The Hypergeometric Function and the Riemann Zeta Function

1. The hypergeometric series

1° Hypergeometric function The hypergeometric function \(F(a, b, c, z)\) is defined by the series \[ F(a, b, c, z) = 1 + \frac{ab}{1!\,c}z + \frac{a(a + 1)\,b(b + 1)}{2!\,c(c + 1)}z^2 + \frac{a(a + 1)(a + 2)\,b(b + 1)(b + 2)}{3!\,c(c + 1)(c + 2)}z^3 + \cdots \] It converges for \( |z| < 1 \), and also for \( |z| = 1 \) if \( \operatorname{Re}(c - a - b) > 0 \). For \( |z| > 1 \), \(F\) can be continued analytically.

With the Pochhammer symbol (rising factorial) \( (a)_n = a(a + 1)\cdots(a + n - 1) \), \( (a)_0 = 1 \), the series is compactly

\[ F(a, b, c, z) = \sum_{n=0}^\infty\frac{(a)_n(b)_n}{(c)_n}\,\frac{z^n}{n!}. \]

(It is often written \({}_2F_1(a, b; c; z)\): two parameters "upstairs", one "downstairs".) We need \(c \ne 0, -1, -2, \dots\), otherwise a denominator vanishes.

Why the name and the radius of convergence. The ratio of consecutive terms is a rational function of \(n\):

\[ \frac{t_{n+1}}{t_n} = \frac{(a + n)(b + n)}{(c + n)(n + 1)}\,z \ \longrightarrow\ z \quad (n \to \infty). \]

This is the defining property of a hypergeometric series (it generalizes the geometric series, whose ratio is constant: \(a = 1\), \(b = c\) gives exactly \(1 + z + z^2 + \cdots\)). By the ratio test, the radius of convergence is \(1\). If \(a\) or \(b\) is a negative integer, the series stops and \(F\) is a polynomial.

2. Special cases: one function, many faces

Parameters	Function
\( F(1, b, b, z) \)	\( \frac{1}{1 - z} = 1 + z + z^2 + \cdots \)
\( F(a, b, b, z) \)	\( (1 - z)^{-a} \) (binomial series)
\( zF(1, 1, 2, z) \)	\( -\ln(1 - z) \)
\( zF(\frac12, \frac12, \frac32, z^2) \)	\( \arcsin z \)
\( zF(\frac12, 1, \frac32, -z^2) \)	\( \arctan z \)
\( F(-n, n + 1, 1, \frac{1 - z}{2}) \)	Legendre polynomial \(P_n(z)\)

Example 1 · The logarithm For \(F(1, 1, 2, z)\): \( \frac{(1)_n(1)_n}{(2)_n\,n!} = \frac{n!\,n!}{(n + 1)!\,n!} = \frac{1}{n + 1} \), so \[ F(1, 1, 2, z) = \sum_{n\ge0}\frac{z^n}{n + 1} = \frac1z\Big(z + \frac{z^2}{2} + \frac{z^3}{3} + \cdots\Big) = -\frac{\ln(1 - z)}{z}. \] At \(z = 0.4\) both sides equal \(1.27706\).

The value at \(z = 1\) has a closed form, Gauss's summation theorem (for \( \operatorname{Re}(c - a - b) > 0 \)):

\[ F(a, b, c, 1) = \frac{\Gamma(c)\,\Gamma(c - a - b)}{\Gamma(c - a)\,\Gamma(c - b)}. \]

3. Euler's integral representation

2° 1) Integral representation For \( |z| < 1 \) and \( \operatorname{Re}c > \operatorname{Re}b > 0 \): \[ F(a, b, c, z) = \frac{\Gamma(c)}{\Gamma(b)\,\Gamma(c - b)}\int_0^1 t^{b-1}(1 - t)^{c-b-1}(1 - tz)^{-a}\,dt. \]

Two misprints in the book

The upper limit of the integral must be \(1\), not \(\infty\): the factor \((1 - t)^{c-b-1}\) only makes sense (for general exponents) on \(0 \le t \le 1\), and the derivation below integrates over \([0, 1]\).
The condition reads \( \operatorname{Re}c > \operatorname{Re}b > 0 \) (the book has "\(< 0\)"). It is exactly what makes the integral converge at \(t = 0\) (\(\operatorname{Re}b > 0\)) and at \(t = 1\) (\(\operatorname{Re}(c - b) > 0\)).

Why. Expand \( (1 - tz)^{-a} = \sum_n\frac{(a)_n}{n!}t^nz^n \) and integrate term by term. Each term gives a Beta integral:

\[ \int_0^1 t^{b + n - 1}(1 - t)^{c - b - 1}\,dt = \mathrm B(b + n, c - b) = \frac{\Gamma(b + n)\Gamma(c - b)}{\Gamma(c + n)}. \]

Since \( \frac{\Gamma(b + n)}{\Gamma(b)} = (b)_n \) and \( \frac{\Gamma(c + n)}{\Gamma(c)} = (c)_n \), the prefactor turns this into \( \frac{(b)_n}{(c)_n} \), and the series of Section 1 appears. (Numerical check at \(a = 0.3\), \(b = 0.5\), \(c = 1.7\), \(z = 0.4\): both sides equal \(1.04181\).)

The integral is useful for more than computing: its right-hand side makes sense for all \(z\) off the ray \([1, \infty)\), so it provides the analytic continuation of \(F\) to \(|z| > 1\) mentioned in 1°. Putting \(z = 1\) and using the Beta function gives Gauss's summation theorem directly.

4. Gauss's differential equation

2° 2) Hypergeometric equation \( y = F(a, b, c, z) \) satisfies Gauss's differential equation \[ z(1 - z)\,y'' + \big[c - (a + b + 1)z\big]\,y' - ab\,y = 0. \]

Why: substitute \( y = \sum t_nz^n \) and collect the coefficient of \(z^n\). The equation requires

\[ (n + 1)(n + c)\,t_{n+1} = (n + a)(n + b)\,t_n, \]

which is exactly the ratio of consecutive coefficients computed in Section 1. In other words, the differential equation and the series encode the same recurrence.

The equation is a linear second-order ODE with three singular points, \(z = 0, 1, \infty\). Every such equation (with "regular" singular points) can be transformed into this one, which is why so many functions of physics (Legendre, Chebyshev, Jacobi polynomials, elliptic integrals) are hypergeometric. A second, independent solution near \(0\) is \( z^{1-c}F(a - c + 1, b - c + 1, 2 - c, z) \) (for \(c\) not an integer).

5. The zeta function

3° Zeta function For \( \operatorname{Re}z > 1 \), the Riemann zeta function is defined by \[ \zeta(z) = \sum_{n=1}^\infty\frac{1}{n^z} = 1 + \frac{1}{2^z} + \frac{1}{3^z} + \cdots \] \(\zeta(z)\) can be continued analytically to all other values of \(z\) (except \(z = 1\)).

Here \( n^z = e^{z\ln n} \) and \( |n^{-z}| = n^{-\operatorname{Re}z} \), so the series converges absolutely for \( \operatorname{Re}z > 1 \) by comparison with the real \(p\)-series. At \(z = 1\) it is the harmonic series, which diverges: \(\zeta\) has a simple pole at \(z = 1\) with residue \(1\). The continuation to the rest of the plane is a single-valued meromorphic function whose only pole is \(z = 1\).

(The book writes the sum with index \(k\) but the term \(\frac{1}{n^z}\); the index is the same letter throughout.)

Figure 1. \(\zeta(x)\) for real \(x\). For \(x > 1\) it decreases from \(+\infty\) towards \(1\), passing \(\zeta(2) = \frac{\pi^2}{6} \approx 1.645\). There is a pole at \(x = 1\). To the left, after analytic continuation, \(\zeta(0) = -\frac12\), \(\zeta(-1) = -\frac{1}{12}\), and the "trivial zeros" lie at \(-2, -4, -6, \dots\) (red dots).

6. Properties of the zeta function

The functional equation

4° 1) Riemann's functional equation \[ \zeta(1 - z) = 2^{1-z}\,\pi^{-z}\cos\frac{\pi z}{2}\,\Gamma(z)\,\zeta(z). \]

The factor \(\pi^{-z}\) is missing in the book The book prints \( \zeta(1 - z) = 2^{1-z}\Gamma(z)\cos\frac{\pi z}{2}\,\zeta(z) \). Without \(\pi^{-z}\) the identity is false: e.g. at \(z = 2\), the correct formula gives \( \zeta(-1) = 2^{-1}\pi^{-2}(-1)\cdot1\cdot\frac{\pi^2}{6} = -\frac{1}{12} \), while the book's version would give \(-\frac{\pi^2}{12}\). (A numerical check at \(z = 0.3 + 0.7i\) confirms the corrected form.)

The functional equation relates values at \(z\) and \(1 - z\), symmetric about the line \( \operatorname{Re}z = \frac12 \). It is what carries \(\zeta\) from the half-plane \( \operatorname{Re}z > 1 \), where the series converges, to \( \operatorname{Re}z < 0 \). Consequences:

For \(z = 2k + 1\) (\(k \ge 1\)), \( \cos\frac{\pi z}{2} = 0 \), so \( \zeta(-2k) = 0 \): the trivial zeros \(-2, -4, -6, \dots\)
\( \zeta(0) = -\frac12 \) and \( \zeta(-1) = -\frac{1}{12} \). The latter is the source of the famous "\( 1 + 2 + 3 + \cdots = -\frac{1}{12} \)": not a sum in the ordinary sense, but the value of the analytically continued function at the point where the series would be \(\sum n\).

An integral representation

4° 2) Integral representation For \( \operatorname{Re}z > 1 \): \[ \zeta(z) = \frac{1}{\Gamma(z)}\int_0^\infty\frac{t^{z-1}}{e^t - 1}\,dt. \]

Why: expand \( \frac{1}{e^t - 1} = \sum_{n\ge1}e^{-nt} \) and use \( \int_0^\infty t^{z-1}e^{-nt}\,dt = \frac{\Gamma(z)}{n^z} \) (substitute \(nt = s\) in Euler's integral). Summing over \(n\) gives \( \Gamma(z)\zeta(z) \).

A sign misprint in the book The book has \(e^t + 1\) in the denominator (with \(\operatorname{Re}z > 0\)). That integral gives a different function: with \( \frac{1}{e^t + 1} = \sum_{n\ge1}(-1)^{n-1}e^{-nt} \), one gets the alternating series \[ \frac{1}{\Gamma(z)}\int_0^\infty\frac{t^{z-1}}{e^t + 1}\,dt = \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^z} = \big(1 - 2^{1-z}\big)\zeta(z) \qquad (\operatorname{Re}z > 0). \] So the book's formula is correct only with the factor \(1 - 2^{1-z}\): \( \zeta(z) = \frac{1}{(1 - 2^{1-z})\Gamma(z)}\int_0^\infty\frac{t^{z-1}}{e^t + 1}dt \). (At \(z = 2.5\): \(\zeta = 1.3415\), while the book's integral gives \(0.8672\).) This version is actually useful: it is valid for \(\operatorname{Re}z > 0\), \(z \ne 1\), so it continues \(\zeta\) into the critical strip.

The integral with \(e^t - 1\) appears in physics as Planck's law: the total energy radiated by a black body involves \( \int_0^\infty\frac{t^3}{e^t - 1}dt = \Gamma(4)\zeta(4) = 6\cdot\frac{\pi^4}{90} = \frac{\pi^4}{15} \).

Values at even integers

4° 3) Euler's formula \[ \zeta(2k) = \frac{2^{2k-1}\pi^{2k}B_k}{(2k)!} \qquad (k = 1, 2, 3, \dots), \] where \(B_k\) are the Bernoulli numbers in the convention \( B_1 = \frac16,\ B_2 = \frac1{30},\ B_3 = \frac1{42},\ B_4 = \frac1{30},\ \dots \)

Which Bernoulli numbers? The book uses the older convention, in which \(B_k\) are positive and indexed \(1, 2, 3, \dots\). In most modern texts the same numbers are \( |B_{2k}| \), with \( B_2 = \frac16 \), \( B_4 = -\frac1{30} \), \( B_6 = \frac1{42} \); then the formula reads \( \zeta(2k) = \frac{(-1)^{k+1}B_{2k}(2\pi)^{2k}}{2\,(2k)!} \). Both say the same thing.

Example 2 · Euler's famous sums \[ \zeta(2) = \frac{2\pi^2\cdot\frac16}{2!} = \frac{\pi^2}{6}, \qquad \zeta(4) = \frac{8\pi^4\cdot\frac1{30}}{4!} = \frac{\pi^4}{90}, \qquad \zeta(6) = \frac{32\pi^6\cdot\frac1{42}}{6!} = \frac{\pi^6}{945}. \] The first is the Basel problem, solved by Euler in 1734: \( 1 + \frac14 + \frac19 + \cdots = \frac{\pi^2}{6} \). It also follows from the Fourier series of \(x^2\) (see the post on Fourier series). For odd arguments no such formula is known; \(\zeta(3) \approx 1.202\) was proved irrational only in 1978.

Euler's product over the primes

4° 4) Euler product For \( \operatorname{Re}z > 1 \): \[ \frac{1}{\zeta(z)} = \prod_p\Big(1 - \frac{1}{p^z}\Big), \qquad\text{i.e.}\qquad \zeta(z) = \prod_p\frac{1}{1 - p^{-z}}, \] where the product is taken over all prime numbers \(p = 2, 3, 5, 7, 11, \dots\)

Why: each factor is a geometric series, \( \frac{1}{1 - p^{-z}} = 1 + \frac{1}{p^z} + \frac{1}{p^{2z}} + \cdots \). Multiplying these series for all primes produces every product \( \frac{1}{(p_1^{k_1}p_2^{k_2}\cdots)^z} \), and by the fundamental theorem of arithmetic every natural number \(n\) arises exactly once as such a prime factorization. So the product equals \( \sum\frac{1}{n^z} \). The Euler product is the bridge between analysis and number theory.

Example 3 · Infinitely many primes, analytically If there were only finitely many primes, the product \( \prod_p\frac{1}{1 - p^{-1}} \) would be a finite number. But as \(z \to 1^+\), \(\zeta(z) \to \infty\) (harmonic series). So there must be infinitely many primes. (A further consequence: the probability that two random integers are coprime is \( \prod_p(1 - p^{-2}) = \frac{1}{\zeta(2)} = \frac{6}{\pi^2} \approx 0.61 \).)

7. Zeros and the Riemann hypothesis

The Euler product shows that \(\zeta(z) \ne 0\) for \( \operatorname{Re}z > 1 \) (a convergent product of nonzero factors), and the functional equation then gives no zeros for \( \operatorname{Re}z < 0 \) except the trivial ones. All other zeros, the nontrivial zeros, lie in the critical strip \( 0 \le \operatorname{Re}z \le 1 \). The first one is at \( \frac12 + 14.1347\,i \).

The Riemann hypothesis (1859, still open) All nontrivial zeros of \(\zeta\) lie on the critical line \( \operatorname{Re}z = \frac12 \).

Billions of zeros have been computed, all on the line, but no proof is known. Through the Euler product, the location of the zeros controls how regularly the prime numbers are distributed; the hypothesis is equivalent to the best possible error term in the prime number theorem. It is one of the Clay Millennium Prize Problems.

8. Summary

Formula	Meaning
\( F(a, b, c, z) = \sum\frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!} \)	hypergeometric series, \(\|z\| < 1\)
\( F = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c - b)}\int_0^1 t^{b-1}(1 - t)^{c-b-1}(1 - tz)^{-a}dt \)	Euler's integral, \(\operatorname{Re}c > \operatorname{Re}b > 0\)
\( z(1 - z)y'' + [c - (a + b + 1)z]y' - aby = 0 \)	Gauss's equation
\( \zeta(z) = \sum n^{-z} \), \(\operatorname{Re}z > 1\)	zeta function; pole at \(1\)
\( \zeta(1 - z) = 2^{1-z}\pi^{-z}\cos\frac{\pi z}{2}\Gamma(z)\zeta(z) \)	functional equation
\( \zeta(z) = \frac{1}{\Gamma(z)}\int_0^\infty\frac{t^{z-1}}{e^t - 1}dt \)	integral, \(\operatorname{Re}z > 1\)
\( \zeta(2k) = \frac{2^{2k-1}\pi^{2k}B_k}{(2k)!} \)	Euler's even values
\( \frac{1}{\zeta(z)} = \prod_p(1 - p^{-z}) \)	Euler product, \(\operatorname{Re}z > 1\)

9. Practice problems

Try each one before opening the solution. More solved problems from this topic are in Series.

Problem 1: Show that \( F(a, b, b, z) = (1 - z)^{-a} \).

The factors \((b)_n\) cancel: \( F = \sum\frac{(a)_n}{n!}z^n \), which is the binomial series of \( (1 - z)^{-a} \), since \( \binom{-a}{n}(-1)^n = \frac{(a)_n}{n!} \).

Problem 2: Check that \( zF(\frac12, \frac12, \frac32, z^2) = \arcsin z \) through the term \(z^3\).

The \(n = 1\) term: \( \frac{\frac12\cdot\frac12}{\frac32}z^2 = \frac{z^2}{6} \), so \( zF = z + \frac{z^3}{6} + \cdots \), which matches \( \arcsin z = z + \frac{z^3}{6} + \frac{3z^5}{40} + \cdots \).

Problem 3: Use Gauss's summation theorem to evaluate \( F(a, b, c, 1) \) for \( a = -n \) (a polynomial), \(b = 1\), \(c = 2\).

\( \frac{\Gamma(2)\Gamma(1 + n)}{\Gamma(2 + n)\Gamma(1)} = \frac{n!}{(n + 1)!} = \frac{1}{n + 1} \).

Problem 4: Compute \( 1 + \frac{1}{16} + \frac{1}{81} + \cdots \) and \( 1 - \frac14 + \frac19 - \frac1{16} + \cdots \).

\( \zeta(4) = \frac{\pi^4}{90} \). The alternating sum is \( (1 - 2^{1-2})\zeta(2) = \frac12\cdot\frac{\pi^2}{6} = \frac{\pi^2}{12} \).

Problem 5: Compute \( \int_0^\infty\frac{x}{e^x - 1}\,dx \).

\( \Gamma(2)\zeta(2) = \frac{\pi^2}{6} \).

Problem 6: Using the functional equation, show \(\zeta(-3) = \frac{1}{120}\).

\(z = 4\): \( \zeta(-3) = 2^{-3}\pi^{-4}\cos(2\pi)\,\Gamma(4)\,\zeta(4) = \frac18\cdot\frac{1}{\pi^4}\cdot 6\cdot\frac{\pi^4}{90} = \frac{6}{720} = \frac{1}{120} \).

Final thoughts

The hypergeometric function is the most "generic" special function: a series whose coefficient ratio is a rational function of \(n\), equivalently the solution of Gauss's equation with three singular points. With the right parameters it becomes almost every elementary and classical function, and its integral representation, built from the Beta function, continues it beyond the unit disk. The zeta function starts as a simple series but leads very far: a functional equation linking \(z\) and \(1 - z\), exact values at even integers involving \(\pi\) and Bernoulli numbers, and the Euler product, which ties it to the prime numbers and to the Riemann hypothesis.