Gamma function

Defined as below for $n\gt 0$: $$

$$\Gamma(n)=\int_0^\infty e^{-x}x^{n-1}\,\text{d}x$$

Aka. Eulerian integral of the second kind.

Convergence

$\Gamma(n)$ is convergent iff $n \gt 0$.

Proof Hint

Direct comparison test is used. Proved in 3 cases:

• Case 1: for positive integer $n$ $$

  • Consider the lemma 2’s limit definition
  • Take $\epsilon=1$
  • Use the convergence of $\int_0^\infty e^{-x/2}\,\text{d}x$

• Case 2: for $n \gt 1$ non-integers $$

  • Use $\lfloor n \rfloor \lt n \lt \lfloor n \rfloor + 1$
  • Use $x^{y-1}e^{-x} \le x^{\lfloor n \rfloor}e^{-x}$
  • $\Gamma(\lfloor n \rfloor + 1)$ is converging from case 1

• Case 3: for $0 \lt n \lt 1$. $$

  • Proof is similar to case 1
  • But $\int_0^N e^{-x}x^{n-1}\,\text{d}x$ is an improper
  • Prove that it is also converging

Properties

Proofs are required for each property mentioned below.

• $\Gamma(1)=1$
• $\Gamma(n+1)=n\Gamma(n)$
• $\Gamma(n+1)=n!$
• $\Gamma(n) \Gamma(1-n) = \pi \csc(\pi x)$
• $\Gamma(\frac{n}{2})$ can be extrapolated from $\Gamma(\frac{1}{2})=\sqrt{\pi}$ (see below for explanation)
• $\Gamma(k)$, where $k$ is a rational number (other than integers and half of any integer), cannot be expressed in a closed form value.

Extension of gamma function

$\Gamma(n)$ function can be extended for negative non-integers using: $$

$$\Gamma(n) = \frac{\Gamma(n+1)}{n}$$

This cannot be used to define $\Gamma(0)$ because of the denominator. And through induction, $\Gamma$ function cannot be defined for negative integers.

Lemmas

Lemma 1

$$\forall s \gt 0 \; \int_0^\infty e^{-sx}\,\text{d}x\;\text{converges}$$

Lemma 2

$$\forall n \in\mathbb{Z}^+ \; \lim_{x\to\infty} \frac{x^{n-1}}{e^{x/2}} =0$$

Transformations

Alternate forms of $\Gamma(n)$. This section is intended to be exam-focused. Proofs for the transformations are included in a separate section. $$

Form 0, 1, 4

For $k\in \mathbb{R}$: $$

$$\Gamma(n) = k \int_0^\infty e^{-x^k} x^{kn-1}\,\text{d}x$$

Form 0 (definition) is resulted when setting $k=1$. Form 1 is resulted when setting $k=\frac{1}{n}$. $$

Form 2

$$\int_0^\infty e^{-kx}x^{n-1}\,\text{d}x = \frac{\Gamma(n)}{k^n}$$

Form 3

$$\Gamma(n)= \int_0^1 \bigg(\ln\frac{1}{x}\bigg)^{n-1} \,\text{d}x$$

Transformations Proofs

Form 1

$\forall n \gt 0$: $$

$$\Gamma(n)=\frac{1}{n}\int_0^\infty e^{-x^{1/n}}\,\text{d}x$$

Proof Hint

Use $x^n=t$. $$

Note

One of the most frequently used integrals in mathematics:

$$\int_0^{\infty} e^{-x^2}\,\text{d}x = \frac{\sqrt{\pi}}{2}$$

Form 2

$$\int_0^\infty e^{-kx}x^{n-1}\,\text{d}x = \frac{\Gamma(n)}{k^n}$$

Proof Hint

Use $x=kt$. $$

Form 3

$$\Gamma(n)= \int_0^1 \bigg(\ln\frac{1}{x}\bigg)^{n-1} \,\text{d}x$$

Proof Hint

Use $e^{-x}=t$. If the given integral’s range is from $0$ to $1$ and there is $\ln$, it’s better to try this substitution.

Form 4

For $k\in \mathbb{R}$: $$

$$\Gamma(n) = k \int_0^\infty e^{-x^k} x^{kn-1}\,\text{d}x$$

Proof Hint

Use $x=t^k$. $$