# Establishing zeta(3) as a definite integral and its computation.

I am a 19 yr old student new to all these ideas. I made the transformation $X(z)=\sum_{n=1}^\infty z^n/n^2$. Therefore $X(1)=\pi^2/6$ as we all know (it is $\zeta(2)$). To calculate $X(1)$, I integrated $$\frac{Y(z)}{z}=\sum_{n=1}^\infty \frac{z^{n-1}}{n}$$ between 0 to 1;

We all know $Y(z)=-\log(1-z)$; so doing integral of $Y(z)/z$ from 0 to 1, I get

$$X(1)=\int_0^1\ln\frac{z}{z-1}dz$$ that is $\pi^2/6$; Now integrating $X(z)/z$ between 0 to 1, I get $\sum_{n=1}^\infty 1/n^3$;

therefore performing that integral I get $\sum1/n^3$ as $$\int_0^1 \frac{(\log x)^2 dx}{1-x};$$ which Iam unable to do.

So can anyone give an idea of how to find $$\sum_{n=1}^\infty\frac1{n^3}=\zeta(3)=\int_0^1 \frac{(\log x)^2 dx}{1-x};$$ given $$\zeta(2)=\int_0^1 \frac{\log x\ dx}{1-x}=\frac{\pi^2}{6}.$$ Please take time to read all these things and help me out by giving some suggestion.

• I'm not sure your method is going to lead anywhere, but you may like to know that the function you call $X(z)$ is usually known as the dilogarithm. See en.wikipedia.org/wiki/Dilogarithm . – Robin Chapman Jul 5 '10 at 20:13
• is there any special name to the integral 0 to z (logx)^2dx/(1-x)? – vamsi krishna Jul 5 '10 at 21:16
• It's known as the trilogarithm function, and bot tri- and dilogarithms belongs to the family of polylogarithms. You can find some basic information about them on the wiki and wolfram pages. By the way, you've already "found" your integral as $\zeta(3)$ which is a pretty famous and well studied constant, Apery's constants (see the others' comments and answers). It's conjectured to be algebraically independent with $\pi$, so it can't be a rational multiple of $\pi^3$ (this has a computational evidence). – Wadim Zudilin Jul 6 '10 at 0:27
• so is li3(1)=zeta(3)? – vamsi krishna Jul 6 '10 at 5:31
• Yes, $Li_k(1)=\zeta(k)$. – Robin Chapman Jul 6 '10 at 5:52

Dear Vamsi,

Unlike the special values $\zeta(2n)$ (for $n \geq 1$), which are known to be simple algebraic expressions in $\pi$ (in fact just rational multiples of $\pi^{2n}$), it is conjectured (but not known) that the values $\zeta(2n+1)$ are genuinely new irrationalities (and that in fact each is genuinely different from the other); more precisely, they are conjectured to be algebraically independent of one another, and of $\pi$. There is no prior, classical, name for these numbers, and in particular you should not expect to be able to evaluate your integral in terms of any numbers whose names you already know.

There is a theoretical basis for this conjecture: the kind of integrals that you are computing are callled "period integrals" (if you search, you will find a few other MO questions about periods, in this sense), and a general philosophy is that period integrals should have no more relations between them than those that are implied by elementary manipulations of integrals (of the type that you made to compute $\zeta(2)$; and here I don't mean elementary in a disparaging sense, just in the sense of standard rules for computing integrals). In fact, period integrals are manifestations of underlying geometry (which I won't get into here; all I will say is that the geometry relevant to zeta values is the geometry of "mixed Tate motives"). One can show that the geometric objects underlying the odd zeta values are independent of one another, in a suitable sense, and of the geometric object underlying $\pi$ (which is basically the circle); what is missing is a proof that the period integrals faithfully reflect the underlying geometry (so that independence in geometry implies independence of period integrals). This is one of the big conjectures in contemporary arithmetic geometry and number theory, and so your question, which is a very nice one, is touching on some very fundamental (and difficult) mathematical issues.

Good luck as you continue your studies!

Best wishes,

Matthew Emerton

• Very interesting - is there a nice expository paper for these motives? Why are the $\zeta(2n)$ values so easy compared to the $\zeta(2n+1)$ values, in these geometric terms? – Michael Burge Jul 6 '10 at 2:00
• zeta(2n) was first obtained by Euler.He expanded sinx/x into its infinite series and since sinx/x=0 has roots pi,-pi,2*pi,-2*pi,........he equated the expansion of sinx/x to (1-x/pi)(1+x/pi)(1-x/2*pi)(1+x/2*pi)........he equated the coeff of x^2n on both sides to get zeta(2n);unfortunately,no such calculations leaded to zeta(2n+1) n>=1. – vamsi krishna Jul 6 '10 at 4:33
• Has anyone been able to prove that zeta(3) is transcendental? – vamsi krishna Jul 6 '10 at 4:35
• One reason the even values are accessible is that it is unnatural to sum over the non-negative integers (say, from the perspective of Eisenstein series) and what one is really computing is the sum of 1/n^k over all non-zero integers n (say, by residue methods). This lets you compute the corresponding non-negative integer sums when k is even only, since all the terms cancel out when k is odd. – Qiaochu Yuan Jul 6 '10 at 20:29
• Qiaochu, isn't it more relevant that, by the nature of the functional function, the positive evens go over to the negative odds where the function takes (rational!) nonzero values, whereas the positives odds $> 1$ go over to the negative evens where the functions has zeros? – BCnrd Jul 10 '10 at 4:21

I'd like to mention that one may prove the irrationality of $\pi$, $\ln 2$, $\zeta(2)$, and $\zeta(3)$ in a relatively uniform way using simple integral representations.

Assume that one wants to show the irrationality of a number $\xi$ which can be presented for every $k\in\mathbb N$ in terms of the moments of some function $f$ $$a_k+b_k\xi=\int_{0}^{1}x^k f(x) dx,$$ where $a_k$, $b_k\in\mathbb Q$. If $\xi$ were rational than the equality might be rewritten as $$\frac{c_n}{d_n}=\int_{0}^{1}P_n(x)f(x)dx,\quad P_n(x)=\frac{1}{n!}\frac{d^n}{dx^n}(x^n(1-x)^n),\ n\in\mathbb N,$$ where and $c_n$, $d_n$ are integers. (The choice of the Legendre polynomials $P_n$ allows to perform integrations by parts easily.) Now, if we can show that

$$d_n\left|\int_{0}^{1}\frac{1}{n!}x^n(1-x)^n\frac{d^n}{dx^n}f(x)dx\right|\to 0,$$ this would imply that $c_n\to 0$ which is impossible; so $\xi$ cannot be rational.

The difficult part, of course, is to find the suitable function $f$.

• For $\xi=\pi$ we may take $f(x)=\sin\pi x$ and use the fact that $\int_{0}^{1}x^k\sin(\pi x) dx$ is a polynomial in $\pi$ of degree $k$ divided by $\pi^k$. Assuming $\pi=a/b$ we will get that $$0<|c_n|=\left|a^n\int_{0}^1P_n(x)\sin(\pi x) dx\right|\to 0.$$

• For $\xi=\ln 2$ take $f(x)=1/(1+x)$. If $\ln 2$ were $a/b$, then $$0<|c_n|=\left|bD_n\int_{0}^{1}\frac{P_n(x)}{1+x}dx\right|\to 0$$ (where $D_n={\rm LCM}\{1,2,\dots,n\}$).

• For $\xi=\zeta(2)$ the choice is $$f(x)=\int_{0}^{1}\frac{(1-y)^n}{1-xy}dy,$$ and the assumption $\zeta(2)=a/b$ leads to $$0<|c_n|=\left|D_{n+1}^2\int_{0}^{1}P_n(x)f(x)dx\right|\to 0$$ (where $D_n={\rm LCM}\{1,2,\dots,n\}$).

• Finally, for $\xi=\zeta(3)$ take $$f(x)=\int_{0}^{1}\frac{P_n(y)}{1-xy}\ln xy\ dy.$$ If $\zeta(3)=a/b$, then $$0<|c_n|=\left|D_{n+1}^3\int_{0}^{1}P_n(x)f(x)dx\right|\to 0.$$

The irrationality proofs are contained in the book by J.M. Borwein and P.B. Borwein. There is also a nice summary in the note by D. Huylebrouck (with all four proofs occupying less than five pages).

I don't think you're going to get a nice answer, considering that we don't know very much about $\zeta(3)$. Check out the wikipedia page here.

Let $$F(x) = \displaystyle\int_0^x{-\frac{\ln(1-t)}{t}}dt$$, where $$x \in [0,1]$$. Easy to see that $$F(1)$$ yields $$\zeta(2)$$. $$F(x)$$ is continuous and derivable on $$(0,1)$$ [The integrand being continuous on $$(0,1)$$] and $$F'(x)= -\displaystyle\frac{\ln(1-x)}{x}$$. Again, $$F(x)=x+\frac{x^2}{2^2}+\frac{x^3}{3^2}+...$$.
It can be readily seen that: $$\displaystyle\int_0^x{\frac{F(x)}{x}}dx=\mathrm{Li}_3(x).$$
Integrating by parts:$$\displaystyle\int_0^x{\frac{F(x)}{x}}dx=$$ $$\displaystyle{[{F(x){\ln(x)}-\int{F'(x)\ln(x)dx}}]_0^x=[F(x)\ln(x)+\frac{\ln(1-x)\ln(x)}{x}]_0^x}$$
$$F(0)=0$$ [follows from the continuity of integral functions] and $$\lim_{x \to 0}\ln(1-x)\ln(x)=0$$.
$$\zeta(3)= \displaystyle\sum_{n=1}^\infty{\frac{1}{n^3}}=\displaystyle\int_0^1{\frac{\ln(1-x)\ln(x)}{x}dx}$$