Derivative of Dirichlet Eta function @ 1

Today we will evaluate the following infinite sum which corresponds to the derivative of Dirichlet eta function @ 1

$\sum_{n=1}^{\infty}\frac{(-1)^n\ln(n)}{n}=\gamma\ln(2)-\frac{\ln^2(2)}{2}$

As a bonus we will compute the following integral

$\int_0^\infty \frac{x\ln(x)}{e^{x^2}+1}\,dx=-\frac{\ln^2(2)}{8}$

Lets first introduce a lemma:

Lemma 1:

$\sum_{n=1}^{2N}(-1)^na_n+\sum_{n=1}^{2N}a_n=2\sum_{n=1}^{N}a_{2n}$ (1)

Proof:

$\begin{aligned} \sum_{n=1}^{2N}(-1)^na_n+\sum_{n=1}^{2N}a_n&=\left(-a_1+a_2-a_3+ \cdots-a_{2N-1}+a_{2N}\right)+\left(a_1+a_2+a_3+ \cdots+a_{2N-1}+a_{2N}\right)\\ &=2a_2+2a_4+ \cdots+2a_{2N}\\ &=2\sum_{n=1}^{N}a_{2n} \qquad \blacksquare \end{aligned}$

Claim:

$\sum_{n=1}^{\infty}\frac{(-1)^n\ln(n)}{n}=\gamma\ln(2)-\frac{\ln^2(2)}{2}$ (2)

If we let $a_n=\frac{\ln(n)}{n}$ in (1) we get

$\begin{aligned} \sum_{n=1}^{2N}\frac{(-1)^n\ln(n)}{n}&=2\sum_{n=1}^{N}\frac{\ln(2n)}{2n}-\sum_{n=1}^{2N}\frac{\ln(n)}{n}\\ &=\ln(2)\sum_{n=1}^{N}\frac{1}{n}+\sum_{n=1}^{N}\frac{\ln(n)}{n}-\sum_{n=1}^{2N}\frac{\ln(n)}{n}\\ &=\ln(2)\operatorname{H}_N+\sum_{n=1}^{N}\frac{\ln(n)}{n}-\sum_{n=1}^{2N}\frac{\ln(n)}{n}\\ \end{aligned}$ (3)

Lets now recall the Euler Maclaurin Formula (proved here) to estimate the last two sums above

$\sum_{k=m}^nf(k)=f(m)+\int_m^n f(x)\,dx+\int_m^n f^\prime(x)P_1(x)\,dx$ (4)

Choosing $f(x)=\frac{\ln(x)}{x}$ and $f^\prime(x)=\frac{1-\ln(x)}{x^2}$ in (4), we get for the first sum :

And for the second sum

$\begin{aligned} \sum_{n=1}^{N}\frac{\ln(n)}{n}&=\frac{\ln(1)}{1}+\int_1^N\frac{\ln(x)}{x}\,dx+\int_1^N\frac{\{x\}}{x^2}\left( 1-\ln(x)\right)\,dx\\ &=\frac{\ln^2(2N)}{2}+\int_1^{2N}\frac{\{x\}}{x^2}\left( 1-\ln(x)\right)\,dx\\ &=\frac{\ln^2(2)}{2}+\ln(2)\ln(N)+\frac{\ln^2(N)}{2}+\int_1^{2N}\frac{\{x\}}{x^2}\left( 1-\ln(x)\right)\,dx\\ \end{aligned}$ (6)

Recall also the integral representation of the Stiltjies constant (shown here):

$\gamma_n=\int_1^\infty \frac{\left\{ x\right\}}{x^2} \left(n-\ln(x) \right)\ln^{n-1}(x)\,dx$ (7)

letting n=1 in (7) we obtain

$\gamma_1=\int_1^\infty \frac{\left\{ x\right\}}{x^2} \left(1-\ln(x) \right)\,dx$ (8)

Now, plugging (5) and (6) back in (3) and letting $N \to \infty$ we obtain

$\begin{aligned} \sum_{n=1}^{\infty}\frac{(-1)^n\ln(n)}{n}&=\lim_{N \to \infty}\left[\ln(2)\operatorname{H}_N+ \frac{\ln^2(N)}{2}+\int_1^N\frac{\{x\}}{x^2}\left( 1-\ln(x)\right)\,dx \right] \\ & \qquad \qquad-\frac{\ln^2(2)}{2}-\lim_{N \to \infty}\left[\ln(2)\ln(N)-\frac{\ln^2(N)}{2}-\int_1^{2N}\frac{\{x\}}{x^2}\left( 1-\ln(x)\right)\,dx \right] \\ &=\ln(2)\lim_{N \to \infty}\left[\operatorname{H}_N-\ln(N) \right]-\frac{\ln^2(2)}{2}+\gamma_1-\gamma_1 & (\text{by eq.(8) above}) \\ &=\gamma\ln(2)-\frac{\ln^2(2)}{2} \qquad \blacksquare \end{aligned}$

We can now use (2) to calculate the following integral

$\int_0^\infty \frac{x\ln(x)}{e^{x^2}+1}\,dx=-\frac{\ln^2(2)}{8}$

Proof:

$\begin{aligned} I&=\int_0^\infty \frac{x\ln(x)}{e^{x^2}+1}\,dx\\ &=\frac14\int_0^\infty \frac{\ln(x)}{e^{x}+1}\,dx \qquad (x^2 \to x)\\ &=\frac14\int_0^\infty \frac{e^{-x}\ln(x)}{e^{-x}+1}\,dx\\ &=\frac14\int_0^\infty e^{-x}\ln(x)\left(\sum_{k=0}^\infty(-1)^ke^{-kx}\right)\,dx\\ &=\frac14\sum_{k=1}^\infty(-1)^{k-1}\,\int_0^\infty e^{-kx}\ln(x)\,dx\\ &=\frac14\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\,\int_0^\infty e^{-x}\ln\left(\frac xk\right)\,dx\\ &=\frac14\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\,\int_0^\infty e^{-x}\ln\left( x\right)\,dx-\frac14\sum_{k=1}^\infty \frac{(-1)^{k-1} \ln(k)}{k}\int_0^\infty e^{-x}\,dx\\ &=-\frac{\gamma}{4}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}+\frac14\eta^\prime(1)\\ &=-\frac{\gamma}{4}\ln(2)+\frac14\eta^\prime(1)\\ &=\frac14\left(\gamma\ln(2)-\frac{\ln^2(2)}{2} \right)-\frac{\gamma}{4}\ln(2)\\ &=-\frac{\ln^2(2)}{8} \qquad \blacksquare \end{aligned}$

Search This Blog

MY MATEX NOTES