Gamma Function: Continuation & Reflection Formula

Derivation

Gamma Function: Continuation & Reflection Formula

D-188 Home PU-205 Threads chance · symmetry Depends on Analytic Continuation & the Identity Theorem, The Residue Theorem, Real Integrals by Residues & Jordan's Lemma

Statement

The Euler integral \( \Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\,dt \), convergent for \( \operatorname{Re}z>0 \), extends by analytic continuation to a meromorphic function on \( \mathbb{C} \) whose only singularities are simple poles at \( z=0,-1,-2,\dots \) with residue \( (-1)^n/n! \) at \( z=-n \). This continuation satisfies the reflection formula \[ \Gamma(z)\,\Gamma(1-z)=\frac{\pi}{\sin(\pi z)},\qquad z\notin\mathbb{Z}. \]

Why it matters

The factorial \( n! \) is defined only on the non-negative integers, yet an astonishing amount of mathematics — probability normalizations, the volume of the \( n \)-ball, the functional equation of the Riemann zeta function, dimensional regularization in quantum field theory — needs a factorial evaluated at half-integers or complex arguments. The Gamma function is the essentially unique interpolation, and the reflection formula is the identity that pins down its values away from the integers and exposes its pole–zero structure.

The reflection formula is also the cleanest example of a deep principle in this thread: a function defined by a manifestly real, positive integral secretly encodes the poles of \( 1/\sin(\pi z) \). It converts a hard question about \( \Gamma \) into an elementary one about a trigonometric function, and it is the engine behind results such as \( \Gamma(\tfrac12)=\sqrt{\pi} \) and the duplication and multiplication theorems.

Assumptions

The Euler integral converges and is holomorphic on \( \operatorname{Re}z>0 \).If dropped, there is no analytic seed to continue; the whole construction rests on \( \Gamma \) being a genuine holomorphic function on a right half-plane, differentiable under the integral sign.

The functional relation \( \Gamma(z+1)=z\,\Gamma(z) \) holds on \( \operatorname{Re}z>0 \).If dropped, we lose the recursion that propagates \( \Gamma \) leftward across each vertical strip; the poles at the non-positive integers would not be forced.

Analytic continuation is unique (identity theorem on a connected domain).If dropped, the continuation off the half-plane would not be well defined, and identities proved on \( \operatorname{Re}z\in(0,1) \) could not be extended to all non-integer \( z \) by uniqueness.

The beta integral \( B(p,q)=\int_0^1 t^{p-1}(1-t)^{q-1}dt=\dfrac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} \) is valid for \( \operatorname{Re}p,\operatorname{Re}q>0 \).If dropped, we lose the bridge that turns \( \Gamma(z)\Gamma(1-z) \) into a single real integral amenable to contour evaluation; an independent route (Weierstrass product) would be required.

Derivation

\[ \Gamma(z+1)=\int_0^\infty t^{z}e^{-t}\,dt=\Big[-t^{z}e^{-t}\Big]_0^\infty+z\int_0^\infty t^{z-1}e^{-t}\,dt \]

Integration by parts with \( u=t^z,\ dv=e^{-t}dt \); the boundary term vanishes at both ends when \( \operatorname{Re}z>0 \). A

\[ \Gamma(z+1)=z\,\Gamma(z)\quad\Longrightarrow\quad \Gamma(z)=\frac{\Gamma(z+1)}{z} \]

The surviving integral is exactly \( \Gamma(z) \); rearranging gives the recursion in the form that expresses \( \Gamma(z) \) via a value one strip to the right. A

\[ \Gamma(z)\ \text{holomorphic on }\operatorname{Re}z>-1,\ z\neq0;\qquad \operatorname{Res}_{z=0}\Gamma=\lim_{z\to0}z\,\Gamma(z)=\Gamma(1)=1 \]

The right-hand side \( \Gamma(z+1)/z \) is holomorphic wherever \( \Gamma(z+1) \) is (i.e. \( \operatorname{Re}z>-1 \)) except for a simple pole at \( z=0 \) from the denominator. This defines the continuation onto the next strip. B

\[ \Gamma(z)=\frac{\Gamma(z+n+1)}{z(z+1)\cdots(z+n)},\qquad \operatorname{Res}_{z=-n}\Gamma=\frac{(-1)^n}{n!} \]

Iterating Step 2 \( n+1 \) times continues \( \Gamma \) to \( \operatorname{Re}z>-(n+1) \); the pole at \( z=-n \) has residue \( \Gamma(1)/\prod_{k\neq n}(k-n)=(-1)^n/n! \). By the uniqueness of analytic continuation this defines a single meromorphic function on \( \mathbb{C} \). C

\[ \Gamma(z)\Gamma(1-z)=\frac{\Gamma(z)\Gamma(1-z)}{\Gamma(1)}=B(z,1-z)=\int_0^1 t^{z-1}(1-t)^{-z}\,dt \]

Beta–Gamma identity with \( p=z,\ q=1-z \) so \( p+q=1 \) and \( \Gamma(p+q)=\Gamma(1)=1 \); valid for \( 0<\operatorname{Re}z<1 \). Now everything is one real integral. B

\[ t=\frac{u}{1+u},\quad 1-t=\frac{1}{1+u},\quad dt=\frac{du}{(1+u)^2}\ \Longrightarrow\ \Gamma(z)\Gamma(1-z)=\int_0^\infty \frac{u^{z-1}}{1+u}\,du \]

Substitution mapping \( t\in(0,1) \) to \( u\in(0,\infty) \); the powers combine as \( t^{z-1}(1-t)^{-z}=u^{z-1}(1+u)^{1-z}\cdot(1+u)^{z}\cdots \) collapsing to \( u^{z-1}/(1+u) \). B

\[ I(z)=\oint_C \frac{w^{z-1}}{1+w}\,dw,\qquad w^{z-1}=e^{(z-1)\log w},\ \ 0\le\arg w<2\pi \]

Introduce a keyhole contour \( C \) about the branch cut on the positive real axis, enclosing the single pole of \( 1/(1+w) \) at \( w=-1=e^{i\pi} \). The branch of \( w^{z-1} \) is fixed by the cut. C

\[ \oint_C=\big(1-e^{2\pi i(z-1)}\big)\int_0^\infty\frac{u^{z-1}}{1+u}\,du=2\pi i\,\operatorname{Res}_{w=-1}\frac{w^{z-1}}{1+w} \]

The two straight edges of the keyhole run just above and below the cut; the phase of \( w^{z-1} \) differs by \( e^{2\pi i(z-1)} \) between them. The large and small circular arcs vanish because \( 0<\operatorname{Re}z<1 \) makes the integrand decay. Residue theorem gives the right side. C

\[ \operatorname{Res}_{w=-1}\frac{w^{z-1}}{1+w}=(-1)^{z-1}=e^{i\pi(z-1)}=-e^{i\pi z} \]

Simple pole: residue is the numerator at \( w=-1=e^{i\pi} \), so \( w^{z-1}=e^{i\pi(z-1)} \). A

\[ \big(1-e^{2\pi i(z-1)}\big)\,I_0=-2\pi i\,e^{i\pi z},\qquad I_0=\int_0^\infty\frac{u^{z-1}}{1+u}\,du \]

Substitute the residue from Step 9 into Step 8; \( e^{2\pi i(z-1)}=e^{2\pi i z} \). A

\[ I_0=\frac{-2\pi i\,e^{i\pi z}}{1-e^{2\pi i z}}=\frac{-2\pi i}{e^{-i\pi z}-e^{i\pi z}}=\frac{-2\pi i}{-2i\sin(\pi z)}=\frac{\pi}{\sin(\pi z)} \]

Divide numerator and denominator by \( e^{i\pi z} \); use \( e^{i\pi z}-e^{-i\pi z}=2i\sin(\pi z) \). B

\[ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}\quad\text{on }0<\operatorname{Re}z<1,\ \text{hence on all }z\notin\mathbb{Z}. \]

Combine Steps 6 and 11. Both sides are meromorphic; they agree on an open strip, so by the identity theorem they agree on the common domain of holomorphy, all non-integer \( z \). C

Result

\[ \Gamma(z)\,\Gamma(1-z)=\frac{\pi}{\sin(\pi z)},\qquad z\in\mathbb{C}\setminus\mathbb{Z} \]

Reading. The product of the Gamma function at a point and at its reflection through \( z=\tfrac12 \) is a pure trigonometric object. The poles of \( 1/\sin(\pi z) \) at every integer are supplied on the left by \( \Gamma(z) \) at \( z=0,-1,-2,\dots \) and by \( \Gamma(1-z) \) at \( z=1,2,3,\dots \), so each integer pole comes from exactly one factor. Because \( \pi/\sin(\pi z) \) never vanishes, neither \( \Gamma(z) \) nor \( \Gamma(1-z) \) has any zero — the Gamma function is zero-free on all of \( \mathbb{C} \).

Units check. The argument \( z \) is dimensionless (a complex number), and \( \Gamma \) returns a dimensionless number, as does \( \pi/\sin(\pi z) \). Both sides are pure numbers; the identity is dimensionally consistent by inspection. As a numerical check at \( z=\tfrac12 \): \( \Gamma(\tfrac12)^2=\pi/\sin(\pi/2)=\pi \), giving \( \Gamma(\tfrac12)=\sqrt\pi\approx1.7725 \).

Limiting cases

\( z\to\tfrac12 \): \( \Gamma(\tfrac12)^2=\pi \Rightarrow \Gamma(\tfrac12)=\sqrt\pi \), the Gaussian normalization.
\( z\to n\in\mathbb{Z} \): the right side blows up; \( \Gamma \) has a pole in exactly one factor, confirming poles at all integers with no cancellation.
\( z\to\tfrac12+iy,\ y\to\infty \): \( |\Gamma(\tfrac12+iy)|^2=\pi/\cosh(\pi y)\to0 \) exponentially, the sharp decay of \( \Gamma \) along vertical lines.
\( z\to0^+ \): \( \Gamma(z)\sim 1/z \) while \( \Gamma(1-z)\to1 \), and \( \pi/\sin(\pi z)\sim1/z \) — the leading singular behaviour matches, giving residue \( 1 \).
Real \( z\in(0,1) \): both \( \Gamma(z),\Gamma(1-z)>0 \) and \( \sin(\pi z)>0 \), so the identity is a statement between positive reals — a useful sanity constraint.

Breaks when

\( z \) is an integer. The formula is meaningless as stated: \( \sin(\pi z)=0 \) so the right side is undefined, matching the pole of the Gamma factor on the left. The identity holds only on \( \mathbb{C}\setminus\mathbb{Z} \).
Outside the strip \( 0<\operatorname{Re}z<1 \), before continuation. The keyhole evaluation (Steps 7–11) requires \( 0<\operatorname{Re}z<1 \) for the circular arcs to vanish; the extension to all non-integer \( z \) is legitimate only because both sides are already known to be meromorphic and the identity theorem applies. Skipping the continuation argument leaves the proof valid only on the strip.
Naive interchange of limit and integral / arc estimates. If \( \operatorname{Re}z\le0 \) or \( \operatorname{Re}z\ge1 \), the small or large arc contribution no longer vanishes, so the residue bookkeeping in Step 8 fails and one cannot conclude \( I_0=\pi/\sin(\pi z) \) directly.

Failure modes

Branch-cut amnesia. Treating \( w^{z-1} \) as single-valued and forgetting the \( 1-e^{2\pi i(z-1)} \) phase difference between the two edges — this collapses the whole integral to zero and loses the factor entirely.
Wrong pole phase. Writing \( (-1)^{z-1}=1 \) or \( e^{-i\pi(z-1)} \) instead of \( e^{+i\pi(z-1)} \); the branch \( 0\le\arg w<2\pi \) forces \( -1=e^{i\pi} \), not \( e^{-i\pi} \).
Residue sign at \( z=-n \). Miscounting the \( n \) sign flips in \( z(z+1)\cdots(z+n) \) and getting \( 1/n! \) instead of \( (-1)^n/n! \).
Claiming a zero of \( \Gamma \). Concluding \( \Gamma \) can vanish somewhere; the reflection formula's non-vanishing right side forbids it — a common false step when reasoning about \( 1/\Gamma \).
Using the formula at integers. Plugging \( z=1 \) to "get" \( \Gamma(1)\Gamma(0) \) and dividing by \( \sin(\pi)=0 \) without taking the residue limit.

Discussion

The construction shows two faces of the Gamma function held together by uniqueness. The Euler integral gives a concrete, computable object on a half-plane; the recursion \( \Gamma(z+1)=z\Gamma(z) \) is a functional equation that, read backwards, drags the function across each vertical strip and manufactures a simple pole every time the denominator \( z(z+1)\cdots \) hits a non-positive integer. Neither ingredient alone gives a function on all of \( \mathbb{C} \); it is the identity theorem — a continuation, if it exists, is unique — that lets us glue the strips into a single meromorphic \( \Gamma \).

The reflection formula belongs to the "symmetry" thread because it is a statement about the involution \( z\mapsto1-z \), reflection through \( z=\tfrac12 \). The right-hand side \( \pi/\sin(\pi z) \) is itself invariant under this reflection (since \( \sin(\pi(1-z))=\sin(\pi z) \)), so the identity is compatible with swapping the two factors. That the pole structure of \( 1/\sin \) — one simple pole per integer — is reproduced exactly by the poles of \( \Gamma \) is not a coincidence but the whole content: the reflection formula is the bridge between the additive periodicity of \( \sin \) and the multiplicative recursion of \( \Gamma \).

The "chance" thread enters through the Beta integral. \( B(z,1-z) \) is, up to normalization, the total mass of a Beta\( (z,1-z) \) density, and the whole calculation can be read as computing the normalizing constant of a probability distribution on \( (0,1) \). The substitution \( t=u/(1+u) \) is precisely the map to the standard \( F \)-type variable on \( (0,\infty) \); the resulting \( \int_0^\infty u^{z-1}/(1+u)\,du \) is a Mellin transform, and reflection is a shadow of the Mellin-transform pair for \( 1/(1+u) \).

At the level of complex analysis, the reflection formula plus the recursion essentially characterize \( \Gamma \). Combining reflection with the Weierstrass product \( 1/\Gamma(z)=z\,e^{\gamma z}\prod_{n\ge1}(1+z/n)e^{-z/n} \) reproduces the Euler product for \( \sin(\pi z)/\pi z=\prod(1-z^2/n^2) \); conversely, knowing that product one recovers reflection. The identity is also the seed of the Legendre duplication formula \( \Gamma(z)\Gamma(z+\tfrac12)=2^{1-2z}\sqrt\pi\,\Gamma(2z) \) and, through the functional equation \( \zeta(s)=2^s\pi^{s-1}\sin(\tfrac{\pi s}{2})\Gamma(1-s)\zeta(1-s) \), sits at the heart of analytic number theory — the very \( \sin \) and \( \Gamma \) factors there are the reflection formula in disguise.

Common misconceptions. A frequent error is to think analytic continuation "chooses" values freely — it does not; on a connected domain the continuation is forced, which is exactly why an identity proved on a thin strip propagates everywhere. Another is to imagine \( \Gamma \) has zeros "between" its poles because it oscillates in sign on the negative axis; in fact \( \Gamma \) is nowhere zero, and the sign changes happen only by passing through poles, not through zeros.

Worked examples

Example 1 — Evaluate \( \Gamma(\tfrac14)\Gamma(\tfrac34) \).

\[ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)},\qquad z=\tfrac14,\ 1-z=\tfrac34 \]

Apply reflection with the symbolic value before substituting numbers. A

\[ \Gamma(\tfrac14)\Gamma(\tfrac34)=\frac{\pi}{\sin(\pi/4)}=\frac{\pi}{\tfrac{\sqrt2}{2}}=\pi\sqrt2 \]

Insert \( \sin(\pi/4)=\sqrt2/2 \) and simplify. A

\[ \pi\sqrt2=3.14159\times1.41421\approx4.4429 \]

Numerical value (dimensionless). A

\[ \Gamma(\tfrac14)\Gamma(\tfrac34)=\pi\sqrt2\approx4.4429 \]

Reading. Neither factor is elementary in closed form, but their product is. Cross-check: \( \Gamma(\tfrac14)\approx3.6256 \), \( \Gamma(\tfrac34)\approx1.2254 \), and \( 3.6256\times1.2254\approx4.443 \). Units: dimensionless.

Example 2 — Residue of \( \Gamma \) at \( z=-3 \), and a value of \( \Gamma \) at a negative half-integer.

\[ \operatorname{Res}_{z=-n}\Gamma(z)=\frac{(-1)^n}{n!},\qquad n=3 \]

Use the residue formula derived in Step 4 with the symbol \( n \) first. A

\[ \operatorname{Res}_{z=-3}\Gamma=\frac{(-1)^3}{3!}=\frac{-1}{6}\approx-0.1667 \]

Substitute \( n=3 \). A

\[ \Gamma(-\tfrac12)\Gamma(\tfrac32)=\frac{\pi}{\sin(-\pi/2)}=-\pi,\qquad \Gamma(\tfrac32)=\tfrac12\Gamma(\tfrac12)=\tfrac{\sqrt\pi}{2} \]

Reflection at \( z=-\tfrac12 \) (so \( 1-z=\tfrac32 \)); evaluate \( \Gamma(\tfrac32) \) by the recursion. B

\[ \Gamma(-\tfrac12)=\frac{-\pi}{\Gamma(3/2)}=\frac{-\pi}{\sqrt\pi/2}=-2\sqrt\pi\approx-3.5449 \]

Solve for the unknown factor. B

\[ \operatorname{Res}_{z=-3}\Gamma=-\tfrac16,\qquad \Gamma(-\tfrac12)=-2\sqrt\pi\approx-3.5449 \]

Reading. The negative residue and negative \( \Gamma(-\tfrac12) \) reflect the alternating sign of \( \Gamma \) between consecutive poles on the negative axis. Cross-check via recursion: \( \Gamma(\tfrac12)=(-\tfrac12)\Gamma(-\tfrac12)\Rightarrow\Gamma(-\tfrac12)=-2\Gamma(\tfrac12)=-2\sqrt\pi \). Units: dimensionless.

Problems

(A) Evaluate \( \Gamma(\tfrac13)\Gamma(\tfrac23) \) in closed form and numerically.
Solution
By reflection with \( z=\tfrac13 \): \( \Gamma(\tfrac13)\Gamma(\tfrac23)=\pi/\sin(\pi/3)=\pi/(\sqrt3/2)=2\pi/\sqrt3=\tfrac{2\pi\sqrt3}{3}\approx3.6276 \). Dimensionless.
(A) Use the reflection formula to prove \( \Gamma(\tfrac12)=\sqrt\pi \).
Solution
Set \( z=\tfrac12 \): \( \Gamma(\tfrac12)\Gamma(\tfrac12)=\pi/\sin(\pi/2)=\pi/1=\pi \). Since \( \Gamma(\tfrac12)=\int_0^\infty t^{-1/2}e^{-t}dt>0 \), take the positive root: \( \Gamma(\tfrac12)=\sqrt\pi\approx1.7725 \).
(B) Find the residue of \( \Gamma(z) \) at \( z=-5 \), and state the sign of \( \Gamma \) just to the right of that pole.
Solution
\( \operatorname{Res}_{z=-5}\Gamma=(-1)^5/5!=-1/120\approx-0.00833 \). A simple pole with negative residue means \( \Gamma(z)\to-\infty \) as \( z\to-5^+ \), so \( \Gamma \) is negative immediately to the right of \( z=-5 \) (on \( (-5,-4) \)). Dimensionless.
(B) Compute \( \Gamma(-\tfrac32) \) using reflection (or the recursion) and verify consistency.
Solution
Recursion: \( \Gamma(-\tfrac12)=(-\tfrac32)\Gamma(-\tfrac32)\Rightarrow\Gamma(-\tfrac32)=\Gamma(-\tfrac12)/(-\tfrac32)=(-2\sqrt\pi)/(-\tfrac32)=\tfrac{4}{3}\sqrt\pi\approx2.3633 \). Check by reflection at \( z=-\tfrac32 \): \( \Gamma(-\tfrac32)\Gamma(\tfrac52)=\pi/\sin(-3\pi/2)=\pi/(1)=\pi \). With \( \Gamma(\tfrac52)=\tfrac32\cdot\tfrac12\sqrt\pi=\tfrac{3\sqrt\pi}{4} \): \( \Gamma(-\tfrac32)=\pi/(\tfrac{3\sqrt\pi}{4})=\tfrac{4\sqrt\pi}{3}\approx2.3633 \). Consistent. Dimensionless.
(C) Show \( |\Gamma(iy)|^2=\dfrac{\pi}{y\sinh(\pi y)} \) for real \( y\neq0 \).
Solution
For real \( y \), \( \overline{\Gamma(iy)}=\Gamma(-iy) \) (Schwarz reflection, since \( \Gamma \) is real on the real axis), so \( |\Gamma(iy)|^2=\Gamma(iy)\Gamma(-iy) \). Use the recursion to relate \( \Gamma(-iy) \) to \( \Gamma(1-iy) \): \( \Gamma(1-iy)=(-iy)\Gamma(-iy) \Rightarrow \Gamma(-iy)=\Gamma(1-iy)/(-iy) \). Then \( \Gamma(iy)\Gamma(-iy)=\dfrac{\Gamma(iy)\Gamma(1-iy)}{-iy}=\dfrac{1}{-iy}\cdot\dfrac{\pi}{\sin(\pi i y)} \) by reflection with \( z=iy \). Since \( \sin(\pi i y)=i\sinh(\pi y) \), \( \dfrac{\pi}{-iy\cdot i\sinh(\pi y)}=\dfrac{\pi}{y\sinh(\pi y)} \). Hence \( |\Gamma(iy)|^2=\dfrac{\pi}{y\sinh(\pi y)} \), which is positive and decays like \( 2\pi\,e^{-\pi|y|}/|y| \) for large \( |y| \). Dimensionless.