Existence of derivative of series at $x = 0$

Question

I want to determine if the following function is differentiable for $x \in [0,\infty)$

$$F(x) = \sum_{j=0}^\infty \frac{e^{-jx}}{j^2 +1}$$

The series is uniformly convergent on $[0,\infty$) by the M-test since $\displaystyle \left|\frac{e^{-jx}}{j^2 +1} \right| < \frac{1}{j^2}$ and the series of termwise derivatives has uniform convergence on any interval $[\alpha,\infty)$ since $\displaystyle \left|\frac{je^{-jx}}{j^2 +1}\right|< \frac{e^{-\alpha j}}{j^2}.$ This proves that $F$ is differentiable for any point $x \in (0,\infty)$.

I know that the series of termwise derivatives does not converge for $x = 0$, so the derivative $F'(0)$ may not exist, but I would like to know how to show this directly. There are examples where a derivative of a series may exist even though it can't be obtained by termwise differentiation.

Answer 1

We write the difference quotient

$$\frac{F(h)-F(0)}{h} = \frac{1}{h}\left(\sum_{j=0}^{\infty}\frac{e^{-jh}-1}{j^2+1}\right)$$

where $h>0$. Notice that $e^{-jh}-1\leq 0$ for every $j$ and therefore

$$\frac{F(h)-F(0)}{h}\leq \frac{1}{h}\left(\sum_{j=0}^{N}\frac{e^{-jh}-1}{j^2+1}\right).$$

Then

$$\limsup_{h\rightarrow0^+}\frac{F(h)-F(0)}{h}\leq \sum_{j=0}^{N}\left(\lim_{h\rightarrow 0^+}\frac{e^{-jh}-1}{h}\right)\frac{1}{j^2+1} = \sum_{j=0}^{N}\frac{-j}{j^2+1} = -\sum_{j=0}^{N}\frac{j}{j^2+1}.$$

Since the right-hand side tends to $-\infty$ as $N\rightarrow \infty$ we can conclude that the limit does not exist.

Answer 2

Alternative approach: since $\frac{1}{j^2+1}=\int_{0}^{+\infty}\sin(z) e^{-jz}\,dz$, by the dominated convergence theorem

$$ F(x)=\sum_{j\geq 0}\frac{e^{-jx}}{j^2+1} = \int_{0}^{+\infty}\sin(z)\sum_{j\geq 0}e^{-j(x+z)}\,dz =\int_{0}^{+\infty}\frac{\sin(z)}{1-e^{-(x+z)}}\,dz$$ as well as $$ \frac{d}{dx}F(x) = -\frac{1}{4}\int_{0}^{+\infty}\frac{\sin(z)}{\sinh^2\left(\frac{x+z}{2}\right)}\,dz $$ for any $x>0$. If $x=0$, the function $\frac{\sin(z)}{\sinh^2\left(\frac{x+z}{2}\right)}$ has a simple pole at the origin, from which $$ \lim_{x\to 0^+} F'(x)=-\infty $$ easily follows. $\lim_{x\to 0^+} F(x)=\sum_{j\geq 0}\frac{1}{j^2+1}=\frac{1+\pi\coth \pi}{2}$ is well-known since Euler's times.

Answer 3

Note that $$ 1-e^{-jx}\ge\frac12\min(jx,1) $$

Therefore, letting $x=1/n$, $$ \begin{align} \frac1x\sum_{j=0}^\infty\frac{1-e^{-jx}}{j^2+1} &=\frac1x\sum_{j=0}^n\frac{1-e^{-jx}}{j^2+1}+\frac1x\sum_{j=n+1}^\infty\frac{1-e^{-jx}}{j^2+1}\\ &\ge\frac12\left(\sum_{j=0}^n\frac{j}{j^2+1}+n\sum_{j=n+1}^\infty\frac1{j^2+1}\right)\\[6pt] &\sim\frac12\log(n)+C+\frac1{8n^2}-\frac{19}{240n^4}+\frac{53}{504n^4} \end{align} $$ where $C=0.45267484$ .

Thus, the derivative at $0$ is $-\infty$.