§12.1 The $Γ$ -Function

12.1 The $Γ$ -Function

Definition 12.1.1.

The Gamma function is defined by

Γ (𝑧) = \int_{0}^{\infty} e^{- 𝑡} 𝑡^{𝑧 - 1} d 𝑡, (12.1 . 1)

where

𝑧 \in ℂ

By letting $𝑧 = 𝑥 + i 𝑦$ where $𝑥, 𝑦 \in ℝ$ , we have $| e^{- 𝑡} 𝑡^{𝑧 - 1} | = e^{- 𝑡} 𝑡^{𝑥 - 1}$ . Notice that for $𝑥 > 0$ ,

\begin{aligned} | Γ (𝑥) | & = \int_{0}^{1} e^{- 𝑡} 𝑡^{𝑥 - 1} d 𝑡 + \int_{1}^{\infty} e^{- 𝑡} 𝑡^{𝑥 - 1} d 𝑡 \\ \leq \int_{0}^{1} 𝑡^{𝑥 - 1} d 𝑡 + \int_{1}^{\infty} e^{- 𝑡} 𝑡^{𝑥 - 1} d 𝑡 = \frac{1}{𝑥} + \int_{1}^{\infty} e^{- 𝑡} 𝑡^{𝑥 - 1} d 𝑡 . \end{aligned}

Since $\int_{1}^{\infty} \frac{d 𝑡}{𝑡^{2}}$ is convergent and ${lim}_{𝑡 \to \infty} \frac{e^{- 𝑡} 𝑡^{𝑧 - 1}}{𝑡^{- 2}} = 0$ , then by comparison, the second integral is convergent.

Therefore, $Γ (𝑥)$ is convergent on $ℝ_{> 0}$ . It follows that $Γ (𝑧)$ is absolutely convergent on the right half-plane ${𝑧 \in ℂ : ℜ𝔢 (𝑧) > 0}$ .

Theorem 12.1.2.

The

Γ

-function is holomorphic on

{𝑧 \in ℂ : ℜ𝔢 (𝑧) > 0}

Proof.

Let $𝛾 \subset {𝑧 \in ℂ : ℜ𝔢 (𝑧) > 0}$ be an arbitrary simple, closed, rectifiable curve. By Morera’s Theorem (Theorem 3.2.6 $Theorem 3.2.6 Theorem 3.2.6$ ), it suffices to show that $\oint_{𝛾} Γ (𝑧) d 𝑧 = 0$ . We have

\begin{aligned} \oint_{𝛾} Γ (𝑧) d 𝑧 & = \oint_{𝛾} (\int_{0}^{\infty} e^{- 𝑡} 𝑡^{𝑧 - 1} d 𝑡) d 𝑧 \\ = \int_{0}^{\infty} e^{- 𝑡} (\oint_{𝛾} 𝑡^{𝑧 - 1} d 𝑧) d 𝑡 = 0, \end{aligned}

where the integral exchange is justified by uniform convergence of

Γ

on compact subsets of

{𝑧 \in ℂ : ℜ𝔢 (𝑧) > 0}

(by the Weierstrass

𝑀

–Test, Theorem 2.3.3

Theorem 2.3.3 Theorem 2.3.3

). ⁠

□

From integration by parts, we obtain

Γ (𝑧 + 1) = \int_{0}^{\infty} e^{- 𝑡} 𝑡^{𝑧} d 𝑡 = - {e^{- 𝑡} 𝑡^{𝑧} |}_{0}^{\infty} + 𝑧 \int_{0}^{\infty} e^{- 𝑡} 𝑡^{𝑧 - 1} d 𝑡 = 𝑧 Γ (𝑧) .

Additionally,

Γ (1) = \int_{0}^{\infty} e^{- 𝑡} d 𝑡 = - {e^{- 𝑡} |}_{0}^{\infty} = 1 .

Hence, we have $Γ (𝑧 + 1) = 𝑧!$ for nonnegative integers, and the $Γ$ -function generalizes the factorial. We also have

Γ (𝑧 + 𝑛) = Γ (𝑧) \prod_{𝑘 = 0}^{𝑛 - 1} (𝑧 + 𝑘), ℜ𝔢 (𝑧) > 0, 𝑛 \in ℕ .

Therefore, can derive its analytic continuation via

Γ (𝑧) = \frac{Γ (𝑧 + 𝑛)}{\prod_{𝑘 = 0}^{𝑛 - 1} (𝑧 + 𝑘)}, ℜ𝔢 (𝑧) > - 𝑛 .

Since the numerator is holomorphic on $ℜ𝔢 (𝑧) > - 𝑛$ and $𝑛$ was arbitrary, the analytic continuation of $Γ$ has simple poles at each of $ℤ_{\leq 0}$ . Hence, $Γ (𝑧)$ is meromorphic on $ℂ$ .

By (4.5.2) $(4.5.2) (4.5.2)$ , the residue at each pole is equal to

\underset{𝑧 = - 𝑛}{Res} Γ (𝑧) = lim_{𝑧 \to - 𝑛} \frac{Γ (𝑧 + 𝑛 + 1)}{\prod_{𝑘 = 0}^{𝑛 - 1} (𝑧 + 𝑘)} = \frac{1}{\prod_{𝑘 = 1}^{𝑛} (- 𝑘)} = \frac{{(- 1)}^{𝑛}}{𝑛!} .

We will now study two representations for the Gamma function.

Theorem 12.1.3 (Gauss).

The Gamma function satisfies

Γ (𝑧) = lim_{𝑛 \to \infty} \frac{𝑛^{𝑧} 𝑛!}{\prod_{𝑘 = 0}^{𝑛} (𝑧 + 𝑘)}, ℜ𝔢 𝑧 > 0 . (12.1 . 2)

Proof.

Define the sequence of functions

𝑓_{𝑛} (𝑧) = \int_{0}^{𝑛} {(1 - \frac{𝑡}{𝑛})}^{𝑛} 𝑡^{𝑧 - 1} d 𝑡 = 𝑛^{𝑧} \int_{0}^{1} {(1 - 𝑡)}^{𝑛} 𝑡^{𝑧 - 1} d 𝑡, ℜ𝔢 (𝑧) > 0 .

By integration by parts, we have

\begin{aligned} 𝑓_{𝑛} (𝑧) & = 𝑛^{𝑧} [{\frac{𝑡^{𝑧}}{𝑧} {(1 - 𝑡)}^{𝑛} |}_{0}^{1} + \frac{𝑛}{𝑧} \int_{0}^{1} {(1 - 𝑡)}^{𝑛 - 1} 𝑡^{𝑧} d 𝑡] \\ = {(\frac{𝑛}{𝑛 - 1})}^{𝑧 + 1} 𝑓_{𝑛 - 1} \frac{𝑧 + 1}{𝑧} = [𝑛^{𝑧 + 1} \frac{𝑛 - 1}{{(𝑛 - 2)}^{𝑧 + 2}}] 𝑓_{𝑛 - 2} \frac{𝑧 + 2}{𝑧 (𝑧 + 1)} \\ = 𝑛^{𝑧 + 1} (𝑛 - 1)! 𝑓_{1} \frac{𝑧 + 𝑛 - 1}{\prod_{𝑘 = 0}^{𝑛 - 2} (𝑧 + 𝑘)} \\ = \frac{𝑛^{𝑧} 𝑛!}{\prod_{𝑘 = 0}^{𝑛} (𝑧 + 𝑘)} . (12.1 . 3) \end{aligned}

Let us now analyze the difference

lim_{𝑛 \to \infty} [\int_{0}^{𝑛} e^{- 𝑡} 𝑡^{𝑧 - 1} d 𝑡 - 𝑓_{𝑛} (𝑧)] = lim_{𝑛 \to \infty} \int_{0}^{𝑛} e^{- 𝑡} 𝑡^{𝑧 - 1} [1 - e^{𝑡} {(1 - \frac{𝑡}{𝑛})}^{𝑛}] d 𝑡 . (12.1 . 4)

Since

\frac{d e^{𝑡} {(1 - \frac{𝑡}{𝑛})}^{𝑛}}{d 𝑡} = e^{𝑡} {(1 - \frac{𝑡}{𝑛})}^{𝑛} - e^{𝑡} {(1 - \frac{𝑡}{𝑛})}^{𝑛 - 1} = - e^{𝑡} \frac{𝑡}{𝑛} {(1 - \frac{𝑡}{𝑛})}^{𝑛 - 1},

we have

1 - e^{𝑡} {(1 - \frac{𝑡}{𝑛})}^{𝑛} = \frac{1}{𝑛} \int_{0}^{𝑡} 𝑢 e^{𝑢} {(1 - \frac{𝑢}{𝑛})}^{𝑛 - 1} d 𝑢 . (12.1 . 5)

Additionally, since

\begin{aligned} \frac{d e^{𝑢} {(1 - \frac{𝑢}{𝑛})}^{𝑛 - 1}}{d 𝑢} & = e^{𝑢} {(1 - \frac{𝑢}{𝑛})}^{𝑛 - 1} - \frac{𝑛 - 1}{𝑛} e^{𝑢} {(1 - \frac{𝑢}{𝑛})}^{𝑛 - 2} \\ = \frac{e^{𝑢}}{𝑛} {(1 - \frac{𝑢}{𝑛})}^{𝑛 - 2} (1 - 𝑢) \end{aligned}

has zeros at $𝑢 = 1$ and at $𝑢 = 𝑛$ , and

\frac{d^{2} e^{𝑢} {(1 - \frac{𝑢}{𝑛})}^{𝑛 - 1}}{d 𝑢^{2}} = \frac{e^{𝑢}}{𝑛^{2}} {(1 - \frac{𝑢}{𝑛})}^{𝑛 - 3} (𝑢^{2} - 2 𝑢 - 𝑛 + 2),

evaluates to $- \frac{e {(𝑛 - 1)}^{𝑛 - 2}}{𝑛^{𝑛 - 1}} < 0$ at $𝑢 = 1$ and evaluates to

\frac{e^{𝑛}}{𝑛^{𝑛 - 1}} {(𝑛 - 𝑢)}^{𝑛 - 3} (𝑛 - 2) (𝑛 - 1) \to 0^{+}

as $𝑢 \to 𝑛^{-}$ , $e^{𝑢} {(1 - \frac{𝑢}{𝑛})}^{𝑛 - 1}$ attains its maximum of $e {(\frac{𝑛 - 1}{𝑛})}^{𝑛 - 1}$ at $𝑢 = 1$ . For $𝑛 > 1$ , $e {(\frac{𝑛 - 1}{𝑛})}^{𝑛 - 1} \leq e$ . From (12.1.5) $(12.1.5) (12.1.5)$ , we have

1 - e^{𝑡} {(1 - \frac{𝑡}{𝑛})}^{𝑛} \leq \frac{e 𝑡^{2}}{2 𝑛} .

Moreover, since for $0 \leq 𝑢 \leq 𝑡 \leq 𝑛$ ,

𝑢 e^{𝑢} {(1 - \frac{𝑢}{𝑛})}^{𝑛 - 1} > 0,

it follows that $1 - e^{𝑡} {(1 - \frac{𝑡}{𝑛})}^{𝑛}$ is positive. By (12.1.4) $(12.1.4) (12.1.4)$ , we have

| \int_{0}^{𝑛} e^{- 𝑡} 𝑡^{𝑧 - 1} [1 - e^{𝑡} {(1 - \frac{𝑡}{𝑛})}^{𝑛}] d 𝑡 | \leq \frac{e}{2 𝑛} | \int_{0}^{𝑛} e^{- 𝑡} 𝑡^{𝑧 + 1} d 𝑡 | < \frac{1}{2 𝑛} | Γ (𝑧 + 2) | \to 0

𝑛 \to \infty

. From (12.1.3)

(12.1.3) (12.1.3)

, we have

Γ (𝑧) = {lim}_{𝑛 \to \infty} \frac{𝑛^{𝑧} 𝑛!}{\prod_{𝑘 = 0}^{𝑛} (𝑧 + 𝑘)}

, or (12.1.2)

(12.1.2) (12.1.2)

⁠

□

The Weierstrass formula is a direct consequence of the Gauss formula.

Theorem 12.1.4 (Weierstrass).

The reciprocal $Γ$ -function has the entire Weierstrass factorization of

\frac{1}{Γ (𝑧)} = 𝑧 \prod_{𝑘 = 1}^{\infty} [(1 + \frac{𝑧}{𝑘}) e^{- \frac{𝑧}{𝑘}}] e^{𝑧 𝛾}, (12.1 . 6)

where

𝛾 = \int_{1}^{\infty} (\frac{1}{⌊ 𝑥 ⌋} - \frac{1}{𝑥}) d 𝑥

Proof.

Since the Gauss formula agrees with (12.1.1) $(12.1.1) (12.1.1)$ on the right half-plane, the analytic continuation of $Γ (𝑧)$ is unique on the entire complex plane except for the poles at $ℤ_{\leq 0}$ by the Identity Theorem (Theorem 3.3.3 $Theorem 3.3.3 Theorem 3.3.3$ ). Since

\begin{aligned} \frac{𝑛^{𝑧} 𝑛!}{\prod_{𝑘 = 0}^{𝑛} (𝑧 + 𝑘)} & = \frac{exp (𝑧 log (𝑛))}{𝑧 \prod_{𝑘 = 1}^{𝑛} (1 + \frac{𝑧}{𝑘})} \\ = \frac{exp (𝑧 \int_{1}^{𝑛} \frac{d 𝑥}{𝑥})}{𝑧 \prod_{𝑘 = 1}^{𝑛} (1 + \frac{𝑧}{𝑘})} \frac{exp (- 𝑧 \sum_{𝑘 = 1}^{𝑛} \frac{1}{𝑘})}{\prod_{𝑘 = 1}^{𝑛} exp (- \frac{𝑧}{𝑘})} \\ = \frac{1}{𝑧 \prod_{𝑘 = 1}^{𝑛} [(1 + \frac{𝑧}{𝑘}) e^{- \frac{𝑧}{𝑘}}]} exp (- 𝑧 (\int_{1}^{𝑛} (\frac{1}{⌊ 𝑥 ⌋} - \frac{1}{𝑥}) d 𝑥)) . \end{aligned}

Therefore,

\begin{aligned} \frac{1}{Γ (𝑧)} & = lim_{𝑛 \to \infty} \frac{\prod_{𝑘 = 0}^{𝑛} (𝑧 + 𝑘)}{𝑛^{𝑧} 𝑛!} \\ = 𝑧 \prod_{𝑘 = 1}^{𝑛} [(1 + \frac{𝑧}{𝑘}) e^{- \frac{𝑧}{𝑘}}] lim_{𝑛 \to \infty} exp (𝑧 (\int_{1}^{𝑛} (\frac{1}{⌊ 𝑥 ⌋} - \frac{1}{𝑥}) d 𝑥)) \\ = 𝑧 \prod_{𝑘 = 1}^{𝑛} [(1 + \frac{𝑧}{𝑘}) e^{- \frac{𝑧}{𝑘}}] exp (𝑧 𝛾) . \end{aligned}

The constant $𝛾 = \int_{1}^{\infty} (\frac{1}{⌊ 𝑥 ⌋} - \frac{1}{𝑥}) d 𝑥$ is known as the Euler–Mascheroni constant. For each $𝑥$ , $0 \leq \frac{1}{⌊ 𝑥 ⌋} - \frac{1}{𝑥} < \frac{1}{⌊ 𝑥 ⌋} - \frac{1}{⌊ 𝑥 ⌋ + 1}$ , and the integral is bounded by $\sum_{𝑛 = 1}^{\infty} (\frac{1}{𝑛} - \frac{1}{𝑛 + 1}) < 1$ , and hence the integral is convergent. The properties of $𝛾$ still largely remain unknown.

By the Weierstrass Factorization Theorem (Theorem 4.4.3.5 $Theorem 4.4.3.5 Theorem 4.4.3.5$ ), if we let $𝑎_{𝑛} = - 𝑛$ and $𝑝_{𝑛} = 1$ , it follows that

\sum_{𝑛 = 1}^{\infty} {| \frac{𝑅}{𝑎_{𝑛}} |}^{𝑝_{𝑛} + 1} = 𝑅^{2} \sum_{𝑛 = 1}^{\infty} \frac{1}{𝑛^{2}} = \frac{𝑅^{2} π^{2}}{6}

is convergent. Thus, the Weierstrass formula defines an entire function with zeros at each of $ℤ_{\leq 0}$ . ⁠ $□$

We have two famous identities on the $Γ$ -function:

Theorem 12.1.5 (Euler’s Reflection Formula).

The $Γ$ -function can be analytically continued to the left half-plane with the functional equation of

Γ (𝑧) Γ (1 - 𝑧) = π csc (π 𝑧) (12.1 . 7)

for

𝑧 \in ℂ ∖ ℤ

Proof.

By the Weierstrass Formula (Theorem 12.1.4 $Theorem 12.1.4 Theorem 12.1.4$ ), we have

\frac{1}{Γ (𝑧)} = 𝑧 \prod_{𝑘 = 1}^{𝑛} [(1 + \frac{𝑧}{𝑘}) e^{- \frac{𝑧}{𝑘}}] e^{𝑧 𝛾}, \frac{1}{Γ (- 𝑧)} = - 𝑧 \prod_{𝑘 = 1}^{𝑛} [(1 - \frac{𝑧}{𝑘}) e^{\frac{𝑧}{𝑘}}] e^{- 𝑧 𝛾} .

Since the Weierstrass elementary factors form an absolutely convergent infinite product, we may rearrange its terms. Hence, by Theorem 4.4.5.4 $Theorem 4.4.5.4 Theorem 4.4.5.4$ , we have

\frac{1}{Γ (𝑧) Γ (1 - 𝑧)} = - \frac{1}{𝑧 Γ (𝑧) Γ (- 𝑧)} = 𝑧 \prod_{𝑘 = 1}^{𝑛} (1 - \frac{𝑧^{2}}{𝑘^{2}}) = \frac{sin (π 𝑧)}{π},

which confirms (12.1.7)

(12.1.7) (12.1.7)

. ⁠

□

Example 12.1.6.

Evaluate

Γ (\frac{1}{2})

Solution to Example 12.1.6 $Example 12.1.6 Example 12.1.6$ .

By the Reflection Formula (Theorem 12.1.5 $Theorem 12.1.5 Theorem 12.1.5$ ), we have that

{Γ (\frac{1}{2})}^{2} = π csc (\frac{π}{2}) = π,

and it follows that

Γ (\frac{1}{2}) = \sqrt{π}

as it is positive. ⁠

□

Theorem 12.1.7 (Legendre’s Duplication Formula).

For any $𝑧 \in ℂ ∖ (- \frac{ℕ}{2})$ , we have

Γ (𝑧) Γ (𝑧 + \frac{1}{2}) = 2^{1 - 2 𝑧} \sqrt{π} Γ (2 𝑧) . (12.1 . 8)

Proof.

From Theorem 12.1.3 $Theorem 12.1.3 Theorem 12.1.3$ , we have

Γ (𝑧) Γ (𝑧 + \frac{1}{2}) = lim_{𝑛 \to \infty} \frac{𝑛^{2 𝑧 + \frac{1}{2}} {𝑛!}^{2}}{\prod_{𝑘 = 0}^{𝑛} (𝑧 + 𝑘) (𝑧 + 𝑘 + \frac{1}{2})} = lim_{𝑛 \to \infty} \frac{2^{2 𝑛 + 2} 𝑛^{2 𝑧 + \frac{1}{2}} {𝑛!}^{2}}{\prod_{𝑘 = 0}^{2 𝑛 + 1} (2 𝑧 + 𝑘)}

where the left-hand side is defined since $𝑧 \in ℂ ∖ (- \frac{ℕ}{2})$ . By expansion of the value, we have

\begin{aligned} Γ (𝑧) Γ (𝑧 + \frac{1}{2}) & = lim_{𝑛 \to \infty} \frac{{(2 𝑛)}^{2 𝑧} (2 𝑛)!}{\prod_{𝑘 = 0}^{2 𝑛} (2 𝑧 + 𝑘)} \cdot \frac{𝑛^{\frac{1}{2}} {𝑛!}^{2} 2^{2 𝑛 + 2 - 2 𝑧}}{(2 𝑧 + 2 𝑛 + 1) (2 𝑛)!} \\ = Γ (2 𝑧) lim_{𝑛 \to \infty} \frac{𝑛^{\frac{1}{2}} {𝑛!}^{2} 2^{2 - 2 𝑧}}{(2 𝑧 + 2 𝑛 + 1) \prod_{𝑘 = 0}^{𝑛 - 1} (𝑘 + \frac{1}{2}) \prod_{𝑘 = 1}^{𝑛} 𝑘} \\ = 2^{2 - 2 𝑧} Γ (2 𝑧) lim_{𝑛 \to \infty} \frac{𝑛^{\frac{1}{2}} 𝑛!}{\prod_{𝑘 = 0}^{𝑛} (𝑘 + \frac{1}{2})} \cdot \frac{𝑛 + \frac{1}{2}}{2 𝑧 + 2 𝑛 + 1} \\ = 2^{1 - 2 𝑧} Γ (2 𝑧) Γ (\frac{1}{2}) \\ = 2^{1 - 2 𝑧} Γ (2 𝑧) \sqrt{π}, \end{aligned}

where the last step is derived from Example 12.1.6

Example 12.1.6 Example 12.1.6

. ⁠

□

The identity above is a special case of the following result:

Theorem 12.1.8 (Gauss Multiplication Theorem).

Suppose $𝑚 \in ℕ_{\geq 2}$ . Let $𝑧 \in ℂ ∖ (- \frac{ℕ}{𝑚})$ . Then we have

Γ (𝑧) Γ (𝑧 + \frac{1}{𝑚}) \dots Γ (𝑧 + \frac{𝑚 - 1}{𝑚}) = {(2 π)}^{\frac{𝑚 - 1}{2}} 𝑚^{\frac{1}{2} - 𝑚 𝑧} Γ (𝑚 𝑧) . (12.1 . 9)

The Gamma function as in (12.1.1) $(12.1.1) (12.1.1)$ is commonly referred to as the Euler Integral of the Second Kind.

Definition 12.1.9 (Beta Function).

The Euler Integral of the First Kind is also known as the Beta function, and is defined by

Β (𝑧_{1}, 𝑧_{2}) = \int_{0}^{1} 𝑡^{𝑧_{1} - 1} {(1 - 𝑡)}^{𝑧_{2} - 1} d 𝑡 .

By a change of variables (by letting $𝜏 = 1 - 𝑡$ ), we derive the symmetry of the Beta function:

Β (𝑧_{1}, 𝑧_{2}) = \int_{0}^{1} 𝜏^{𝑧_{2} - 1} {(1 - 𝜏)}^{𝑧_{1} - 1} d 𝜏 = Β (𝑧_{2}, 𝑧_{1}) .

The Beta function is commonly treated as an auxiliary function in many cases of integral evaluation due to its connection with the Gamma function:

Theorem 12.1.10.

For any $ℜ𝔢 (𝑧_{1}), ℜ𝔢 (𝑧_{2}) > 0$ , we have

Β (𝑧_{1}, 𝑧_{2}) = \frac{Γ (𝑧_{1}) Γ (𝑧_{2})}{Γ (𝑧_{1} + 𝑧_{2})} .

Proof.

Consider the product $Γ (𝑧_{1}) Γ (𝑧_{2})$ . By letting $𝑠 = 𝑢 𝑡$ and $𝑣 = 𝑡 (𝑢 + 1)$ , we have

\begin{aligned} Γ (𝑧_{1}) Γ (𝑧_{2}) & = \int_{0}^{\infty} e^{- 𝑠} 𝑠^{𝑧_{2} - 1} [\int_{0}^{\infty} e^{- 𝑡} 𝑡^{𝑧_{1} - 1} d 𝑡] d 𝑠 \\ = \int_{0}^{\infty} 𝑢^{𝑧_{2} - 1} [\int_{0}^{\infty} e^{- 𝑣} {(\frac{𝑣}{𝑢 + 1})}^{𝑧_{1} + 𝑧_{2} - 1} d (\frac{𝑣}{𝑢 + 1})] d 𝑢 \\ = \int_{0}^{\infty} \frac{𝑢^{𝑧_{2} - 1}}{{(𝑢 + 1)}^{𝑧_{1} + 𝑧_{2}}} [\int_{0}^{\infty} e^{- 𝑣} 𝑣^{𝑧_{1} + 𝑧_{2} - 1} d 𝑣] d 𝑢 . (12.1 . 10) \end{aligned}

Let $𝑟 = \frac{𝑢}{𝑢 + 1}$ , $𝑢 = \frac{𝑟}{1 - 𝑟}$ , and $d 𝑢 = \frac{1}{{(1 - 𝑟)}^{2}} d 𝑟$ . Then we have

Γ (𝑧_{1}) Γ (𝑧_{2}) = Γ (𝑧_{1} + 𝑧_{2}) \int_{0}^{1} 𝑟^{𝑧_{2} - 1} {(1 - 𝑟)}^{𝑧_{1} - 1} d 𝑟 = Γ (𝑧_{1} + 𝑧_{2}) Β (𝑧_{1}, 𝑧_{2}) . □

Example 12.1.11 (MIT Integration Bee 2023 Finals #1).

Evaluate

\int_{0}^{\frac{π}{2}} \frac{{tan (𝑥)}^{\frac{1}{3}}}{{(sin (𝑥) + cos (𝑥))}^{2}} d 𝑥 .

Solution to Example 12.1.11 $Example 12.1.11 Example 12.1.11$ .

By rewriting the integral, and letting $𝑢 = tan (𝑥)$ , we have

\begin{aligned} 𝐼 = \int_{0}^{\frac{π}{2}} \frac{{tan (𝑥)}^{\frac{1}{3}}}{{(sin (𝑥) + cos (𝑥))}^{2}} d 𝑥 & = \int_{0}^{\infty} \frac{𝑢^{\frac{1}{3}} {sec}^{2} (𝑥) d 𝑥}{{(𝑢 + 1)}^{2}} \\ = \int_{0}^{\infty} 𝑢^{\frac{1}{3}} d \frac{𝑢}{{(𝑢 + 1)}^{2}} \\ = Β (\frac{4}{3}, \frac{2}{3}), \end{aligned}

where the last step recognizes the form of (12.1.10) $(12.1.10) (12.1.10)$ . Theorem 12.1.10 $Theorem 12.1.10 Theorem 12.1.10$ then gives

𝐼 = \frac{Γ (\frac{2}{3}) Γ (\frac{4}{3})}{Γ (2)} = \frac{1}{3} Γ (\frac{1}{3}) Γ (\frac{2}{3}) .

Lastly, the Reflection Formula (Theorem 12.1.5 $Theorem 12.1.5 Theorem 12.1.5$ ) gives that

𝐼 = \frac{π}{3 sin (\frac{π}{3})} = \frac{2 π \sqrt{3}}{9} . □

Example 12.1.12.

Evaluate

𝐼 = \int_{0}^{\frac{π}{2}} 𝑥 \sqrt{tan 𝑥} d 𝑥 .

Proof of Example 12.1.12 $Example 12.1.12 Example 12.1.12$ .

Apply the substitution $𝑢 = tan 𝑥$ , $𝑠 = 𝑢^{2}$ and the integral representation $arctan 𝑢 = \int_{0}^{𝑢} \frac{d 𝑣}{1 + 𝑣^{2}} = \int_{0}^{1} 𝑢 \frac{d 𝑣}{1 + 𝑢^{2} 𝑣^{2}}$ to get

\begin{aligned} 𝐼 & = \int_{0}^{\infty} arctan 𝑢 \frac{\sqrt{𝑢} d 𝑢}{1 + 𝑢^{2}} = \int_{0}^{\infty} (\int_{0}^{1} \frac{𝑢}{1 + 𝑢^{2} 𝑣^{2}} d 𝑣) \frac{\sqrt{𝑢} d 𝑢}{1 + 𝑢^{2}} \\ = \int_{0}^{1} \int_{0}^{\infty} \frac{𝑢^{\frac{3}{2}}}{(1 + 𝑢^{2} 𝑣^{2}) (1 + 𝑢^{2})} d 𝑢 d 𝑣 = \int_{0}^{1} \int_{0}^{\infty} \frac{𝑠^{\frac{1}{4}}}{2 (1 + 𝑠 𝑣^{2}) (1 + 𝑠)} d 𝑠 d 𝑣 . \end{aligned}

The integral switch is justified since the integrand $\geq 0$ throughout. Consider the integral for

\int_{0}^{\infty} \frac{𝑠^{\frac{1}{4}} d 𝑠}{(1 + 𝑠 𝑣^{2}) (𝑠 + 1)} = \frac{1}{1 - 𝑣^{2}} \int_{0}^{\infty} 𝑠^{𝛽} (\frac{1}{1 + 𝑠} - \frac{1}{𝑐 + 𝑠}) d 𝑠,

where $𝑐 = 𝑣^{- 2}$ and $𝛽 = \frac{1}{4}$ . Let the integral in the last expression be a complex function of $𝛽$ , more explicitly $𝐽 (𝛽)$ , defined for $- 1 < ℜ𝔢 𝛽 < 1$ .

Notice that the integral maybe written as the difference of two convergent integrals for $- 1 < ℜ𝔢 𝛽 < 0$ :

\begin{aligned} 𝐽 (𝛽) & = \int_{0}^{\infty} 𝑠^{𝛽} \frac{d 𝑠}{1 + 𝑠} - \int_{0}^{\infty} 𝑠^{𝛽} 𝑐^{𝛽} \frac{d 𝑠}{1 + 𝑠} \\ = (1 - 𝑐^{𝛽}) Β (- 𝛽, 𝛽 + 1) = (𝑐^{𝛽} - 1) π csc (π 𝛽) \end{aligned}

by the reflection formula (Theorem 12.1.5 $Theorem 12.1.5 Theorem 12.1.5$ ), which derives an analytic continuation of $𝐽 |_{- 1 < ℜ𝔢 𝛽 < 1}$ to everywhere in $ℂ ∖ ℤ$ .

One can show by differentiation under the integral sign, that the integral $𝐽 (𝛽)$ is holomorphic for $- 1 < ℜ𝔢 𝛽 < 1$ . The identity theorem (Theorem 3.3.3 $Theorem 3.3.3 Theorem 3.3.3$ ) then allows the computed cosecant expression to be assigned to $𝐽 (𝛽)$ :

\begin{aligned} 𝐼 & = \int_{0}^{1} \frac{(𝑣^{- \frac{1}{2}} - 1) π csc (\frac{π}{4})}{2 - 2 𝑣^{2}} d 𝑣 = \frac{π \sqrt{2}}{2} \int_{0}^{1} \frac{𝑣^{- \frac{1}{2}} - 1}{1 - 𝑣^{2}} d 𝑣 = π \sqrt{2} \int_{0}^{1} \frac{1 - 𝑢}{1 - 𝑢^{4}} d 𝑢 \\ = π \sqrt{2} \int_{0}^{1} \frac{d 𝑢}{(1 + 𝑢) (1 + 𝑢^{2})} = \frac{π \sqrt{2}}{2} \int_{0}^{1} (\frac{1}{1 + 𝑢} + \frac{1 - 𝑢}{1 + 𝑢^{2}}) d 𝑢 \\ = \frac{π \sqrt{2}}{2} (log 2 + arctan 1 - \frac{1}{2} log 2) = \frac{π^{2} \sqrt{2}}{8} + \frac{π \sqrt{2}}{4} log 2 . □ \end{aligned}

§12.1 The Γ-Function

12.1 The Γ-Function

§12.1 The $Γ$ -Function

12.1 The $Γ$ -Function