Beta and Gamma Function Calculator
Evaluate the Gamma and Beta functions, ∫₀^(π/2) sinᵖθ cosᵠθ dθ and ∫₀^∞ xᵐ e^(−axⁿ) dx, exactly for whole numbers and halves (with √π and π), step by step.
About the Beta and Gamma Function Calculator
The Gamma function Γ(x) = ∫₀^∞ t^(x − 1) e^(−t) dt extends the factorial: Γ(n) = (n − 1)! for whole numbers, and Γ(x + 1) = xΓ(x) for every x. With Γ(1/2) = √π, every half has an exact value too: Γ(7/2) = (5/2)(3/2)(1/2)√π = 15√π/8.
The Beta function B(m, n) = ∫₀¹ x^(m − 1)(1 − x)^(n − 1) dx equals Γ(m)Γ(n)/Γ(m + n). Its trigonometric form gives ∫₀^(π/2) sinᵖθ cosᵠθ dθ = ½B((p + 1)/2, (q + 1)/2), the source of the reduction (Wallis) formulas.
The formula
Γ(x + 1) = xΓ(x), Γ(1/2) = √π, B(m, n) = Γ(m)Γ(n)/Γ(m + n)
- Γ(n)
- (n − 1)! for a whole number n
- sin–cos integral
- ∫₀^(π/2) sinᵖθ cosᵠθ dθ = Γ((p + 1)/2)Γ((q + 1)/2) / (2Γ((p + q + 2)/2))
Worked example
∫₀^(π/2) sin⁴θ cos³θ dθ
- = ½B(5/2, 2) = Γ(5/2)Γ(2) / (2Γ(9/2))
- Γ(5/2) = 3√π/4, Γ(9/2) = 105√π/16
- = (3√π/4)(1) / (2 · 105√π/16) = 2/35
Questions people ask
Why is Γ(1/2) = √π?
Γ(1/2) = ∫₀^∞ t^(−1/2) e^(−t) dt; putting t = x² gives 2∫₀^∞ e^(−x²) dx = √π, the Gaussian integral.
What about negative x?
Γ(x) = Γ(x + 1)/x carries it below 0, so Γ(−1/2) = −2√π. It has no value at 0, −1, −2, … (poles).