The Gamma Function and Its Recurrence

Why the Gamma Function Matters

The factorial is a central quantity in probability, but its familiar definition applies only to nonnegative integers. The gamma function provides a closely related integral for every positive real argument. Its recurrence reproduces factorials at integer inputs and, together with one Gaussian integral, gives exact values at half-integers. These facts will provide useful constants in later distribution theory.

The defining integral is improper at both zero and infinity, so it is important to check that it exists before using it. We will do that first, then derive the recurrence and the two families of special values.

Definition and Convergence

To verify convergence, split the integral at \(1\). On \((0,1)\), \(e^{-t}\leq 1\), so $$ 0\leq \int_0^1 t^{x-1}e^{-t}\,dt \leq \int_0^1 t^{x-1}\,dt =\frac{1}{x}<\infty. $$ This is where the condition \(x>0\) is essential.

For the tail, choose a nonnegative integer \(m\) such that \(m\geq x-1\). For \(t\geq1\), \(t^{x-1}\leq t^m\). The function \(t^m e^{-t/2}\) is bounded on \([1,\infty)\): for \(m>0\), its derivative is \(t^{m-1}e^{-t/2}(m-t/2)\), so it decreases once \(t>2m\); the case \(m=0\) is immediate. Thus there is a finite constant \(C\) such that \(t^m e^{-t}\leq C e^{-t/2}\) for \(t\geq1\). Since \(\int_1^\infty e^{-t/2}\,dt<\infty\), the tail converges. Therefore \(\Gamma(x)\) is finite for every \(x>0\).

The integrand is positive for \(t>0\), so \(\Gamma(x)>0\). The definition does not include \(x=0\) or negative arguments: the integral near zero would not converge there.

The Recurrence Relation

To prove the identity, first integrate over a finite interval \([\varepsilon,R]\), where \(0<\varepsilon<R\). Integration by parts with \(u=t^x\) and \(dv=e^{-t}\,dt\) gives $$ \int_\varepsilon^R t^x e^{-t}\,dt =\left[-t^x e^{-t}\right]_\varepsilon^R +x\int_\varepsilon^R t^{x-1}e^{-t}\,dt. $$ We must check that the boundary term vanishes as \(\varepsilon\downarrow0\) and \(R\to\infty\). At zero, \(\varepsilon^x e^{-\varepsilon}\to0\) because \(x>0\) and \(e^{-\varepsilon}\leq1\). At infinity, choose an integer \(m>x\). The exponential series gives \(e^R\geq R^m/m!\) for \(R>0\), and hence $$ 0\leq R^x e^{-R}\leq m!R^{x-m}\longrightarrow0. $$ Both integrals in the finite-interval identity converge to their defining improper integrals, as established above for the positive arguments \(x+1\) and \(x\). Taking the limits therefore yields \(\Gamma(x+1)=x\Gamma(x)\). \(\square\)

The recurrence steps the argument upward by one and multiplies by the current argument. It is the key link between the integral definition and factorial arithmetic.

Values at Positive Integers

For \(n=1\), the formula states \(\Gamma(1)=0!=1\), already established. Suppose it holds for a positive integer \(n\). The recurrence then gives $$ \Gamma(n+1)=n\Gamma(n)=n(n-1)!=n!. $$ Thus the formula holds for \(n+1\) whenever it holds for \(n\). By induction, it is true for every positive integer \(n\). \(\square\)

This identity explains the one-unit shift that often causes errors: the gamma function at \(n\) equals the factorial of \(n-1\). Equivalently, \(k!=\Gamma(k+1)\) for every nonnegative integer \(k\).

The First Half-Integer Value

The recurrence alone cannot determine values at half-integers from integer values. We first need \(\Gamma(1/2)\), which is connected to the Gaussian integral. Substituting \(t=u^2\), so \(dt=2u\,du\), gives $$ \Gamma\left(\frac12\right) =\int_0^\infty t^{-1/2}e^{-t}\,dt =2\int_0^\infty e^{-u^2}\,du. $$ To evaluate the last integral, set \(I=\int_0^\infty e^{-u^2}\,du\). It is finite: on \([0,1]\) the integrand is at most \(1\), while for \(u\geq1\), \(e^{-u^2}\leq e^{-u}\), whose tail integral is finite. Since the integrand is nonnegative, Tonelli's theorem allows the product integral to be written as an integral over the first quadrant: $$ I^2=\int_0^\infty\int_0^\infty e^{-(u^2+v^2)}\,du\,dv. $$ Use polar coordinates \(u=r\cos\theta,\ v=r\sin\theta\) on this quadrant, where \(r\geq0\), \(0\leq\theta\leq\pi/2\), and the area element is \(r\,dr\,d\theta\). Then $$ I^2 =\int_0^{\pi/2}\int_0^\infty e^{-r^2}r\,dr\,d\theta =\frac{\pi}{2}\left[-\frac12 e^{-r^2}\right]_{0}^{\infty} =\frac{\pi}{4}. $$ Because \(I>0\), \(I=\sqrt{\pi}/2\). Substitution into the expression for \(\Gamma(1/2)\) proves the value below.

All Positive Half-Integer Values

Repeated application of the recurrence gives $$ \Gamma\left(n+\frac12\right) =\left(n-\frac12\right)\left(n-\frac32\right)\cdots\left(\frac12\right) \Gamma\left(\frac12\right), $$ with an empty product interpreted as \(1\) when \(n=0\). Each factor in the product can be written as an odd integer divided by \(2\), so $$ \left(n-\frac12\right)\left(n-\frac32\right)\cdots\left(\frac12\right) =\frac{1\cdot3\cdot5\cdots(2n-1)}{2^n}. $$ The even factors in \((2n)!\) have product \(2\cdot4\cdots(2n)=2^n n!\). Therefore $$ (2n)! =\bigl(1\cdot3\cdot5\cdots(2n-1)\bigr)2^n n!, $$ which rearranges the odd product to \((2n)!/(4^n n!)\). Multiplying by \(\Gamma(1/2)=\sqrt{\pi}\) proves the formula, including \(n=0\). \(\square\)

Common Pitfalls and Why the Values Matter

These evaluations give exact constants without requiring numerical integration. They are especially useful when expressions involving gamma functions appear in distribution theory: integer and half-integer arguments can be simplified to factorials and powers of \(\pi\). The recurrence also provides a reliable way to check whether a proposed simplification is consistent.

• Using \(n!\) instead of \((n-1)!\). The recurrence ends at \(\Gamma(1)=1\), so \(\Gamma(n)=(n-1)!\). For example, \(\Gamma(6)=120\), not \(720\).
• Applying the recurrence with the wrong factor. The identity is \(\Gamma(x+1)=x\Gamma(x)\), so the multiplier is the lower argument \(x\).
• Ignoring the domain. The defining integral requires \(x>0\); its behavior near zero cannot be omitted when deciding whether the integral exists.
• Assuming the recurrence supplies every value by itself. It relates arguments one unit apart, but the half-integer sequence needs the separate starting value \(\Gamma(1/2)=\sqrt{\pi}\).

A compact check is to substitute a computed value back into the recurrence. For example, \(\Gamma(7/2)\) must equal \((5/2)\Gamma(5/2)\). If the claimed values fail this identity, a factor or argument shift has been mishandled.