Beta and Gamma functions

The two defining integrals. The Beta function is

where: $m,n$ are the two positive parameters and the domain of integration is fixed at $[0,1]$. Swapping $x\to 1-x$ inside the integral shows it is symmetric, $B(m,n)=B(n,m)$. The Gamma function is

where: $n$ is the single positive parameter and $e^{-x}$ is the ordinary exponential decay. Both are improper integrals — $\Gamma(n)$ has an infinite upper limit, and for $0<n<1$ also a singularity of $x^{n-1}$ at $x=0$ — and the restriction $m,n>0$ is exactly what the earlier convergence tests in this unit certify as safe: the exponential tail always dies fast enough at infinity, and the power singularity at $0$ stays integrable only while the exponent on $x$ exceeds $-1$.

The bridge between them. The relation that does almost all the work in this topic is

where: every symbol is the same $m,n$ as above. It means a Beta value never needs to be attacked as a defining integral — convert it to three Gamma values instead.

Recurrence and the factorial link. Integrating $\Gamma(n)$ by parts gives

where: $n>0$ is any positive real. Since $\Gamma(1)=\int_0^\infty e^{-x}dx=1$, repeated use of the recurrence for a positive integer $n$ collapses to

which is the precise sense in which Gamma "is" the factorial — shifted by one and defined for every positive real in between, not only the integers. The special half-integer seed value is

Trigonometric form. Substituting $x=\sin^2\theta$ (so $1-x=\cos^2\theta$ and $dx = 2\sin\theta\cos\theta\,d\theta$) into the defining Beta integral turns it into

where: $m,n>0$ fix the odd exponents $2m-1$ and $2n-1$ on sine and cosine. This single substitution is why a whole class of trigonometric-power integrals — otherwise handled only by tedious reduction formulae — becomes a two-line Gamma computation.

Why the domain restrictions matter. None of the four boxed results above are arbitrary rules to memorize — they all trace back to the convergence tests for improper integrals covered earlier in this unit. In $\Gamma(n)=\int_0^\infty e^{-x}x^{n-1}dx$, the exponential factor $e^{-x}$ guarantees convergence at the infinite upper limit for every real $n$, no matter how large; the only thing that can go wrong is at the lower limit $x=0$, where $x^{n-1}$ blows up once $n<1$, and the standard $p$-test for integrals near a singular point shows that stays integrable only while $n-1>-1$, i.e. $n>0$. The same logic bounds $B(m,n)$: both endpoints $x=0$ and $x=1$ carry potential singularities, of orders $x^{m-1}$ and $(1-x)^{n-1}$ respectively, and each needs its exponent to exceed $-1$ — exactly $m>0$ and $n>0$. Recognizing a given integral as "really" a Beta or Gamma integral in disguise, then checking its parameters satisfy these domain conditions, is itself an exam skill worth practicing.

Problem. Evaluate $\displaystyle\int_0^{\pi/2}\sin^4\theta\,\cos^2\theta\,d\theta$ using the Beta function, then use the result to evaluate the related integral $\displaystyle\int_0^1 x^3(1-x)^4\,dx$.

Step 1. Match exponents. Comparing $\sin^4\theta\cos^2\theta$ against $\sin^{2m-1}\theta\cos^{2n-1}\theta$ gives $2m-1=4\Rightarrow m=\tfrac52$ and $2n-1=2\Rightarrow n=\tfrac32$, so the integral equals $\tfrac12 B(\tfrac52,\tfrac32)$.

Step 2. Convert to Gamma values. $B(\tfrac52,\tfrac32)=\dfrac{\Gamma(5/2)\Gamma(3/2)}{\Gamma(4)}$. From the recurrence starting at $\Gamma(1/2)=\sqrt\pi$: $\Gamma(3/2)=\tfrac12\sqrt\pi$ and $\Gamma(5/2)=\tfrac32\cdot\tfrac12\sqrt\pi=\tfrac34\sqrt\pi$; and $\Gamma(4)=3!=6$.

Step 3. Substitute and simplify. $$\int_0^{\pi/2}\sin^4\theta\cos^2\theta\,d\theta = \frac12\cdot\frac{\left(\tfrac34\sqrt\pi\right)\left(\tfrac12\sqrt\pi\right)}{6} = \frac12\cdot\frac{\tfrac38\pi}{6} = \frac{3\pi}{96} = \frac{\pi}{32}.$$ So $\displaystyle\int_0^{\pi/2}\sin^4\theta\cos^2\theta\,d\theta = \dfrac{\pi}{32}$.

Competitive stretch

Step 1. The algebraic-power integral $\int_0^1 x^3(1-x)^4\,dx$ matches the Beta definition $\int_0^1 x^{m-1}(1-x)^{n-1}dx$ directly with $m-1=3\Rightarrow m=4$ and $n-1=4\Rightarrow n=5$, so the integral is $B(4,5)$.

Step 2. Convert to Gamma values: $B(4,5)=\dfrac{\Gamma(4)\Gamma(5)}{\Gamma(9)}=\dfrac{3!\,4!}{8!}=\dfrac{6\times24}{40320}=\dfrac{144}{40320}$.

Step 3. Simplify the fraction: $\dfrac{144}{40320}=\dfrac{1}{280}$. So $\displaystyle\int_0^1 x^3(1-x)^4\,dx = \dfrac{1}{280}$ — an integer-power Beta integral that would otherwise need four rounds of integration by parts, done here with two factorial lookups.

• Writing $\Gamma(n)=(n-1)!$ but then substituting $n$ where $n+1$ was meant — keep the shift explicit: $\Gamma(n+1)=n!$, so $\Gamma(n)=(n-1)!$ only for positive integers $n$.
• Dropping the factor of $2$ in the trigonometric form $B(m,n)=2\int_0^{\pi/2}\sin^{2m-1}\theta\cos^{2n-1}\theta\,d\theta$ — the single most common slip, halving an otherwise correct answer.
• Misreading exponents when converting $\sin^p\theta\cos^q\theta$ to Beta form: the correspondence is $m=\tfrac{p+1}{2}$, $n=\tfrac{q+1}{2}$, not $m=p$, $n=q$.
• Attacking a Beta-function value by brute-force integration of its defining integral instead of reaching for $B(m,n)=\Gamma(m)\Gamma(n)/\Gamma(m+n)$, which is almost always faster once $\Gamma$ at integers and half-integers is known.
• Forgetting $\Gamma(1)=1$ as the base case, or omitting $-1$ when converting an algebraic-power integral $\int_0^1 x^a(1-x)^b\,dx$ to $B(a+1,b+1)$.
• Assuming the Beta or Gamma definitions hold for all real parameters — both integrals require strictly positive arguments to converge; a negative or zero parameter is outside the domain covered here, not a special case to compute.

Ask the tutor: if matching a $\sin^p\theta\cos^q\theta$ or $x^a(1-x)^b$ integral to Beta form still feels shaky, ask the AI tutor for a few more conversions to practice the exponent bookkeeping.