The Gamma Function: What Is the Factorial of One Half?

The factorial is one of the first genuinely powerful operations a student meets. Write $5!$ and you mean $5 \times 4 \times 3 \times 2 \times 1 = 120$. It counts the number of ways to arrange five objects in a row, and it shows up everywhere combinations and permutations appear.

But the definition has a built-in limitation. To compute $5!$ you multiply together the whole numbers from 1 to 5. The recipe only makes sense when you start with a whole number. So what could $\left(\tfrac{1}{2}\right)!$ possibly mean? There is no list of whole numbers running from 1 up to one half. The question looks like nonsense.

It is not nonsense. There is a single, natural function that agrees with the factorial at every whole number and also produces a sensible, unique value for every fraction, every irrational, and even every complex number. It is called the gamma function, written $\Gamma$, and the answer it gives is

$\left(\tfrac{1}{2}\right)! = \Gamma\!\left(\tfrac{3}{2}\right) = \frac{\sqrt{\pi}}{2} \approx 0.8862.$

The appearance of $\pi$ — the circle constant — in a question about factorials is one of the small miracles of analysis. This article is about where the gamma function comes from, why it looks the way it does, and why $\pi$ shows up.

Connecting the dots

Start with a picture. Plot the factorials as points: $0! = 1$, $1! = 1$, $2! = 2$, $3! = 6$, $4! = 24$. They rise steeply. The obvious question is whether there is a smooth curve passing through all of them — a curve we could read off at $x = \tfrac{1}{2}$ or $x = \pi$.

The trouble is that infinitely many smooth curves pass through any given set of points. Drawing a wiggly line through five dots proves nothing. What we want is not just a curve but the curve — one singled out by some natural property, so that calling it “the factorial of $\tfrac{1}{2}$” actually means something.

This is the problem Leonhard Euler solved in 1729, at the age of 22, in a letter to Christian Goldbach. He found a formula that reproduces the factorials and extends them in the most natural possible way.

Euler’s integral

Euler’s key idea was to express the factorial not as a product but as an integral — an area under a curve. The modern form of his answer is

$\Gamma(x) = \int_0^\infty t^{\,x-1}\,e^{-t}\,dt.$

This looks intimidating, but every piece earns its place. The integration variable $t$ runs from 0 to infinity. The factor $e^{-t}$ decays rapidly, guaranteeing the area is finite. The factor $t^{x-1}$ is where $x$ enters — and crucially, $x$ here can be any number, whole or not, because raising $t$ to a fractional power is perfectly well defined.

To see that this really is the factorial in disguise, evaluate it at a whole number and use integration by parts. The calculation produces the single most important property of the gamma function, its recurrence relation:

$\Gamma(x+1) = x \cdot \Gamma(x).$

This is exactly the rule the factorial obeys: $5! = 5 \times 4!$. Combined with the starting value $\Gamma(1) = \int_0^\infty e^{-t}\,dt = 1$, the recurrence forces

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

and in general $\Gamma(n) = (n-1)!$. The function lands precisely on the factorials. (The off-by-one shift — $\Gamma(n) = (n-1)!$ rather than $n!$ — is a historical quirk of notation, nothing deeper.)

Why this curve and no other

Hitting the factorial values is necessary but not sufficient; many curves do that. What makes the gamma function the answer is a theorem proved by Harald Bohr and Johannes Mollerup in 1922.

A function is called log-convex if its logarithm curves upward — informally, it grows in a smoothly accelerating way with no dips or wiggles. The factorial values grow exactly like this. The Bohr–Mollerup theorem says:

The gamma function is the only function on the positive reals that satisfies $\Gamma(1) = 1$, obeys the recurrence $\Gamma(x+1) = x\,\Gamma(x)$, and is log-convex.

That is the precise sense in which $\Gamma$ is the natural extension of the factorial. Any other interpolating curve must either break the multiplication rule or wiggle in a way the factorial never does. Once you demand smooth, accelerating growth, the curve is pinned down completely — including its value at $\tfrac{1}{2}$.

Where the √π comes from

Now to the headline result. Evaluate Euler’s integral at $x = \tfrac{1}{2}$:

$\Gamma\!\left(\tfrac{1}{2}\right) = \int_0^\infty t^{-1/2}\,e^{-t}\,dt.$

Substitute $t = u^2$, so $dt = 2u\,du$ and $t^{-1/2} = 1/u$. The integral becomes

$\Gamma\!\left(\tfrac{1}{2}\right) = \int_0^\infty \frac{1}{u}\,e^{-u^2}\,(2u\,du) = 2\int_0^\infty e^{-u^2}\,du.$

That remaining integral is the famous Gaussian integral, the area under the bell curve. Its value is one of the most celebrated results in analysis:

$\int_0^\infty e^{-u^2}\,du = \frac{\sqrt{\pi}}{2}.$

Therefore $\Gamma\!\left(\tfrac{1}{2}\right) = \sqrt{\pi}$, and using the recurrence,

$\left(\tfrac{1}{2}\right)! = \Gamma\!\left(\tfrac{3}{2}\right) = \tfrac{1}{2}\,\Gamma\!\left(\tfrac{1}{2}\right) = \frac{\sqrt{\pi}}{2} \approx 0.8862.$

So $\pi$ enters through the back door. It was never about circles directly — it arrived through the Gaussian, which is itself tied to circles through the trick of squaring the integral and switching to polar coordinates. The factorial of one half, the bell curve, and the circle constant are all the same fact wearing different clothes.

Beyond the positive numbers

The integral formula only converges when $x > 0$. But the recurrence $\Gamma(x) = \Gamma(x+1)/x$ lets us run the function backward into negative territory. For example,

$\Gamma\!\left(-\tfrac{1}{2}\right) = \frac{\Gamma\!\left(\tfrac{1}{2}\right)}{-\tfrac{1}{2}} = -2\sqrt{\pi}.$

Pushing this idea through the whole complex plane — a process called analytic continuation — extends $\Gamma$ to every complex number except the non-positive integers $0, -1, -2, \dots$. At each of those points the division by zero in the recurrence creates a pole: the function blows up to infinity. Between the poles, on the negative axis, the gamma function dips up and down in alternating signs, a shape that surprises everyone seeing it for the first time.

Why anyone should care

The gamma function is not a curiosity dreamed up to answer a trick question. It is one of the workhorses of mathematics, the prototype of what analysts call a special function.

It sets the normalization constants of the most important probability distributions: the chi-squared, the Student $t$, the gamma and beta distributions all carry $\Gamma$ in their formulas, which is why it is impossible to do statistics without it. It gives the volume of a ball in $n$ dimensions — the formula $V_n = \pi^{n/2} / \Gamma(\tfrac{n}{2}+1)$ works for any $n$, including odd and even and, formally, fractional dimensions. It is the foundation of fractional calculus, where one asks what it means to take “half a derivative.” It appears throughout physics, in everything from the quantum harmonic oscillator to string theory’s Veneziano amplitude, which was historically the spark that ignited the whole subject.

And it began with a 22-year-old trying to answer a question a curious student might ask: what is the factorial of a half? The answer, $\sqrt{\pi}/2$, links the counting of arrangements to the geometry of the circle — and the function built to find it turned out to be one of the most useful objects in all of analysis. That is the recurring pattern in mathematics: a small, almost playful question, taken seriously, opens onto an entire landscape.

Frequently asked

Why is the gamma function shifted, with Γ(n) = (n−1)! instead of n!?

Historical convention. Euler originally wrote the function in a form that matched the factorial directly, but Legendre later introduced the notation Γ(n) with the shift Γ(n) = (n−1)!. The shift makes several other formulas cleaner — for instance, it lines up nicely with the beta function and with integral transforms. Mathematicians have grumbled about it ever since, but the convention stuck.

What is the factorial of one half?

Using the gamma function, (1/2)! = Γ(3/2) = √π / 2 ≈ 0.8862. The appearance of √π comes from the Gaussian integral ∫e^(−x²)dx = √π, which is hidden inside the gamma integral when you evaluate it at half-integer points.

Does the gamma function have a value for negative numbers?

Yes, for most of them. Through analytic continuation, Γ(x) is defined for all complex numbers except the non-positive integers (0, −1, −2, …), where it has poles — it shoots off to infinity. Between the negative integers it takes finite values that alternate in sign.

Where is the gamma function actually used?

Everywhere that 'factorial-like' growth appears with non-integer arguments: the volume of a sphere in n dimensions, the chi-squared and t distributions in statistics, the normalization of the Gaussian, fractional calculus, and many integrals in physics. It is one of the most important 'special functions' in all of mathematics.

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Leonhard Euler

Swiss mathematician of extraordinary range and productivity. Formulated Euler's identity, founded graph theory, and produced more mathematical output than any mathematician before or since.

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