Recipe

Evaluate Gamma(z) using the functional equation

Use when: you need \Gamma(z) at an integer, a half-integer, or any point reached from a known value by whole-integer shifts, especially where \operatorname{Re}(z)\le0 so the defining integral doesn't converge

Goal: \Gamma(z)'s value, or confirmation that z is a pole with no finite value, built from a known starting value using only algebra, no integration

Taught in The functional equation and the reflection formula

1. 1

   Pick a value of \Gamma you already know exactly. An integer target starts from \Gamma(1)=1. A half-integer target starts from \Gamma\!\left(\tfrac12\right)=\sqrt\pi. To get that, set z=\tfrac12 in the reflection formula \Gamma(z)\Gamma(1-z)=\dfrac{\pi}{\sin(\pi z)}. Both factors on the left then become the same number.

   Example

   To reach \Gamma(5), start from \Gamma(1)=1. To reach \Gamma\!\left(\tfrac72\right), start from \Gamma\!\left(\tfrac12\right)=\sqrt\pi. That value itself comes from \Gamma\!\left(\tfrac12\right)^2=\pi/\sin(\pi/2)=\pi.

   • the target differs from an integer by a whole number — start from \Gamma(1)=1.
   • the target differs from \tfrac12 by a whole number — start from \Gamma\!\left(\tfrac12\right)=\sqrt\pi.

   Why: the functional equation only moves a value one integer step at a time. So it needs a starting value that is already pinned down before it can go anywhere.

2. 2

   Compare the target z to the known starting point. If the target is larger, you climb up, using \Gamma(z+1)=z\Gamma(z) forward. If it is smaller, you climb down, using the rearranged form \Gamma(z)=\Gamma(z+1)/z. Before you climb down, check whether the path passes through any of 0,-1,-2,\dots.

   Example

   Reaching \Gamma(5) from \Gamma(1) climbs up four steps. It crosses no nonpositive integers. Reaching \Gamma(-3) from \Gamma(1) climbs down. It passes straight through z=0 on the way.

   • target is above the known value — climb up. There is no pole risk, since 0,-1,-2,\dots never lie above a starting value of 1 or \tfrac12.
   • target is below the known value and the path crosses 0,-1,-2,\dots — the chain will demand dividing by zero at that point. Then z, and every nonpositive integer below it, is a pole with no finite value. Go to Step 4 to confirm this, not Step 3.
   • target is below the known value but the path stays above every nonpositive integer — climb down safely, using \Gamma(z)=\Gamma(z+1)/z at each step.

   Why: the rule \Gamma(z)=\Gamma(z+1)/z, read backward, is what continues \Gamma past where the original integral converges. It is undefined at exactly the points where it would divide by zero.

3. 3

   Apply the recursion one integer step at a time. Track the running value until you land exactly on the target z.

   Example

   \Gamma(2)=1\cdot\Gamma(1)=1,\ \Gamma(3)=2\cdot\Gamma(2)=2,\ \Gamma(4)=3\cdot\Gamma(3)=6,\ \Gamma(5)=4\cdot\Gamma(4)=24

   ⚡ Shortcut

   For a positive integer target n, skip the rung-by-rung climb. Compute \Gamma(n)=(n-1)! directly. Then \Gamma(5)=4!=24. This matches the four-step climb above, with no middle values written out.

   Why: each use of \Gamma(z+1)=z\Gamma(z) is one proven algebra step. So chaining several of them is still exact, not an approximation.

4. 4

   Write out the chain until it reaches \Gamma(0) divided by something. That step is undefined. So \Gamma at that point, and at the original target, is a pole. It is not the number 0, and it is not any other finite value.

   Example

   \Gamma(-3) needs \Gamma(-2)/(-3). That needs \Gamma(-1)/(-2), which needs \Gamma(0)/(-1). Then \Gamma(0) is exactly the case the recursion cannot resolve. It demands dividing \Gamma(1)=1 by 0.

   Why: this is the same division-by-zero argument from the technical explanation above. Here it is applied to one target instead of stated in general.