Spence's function

[Tolaso J Kos]

Let \( \displaystyle f(x)=\sum_{n=1}^{\infty} \frac{x^n}{n^2} , \; x \in (0 , 1) \). Prove that:

$$f(x)+f(1-x)=\dfrac{\pi^2}{6}-\ln x \ln (1-x) $$

[Papapetros Vaggelis]

1st part

It's known that \(\displaystyle{\ln\,(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\,\frac{x^{n}}{n}\,\,,-1<x\leq 1}\) .

2nd part

\(\displaystyle{\left(\forall\,x\in\left(0,1\right)\right)\,\left(\forall\,n\in\mathbb{N}\right): \left|\dfrac{x^{n}}{n^2}\right|=\dfrac{\left|x\right|^{n}}{n^2}<\dfrac{1}{n^2}}\) .

So, \(\displaystyle{\forall\,x\in\left(0,1\right): \dfrac{x^{n}}{n^2}\longrightarrow 0}\) and the series converges uniformly. Also, the functions

\(\displaystyle{f_{n}:\left(0,1\right)\longrightarrow \mathbb{R}\,\,,f_{n}(x)=\dfrac{x^{n}}{n^2}\,\,,n\in\mathbb{N}}\) are continuous, and

infinitely many times differentiable at \(\displaystyle{\left(0,1\right)}\) . Then ,

\(\displaystyle{f^\prime(x)=\sum_{n=1}^{\infty}\dfrac{n\,x^{n-1}}{n^2}=\sum_{n=1}^{\infty}\dfrac{x^{n-1}}{n}\,,0<x<1}\)

and cause \(\displaystyle{f(1-x)=\sum_{n=1}^{\infty}\dfrac{\left(1-x\right)^{n}}{n^2}\,,0<x<1}\), we have that :

\(\displaystyle{\left(f(1-x)\right)'=-f^\prime(1-x)=-\sum_{n=1}^{\infty}\dfrac{(1-x)^{n-1}}{n}\,\,,0<x<1}\)

so:

\(\displaystyle{f^\prime(x)+\left(f(1-x)\right)'=\sum_{n=1}^{\infty}\dfrac{x^{n-1}}{n}-\sum_{n=1}^{\infty}\dfrac{\left(1-x\right)^{n-1}}{n}\,,0<x<1}\) .

According to the 1st part and the fact that

\(\displaystyle{0<x<1\implies -1<x-1<0\,,-1<x<0}\), we get :

\(\displaystyle{\begin{aligned}\left(\dfrac{\pi^2}{6}-\ln\,x\,\left(1-x\right)\right)'&=-\dfrac{1}{x}\,\ln\,(1-x)-\ln\,x\,\dfrac{-1}{1-x}\\&=-\dfrac{1}{x}\,\sum_{n=1}^{\infty}(-1)^{n+1}\,\dfrac{(-x)^{n}}{n}+\ln\,(1+(x-1))\,\dfrac{1}{1-x}\\&=\sum_{n=1}^{\infty}\dfrac{x^{n-1}}{n}+\dfrac{1}{1-x}\,\sum_{n=1}^{\infty}(-1)^{n+1}\,\dfrac{\left(x-1\right)^{n}}{n}\\&=\sum_{n=1}^{\infty}\dfrac{x^{n-1}}{n}-\sum_{n=1}^{\infty}\dfrac{\left(1-x\right)^{n-1}}{n}\end{aligned}}\)

and then :

\(\displaystyle{\left(f(x)+f(1-x)\right)'=\left(\frac{\pi^2}{6}-\ln\,x\,\ln\,(1-x)\right)'}\)

\(\displaystyle{\implies \left(\exists\,c\in\mathbb{R}\right)\,\left(\forall\,x\in\left(0,1\right)\right): f(x)+f(1-x)=\dfrac{\pi^2}{6}-\ln\,x\,\ln\,(1-x)+c}\) .

For \(\displaystyle{x=\dfrac{1}{2}}\) we get: \(\displaystyle{c=2\,f\,\left(\dfrac{1}{2}\right)-\dfrac{\pi^2}{6}+\ln^2\,2\stackrel{(\ast)}{=}0}\)

and finally, \(\displaystyle{f(x)+f(1-x)=\dfrac{\pi^2}{6}-\ln\,x\,\ln\,\left(1-x\right)\,,0<x<1}\) .

\(\displaystyle{(\ast)}\) : Left as an exercise .

[Tolaso J Kos]

Hello Vaggelis.. for the \( (*) \) you can do the following by exploiting the identity:

$$\int_{0}^{1} x^n\log x\,dx = -\frac{1}{(n+1)^2}$$

Then by multiplying the previous equation by \( \dfrac{1}{2^{n+1}} \) and summing over \( n \) we get:

$$f\left(1/2 \right) =-\frac{1}{2} \int_0^1 \frac{\ln x }{1-x/2} \, dx $$

Can you take from here?