard Euler's Integral: A Historical Profile of the Gamma Function