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✓ 7 results · all verified · 4 also independently AI-judged

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The Real Gamma and Beta Functions: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

Special values of the real Gamma and Beta functions

Example

The real Gamma and Beta functions have the values

$Γ (1) = 1, Γ (2) = 1, Γ (3) = 2, Γ (1 / 2) = \sqrt{π}, Γ (3 / 2) = \frac{\sqrt{π}}{2}, B (1, 1) = 1, B (2, 3) = \frac{1}{12} .$

Facts & Assumptions

Given: The displayed positive arguments.

[F1]

For every $s > 0$ , $Γ (s + 1) = s Γ (s)$ , and $Γ (1) = 1$ (The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ ).

[F2]

$Γ (1 / 2) = \sqrt{π}$ ( $Γ (1 / 2) = \sqrt{π}$ from the Gaussian integral).

[F3]

For $p, q > 0$ , $B (p, q) = Γ (p) Γ (q) / Γ (p + q)$ (The real Beta--Gamma identity).

Verification

technique · direct

1.1

Repeated use of [F1] gives $Γ (1) = 1$ , $Γ (2) = 1$ , and $Γ (3) = 2$ .

F1algebra

1.2

Fact [F2] and [F1] give $Γ (1 / 2) = \sqrt{π}$ and $Γ (3 / 2) = (1 / 2) \sqrt{π}$ .

F1F2algebra

2.1

Substituting the integer values into [F3] gives $B (1, 1) = 1$ and $B (2, 3) = 1 \cdot 2 / 24 = 1 / 12$ .

step 1.1F3algebra∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-24Open item page →

The unit-ball volumes through dimension eight from the Gamma formula

Example

For dimensions $1$ through $8$ , the unit-ball volumes are

$2, π, \frac{4 π}{3}, \frac{π^{2}}{2}, \frac{8 π^{2}}{15}, \frac{π^{3}}{6}, \frac{16 π^{3}}{105}, \frac{π^{4}}{24} .$

Facts & Assumptions

Given: Positive integer dimensions $1 \leq n \leq 8$ .

[F1]

For every $n \geq 1$ , $V_{n} (1) = π^{n / 2} / Γ (n / 2 + 1)$ (The closed form for the volume of the unit $n$ -ball).

[F2]

For every $s > 0$ , $Γ (s + 1) = s Γ (s)$ (The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ ).

[F3]

$Γ (1 / 2) = \sqrt{π}$ ( $Γ (1 / 2) = \sqrt{π}$ from the Gaussian integral).

Verification

technique · direct

1.1

Substitute $n = 1, 2, 3, 4$ into [F1] and use [F2] and [F3]. This gives $2, π, 4 π / 3, π^{2} / 2$ .

F1F2F3algebra

1.2

The same substitution with [F2] and [F3] for $n = 5, 6, 7, 8$ gives $8 π^{2} / 15, π^{3} / 6, 16 π^{3} / 105, π^{4} / 24$ .

F1F2F3algebra

2.1

Steps 1.1 and 1.2 establish every value in the displayed list.

step 1.1step 1.2∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-24Open item page →

A positive convex function need not be log-convex

Example

The identity function $f (x) = x$ on $(0, \infty)$ is positive and convex, but it is not log-convex.

Facts & Assumptions

Given: The identity function on the positive real axis.

[F1]

A positive function is log-convex when its logarithm is convex (Log-convex positive functions).

[F2]

The natural logarithm is strictly increasing on $(0, \infty)$ (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

Verification

technique · direct

1.1

For $x, y > 0$ and $0 \leq λ \leq 1$ , $f ((1 - λ) x + λ y) = (1 - λ) f (x) + λ f (y)$ , so $f$ is affine and hence convex.

givenalgebra

1.2

The midpoint of $1$ and $4$ is $5 / 2 > 2$ . By [F2], $\log (5 / 2) > \log 2 = (\log 1 + \log 4) / 2$ , so $\log \circ f$ violates the midpoint convexity inequality and [F1] shows that $f$ is not log-convex.

F1F2algebra

2.1

Thus this positive function is convex but not log-convex.

step 1.1step 1.2∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-24Open item page →

A positive non-log-convex solution of the Gamma functional equation

Example

Put $D : = \frac{1}{2} (\log Γ (1) + \log Γ (3 / 2)) - \log Γ (5 / 4)$ , let $A : = 1 + ∣ D ∣$ , and define $F (s) : = Γ (s) e^{A \sin (2 π s)}$ for $s > 0$ . The function $F (s) = Γ (s) e^{A \sin (2 π s)}$ is positive, satisfies $F (1) = 1$ and $F (s + 1) = s F (s)$ , differs from Gamma, and is not log-convex.

Facts & Assumptions

Given: The constants $D, A$ and the function $F$ in the Example.

[F1]

For every $s > 0$ , $Γ (s + 1) = s Γ (s)$ , and $Γ (1) = 1$ (The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ ).

[F2]

Sine has period $2 π$ (The zero sets of sine and cosine and the least positive common period 2 pi).

[F3]

A positive function is log-convex when its logarithm is convex (Log-convex positive functions).

Verification

technique · direct

1.1

By [F2], $e^{A \sin (2 π (s + 1))} = e^{A \sin (2 π s)}$ . Together with [F1], this gives $F (s + 1) = s F (s)$ and $F (1) = 1$ ; positivity is immediate.

F1F2algebra

1.2

At $s = 1, 5 / 4, 3 / 2$ , the sine terms are respectively $0, 1, 0$ . Hence the midpoint value of $\log F$ exceeds the average of its endpoint values exactly when $A > D$ , which holds because $A = 1 + ∣ D ∣$ . By [F3], $F$ is not log-convex.

F3algebra

2.1

Since $A > 0$ , $F (5 / 4) = e^{A} Γ (5 / 4) \neq Γ (5 / 4)$ . Thus $F$ is a different positive normalized solution of the recurrence.

step 1.1step 1.2algebra∎

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

FALSE: normalization and the functional equation determine the real Gamma function

Statement

False claim: Gamma is the only positive function $f : (0, \infty) \to (0, \infty)$ satisfying $f (1) = 1$ and $f (s + 1) = s f (s)$ .

Facts & Assumptions

Given: The periodic perturbation of Gamma from the preceding example.

[F1]

The function $F (s) = Γ (s) e^{A \sin (2 π s)}$ is positive, satisfies $F (1) = 1$ and $F (s + 1) = s F (s)$ , differs from Gamma, and is not log-convex (A positive non-log-convex solution of the Gamma functional equation).

[F2]

Gamma is the unique positive log-convex function with the normalization and recurrence (Bohr--Mollerup characterisation of the real Gamma function).

Refutation

technique · direct

1.1

By [F1], the function $F$ is positive, normalized, recurrent, and differs from Gamma.

2.1

Fact [F2] identifies the missing hypothesis: log-convexity excludes this $F$ and restores uniqueness.

step 1.1F2

3.1

Therefore normalization and the functional equation alone do not determine Gamma.

step 1.1step 2.1∎

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

FALSE: Euler's real Gamma integral converges at the nonpositive integers

Statement

False claim: the Euler integral $\int_{0}^{\infty} t^{s - 1} e^{- t} d t$ converges when $s$ is a nonpositive integer and thereby defines a real Gamma value there.

Facts & Assumptions

Given: A nonpositive integer $s$ .

[F1]

The Euler integral converges if and only if $s > 0$ (Euler's Gamma integral converges exactly for positive real parameters).

Refutation

technique · direct

1.1

Since $s \leq 0$ , the reverse direction of [F1] says that the Euler integral diverges, already at its endpoint $0$ .

F1algebra

2.1

The real Gamma function defined by Euler's integral therefore has no value at this $s$ ; a different continuation would not be convergence of this integral.

step 1.1

3.1

As the argument applies to every nonpositive integer, the claim is false.

step 1.1step 2.1∎

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24Open item page →

FALSE: unit-ball volume increases with dimension

Statement

False claim: the volume $V_{n} (1)$ of the unit ball increases with the positive integer dimension $n$ .

Facts & Assumptions

Given: The sequence of positive-dimensional unit-ball volumes.

[F1]

Among positive integer dimensions, the unit-ball volume is uniquely maximal at $n = 5$ (The unit-ball volume is maximal in dimension five).

[F2]

$V_{n} (1) \to 0$ as $n \to \infty$ (The volume of the unit $n$ -ball tends to zero with dimension).

Refutation

technique · direct

1.1

Fact [F1] gives the explicit strict decrease $V_{6} (1) < V_{5} (1)$ , contradicting monotone increase.

1.2

Independently, [F2] says the positive volumes tend to zero, which is incompatible with an increasing positive sequence.

2.1

Either step refutes the universal monotonicity claim.

step 1.1step 1.2∎

The Real Gamma and Beta Functions: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Special values of the real Gamma and Beta functions

Example

Facts & Assumptions

Verification

The unit-ball volumes through dimension eight from the Gamma formula

Example

Facts & Assumptions

Verification

A positive convex function need not be log-convex

Example

Facts & Assumptions

Verification

A positive non-log-convex solution of the Gamma functional equation

Example

Facts & Assumptions

Verification

FALSE: normalization and the functional equation determine the real Gamma function

Statement

Facts & Assumptions

Refutation

FALSE: Euler's real Gamma integral converges at the nonpositive integers

Statement

Facts & Assumptions

Refutation

FALSE: unit-ball volume increases with dimension

Statement

Facts & Assumptions

Refutation

Sources