MATHEMATICS / INTEGRAL LABORATORY
Gamma & Beta
Download the PDF ↗Two functions that turn families of difficult integrals into a change of variables.
01 / SEE THE AREA
A function defined by an integral
02 / UNDERSTAND THE TWO FUNCTIONS
A factorial extension. An endpoint pairing.
Assume real parameters throughout. You need basic substitution, integration by parts, and improper integrals. The central task is evaluating definite integrals; an elementary antiderivative is not required.
Gamma: powers with exponential decay
Near zero, the integrand behaves like \(t^{s-1}\), which is integrable exactly when \(s>0\). At infinity, exponential decay beats every fixed power.
The shift matters: \(\Gamma(4)=3!=6\). Also, \(\Gamma(1/2)=\sqrt\pi\), so recurrence evaluates every positive half-integer.
Beta: powers at both ends of an interval
The first parameter controls the endpoint at zero; the second controls the endpoint at one. Replacing \(t\) by \(1-t\) proves \(B(a,b)=B(b,a)\).
At positive integers, \(B(m,n)=(m-1)!(n-1)!/(m+n-1)!\). Beta is the unnormalized area: dividing its integrand by \(B(a,b)\) gives a probability density of area one.
Why does Gamma extend factorials?
Integrate by parts with \(u=t^s\) and \(dv=e^{-t}dt\). For \(s>0\), the boundary term vanishes at both ends.
Since \(\Gamma(1)=\int_0^\infty e^{-t}dt=1\), induction gives \(\Gamma(n+1)=n!\). The integral supplies values between those integers. Recurrence alone does not uniquely specify an interpolation.
Where does Γ(1/2) = √π come from?
Set \(t=u^2\) to obtain \(\Gamma(1/2)=2J\), where \(J=\int_0^\infty e^{-u^2}du\). Square \(J\), then use polar coordinates in the first quadrant:
The integrand is nonnegative, so the product and change of variables are justified. As \(J>0\), \(J=\sqrt\pi/2\), giving the result.
Why is Beta a ratio of Gamma functions?
Start with the product of the two Gamma integrals. Because the integrand is nonnegative, Tonelli’s theorem allows a double integral. Use \(r=u+v\) and \(t=u/(u+v)\), so \(u=rt\), \(v=r(1-t)\), and the absolute Jacobian is \(r\).
The first bracket is \(\Gamma(a+b)\); the second is \(B(a,b)\). Divide by the positive number \(\Gamma(a+b)\). The coordinates separate total size \(r\) from fraction \(t\).
03 / CHANGE VARIABLES
Build an integral. See the match.
Choose a family and adjust its parameters. The substitution, convergence conditions, and value update together. This calculator handles the four displayed families.
Decimal values are approximations. The formulas state the exact identities.
The substitution
The match
04 / FOLLOW THE METHOD
Ten worked integrals
Select an example, then move through the substitution, matching, and evaluation. Each example includes an endpoint check.
05 / RECOGNIZE THE SHAPE
The integral field guide
First inspect the bounds and the troublesome expression. Choose the substitution that turns it into an exponential, \(1-t\), or \(1+t\). Then match exponents and simplify.
| Integral | Substitution and result | Conditions |
|---|---|---|
| \(\int_0^\infty x^{p-1}e^{-cx^r}dx\) | \(t=cx^r\) \(\Gamma(p/r)/(r c^{p/r})\) | \(p,c,r>0\) |
| \(\int_0^L x^{a-1}(L-x)^{b-1}dx\) | \(x=Lt\) \(L^{a+b-1}B(a,b)\) | \(a,b,L>0\) |
| \(\int_0^1x^{p-1}(1-x^r)^{q-1}dx\) | \(t=x^r\) \(B(p/r,q)/r\) | \(p,q,r>0\) |
| \(\int_0^\infty\frac{x^{p-1}}{(1+cx^r)^q}dx\) | \(t=cx^r\) \(B(p/r,q-p/r)/(r c^{p/r})\) | \(c,r>0\), \(0 |
| \(\int_0^{\pi/2}\sin^m\theta\cos^n\theta d\theta\) | \(t=\sin^2\theta\) \(\tfrac12 B((m+1)/2,(n+1)/2)\) | \(m,n>-1\) |
| \(\int_0^1x^{p-1}(-\ln x)^{q-1}dx\) | \(t=-\ln x\) \(\Gamma(q)/p^q\) | \(p,q>0\) |
| \(\int_0^\infty\frac{x^{p-1}}{1+x^r}dx\) | Beta, then reflection \(\pi/[r\sin(\pi p/r)]\) | \(r>0\), \(0 |
Derived here by substitution from DLMF §5.9.1§5.12.1–3§5.5.3
Why does Beta also describe an infinite interval?
In \(B(a,b)\), substitute \(t=u/(1+u)\). Then \(1-t=1/(1+u)\), \(dt=du/(1+u)^2\), and \([0,1)\) becomes \([0,\infty)\).
In the rational family above, zero requires \(p>0\). At infinity the integrand behaves like \(x^{p-1-rq}\), so \(p<rq\). This is exactly the condition that both Beta arguments are positive.
Common errors to catch before the last line
Off by one: the power \(t^k\) corresponds to \(\Gamma(k+1)\). Missing scale: transform \(dx\) too. Wrong Beta argument: in the rational form, the denominator exponent is \(a+b\), not \(b\). Wrong range: the trigonometric identity here is for \([0,\pi/2]\). Divergence: a formal Gamma ratio cannot replace an endpoint test.
06 / TRY IT YOURSELF
Eight practice integrals
For each integral, check convergence, identify the family, and write the substitution before opening the solution. One integral diverges.
Sources & scope
The definitions and standard identities were checked against the NIST Digital Library of Mathematical Functions. Worked values and calculator cases were also checked numerically; numerical checks support the calculations but do not replace the convergence arguments.
§5.2: Gamma definition and digamma§5.4: special values§5.5: recurrence and reflection§5.9: integral representations§5.12: Beta integralsThe plots show integrands, not the graph of Γ(s) or a normalized probability density. Cropped endpoints and tails are disclosed beneath each plot. Displayed values use Gamma identities evaluated in logarithmic form. All examples concern ordinary real improper integrals.