Is the dilogarithm Li2(x) differentiable at x = 1? Justify your answer.

Step-by-step solution

1. 1

   Check that the function is even defined at $x=1$. The series has radius of convergence $1$, and at the endpoint $x=1$ it becomes the convergent $p$-series with $p=2$:

   $\operatorname{Li}_{2}(1)=\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}\approx1.644934.$

   So the function is defined there, and by Abel's theorem it is also continuous from the left. Differentiability is a strictly stronger requirement.

2. 2

   Differentiate the series inside the disc. For $|x|<1$ term-by-term differentiation is valid:

   $\operatorname{Li}_{2}'(x)=\sum_{n=1}^{\infty}\frac{n\,x^{n-1}}{n^{2}}=\frac{1}{x}\sum_{n=1}^{\infty}\frac{x^{n}}{n}.$

   The inner sum is the Mercator series $-\ln(1-x)$, so

   $\operatorname{Li}_{2}'(x)=-\frac{\ln(1-x)}{x},\qquad |x|<1.$

   Numerical check at $x=0.99$: a central difference of the series gives $4.651687$ and $-\frac{\ln(0.01)}{0.99}=4.651687$ ✓.

3. 3

   Examine the derivative as $x\to1^{-}$. As $x$ approaches $1$ from below, $1-x\to0^{+}$ so $\ln(1-x)\to-\infty$, hence

   $-\frac{\ln(1-x)}{x}\longrightarrow+\infty.$

   The slope grows without bound: at $x=0.9$ it is $2.56$, at $x=0.99$ it is $4.65$, at $x=0.999$ it is $6.91$ — increasing like $\ln\frac{1}{1-x}$, slowly but without limit.

4. 4

   Conclude with the definition of the derivative. Since $\operatorname{Li}_{2}$ is differentiable on $(0,1)$ with derivative tending to $+\infty$, the mean value theorem forces the difference quotient $\frac{\operatorname{Li}_{2}(1)-\operatorname{Li}_{2}(x)}{1-x}$ to diverge as well. So the left derivative is $+\infty$ and no finite derivative exists:

   $\operatorname{Li}_{2}\text{ is continuous but not differentiable at }x=1.$

   Geometrically the graph has a vertical tangent there.

5. 5

   See it in the local expansion. Landen's identity $\operatorname{Li}_{2}(x)+\operatorname{Li}_{2}(1-x)=\frac{\pi^{2}}{6}-\ln x\ln(1-x)$ gives, with $y=1-x\to0^{+}$,

   $\operatorname{Li}_{2}(x)=\frac{\pi^{2}}{6}+(1-x)\ln(1-x)-(1-x)+O\!\left((1-x)^{2}\right).$

   Numerically at $y=0.001$: the expansion gives $1.63702631$ against the true $1.63702261$ ✓. The $y\ln y$ term is what does the damage — its derivative $\ln y+1$ blows up, confirming the failure of differentiability from a second direction.