Closed form of $\sum_{n=1}^{\infty}(-1)^{n+1} n (\log(n^2+1)-\log(n^2))$

Question

How would you start computing this series?

$$\sum_{n=1}^{\infty}(-1)^{n+1} n (\log(n^2+1)-\log(n^2))$$

One of the ways to think of would be Frullani integral with the exponential function , but it's troublesome due to the power of $n$ under logarithm. What else might I try?

Answer 1

I'm getting the same answer as you.

$$ \sum_{n=1}^{\infty} (-1)^{n+1} n \log \left( \frac{n^{2}+1}{n^{2}}\right) = \sum_{n=1}^{\infty} (-1)^{n+1} n \int_{0}^{1} \frac{1}{n^{2}+x} \ dx$$

Then since $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1}n}{n^{2}+x}$ converges uniformly on $[0,1]$,

$$ \begin{align} &\sum_{n=1}^{\infty} (-1)^{n+1} n \int_{0}^{1} \frac{1}{n^{2}+x} \ dx \\ &= \int_{0}^{1} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}n}{n^{2}+x} \ dx \\ &= \frac{1}{4} \int_{0}^{1} \left[ \psi \left(\frac{i \sqrt{x}}{2} \right)- \psi \left(\frac{1}{2} + \frac{i \sqrt{x}}{2} \right) + \psi \left(-\frac{ i \sqrt{x}}{2} \right)- \psi \left(\frac{1}{2} - \frac{i \sqrt{x}}{2} \right) \right] \ dx .\end{align}$$

The above series can be derived by working backwards and using the series representation of the digamma function (14) .

Now let $t = \sqrt{x}$.

Then $$ \begin{align} &\sum_{n=1}^{\infty} (-1)^{n+1} n \log \left(\frac{n^{2}+1}{n^{2}} \right) \\&= \frac{1}{2} \int_{0}^{1} \left[t \psi \left(\frac{it}{2} \right) - t \psi \left(\frac{1}{2} + \frac{i t}{2} \right) + t \psi \left(-\frac{it}{2} \right) - t \psi \left(\frac{1}{2} - \frac{it}{2} \right) \right] \ dt . \end{align}$$

And integrating by parts,

$$ \begin{align} &\sum_{n=1}^{\infty} (-1)^{n+1} n \log \left(\frac{n^2+1}{n^{2}} \right) \\ &= \frac{1}{2} \Bigg[ 4 \psi^{(-2)} \left(\frac{it}{2} \right) - 2i t \log \Gamma \left( \frac{it}{2}\right) - 4 \psi^{(-2)} \left(\frac{1}{2}+ \frac{it}{2} \right) + 2i t \log \Gamma \left( \frac{1}{2} + \frac{it}{2}\right) \\ &+ 4 \psi^{(-2)} \left(-\frac{it}{2} \right) + 2i t \log \Gamma \left(- \frac{it}{2}\right) - 4 \psi^{(-2)} \left(\frac{1}{2} -\frac{it}{2} \right) - 2i t \log \Gamma \left(\frac{1}{2} - \frac{it}{2}\right)\Bigg] \Bigg|^{1}_{0} \\ &= 2 \psi^{(-2)} \left(\frac{i}{2} \right) - i \log \Gamma \left(\frac{i}{2} \right) - 2 \psi^{(-2)} \left(\frac{1}{2} + \frac{i}{2} \right) + i \log \Gamma \left(\frac{1}{2} + \frac{i}{2} \right) + 2 \psi^{(-2)} \left(-\frac{i}{2} \right) \\ &+ i \log \Gamma \left(-\frac{i}{2} \right) - 2 \psi^{(-2)} \left(\frac{1}{2} - \frac{i}{2} \right) - i \log \Gamma \left(\frac{1}{2} - \frac{i}{2} \right) + 4 \psi^{(-2)} \left(\frac{1}{2} \right). \end{align}$$

In terms of simplification, $\psi^{(-2)} \left( \frac{1}{2}\right)$ can be expressed in terms of the Glaisher-Kinkelin constant.

And further simplification is possible using the Schwarz reflection principle.

$$\begin{align} \sum_{n=1}^{\infty} (-1)^{n+1} n \log \left(\frac{n^{2}+1}{n^{2}} \right) &= 4 \ \text{Re} \ \psi^{(-2)} \left(\frac{i}{2} \right) -4 \ \text{Re} \ \psi^{(-2)} \left(\frac{1}{2} + \frac{i}{2} \right) + 2 \ \text{Im} \ \log \Gamma \left( \frac{i}{2}\right) \\ &-2 \ \text{Im} \ \log \Gamma \left(\frac{1}{2} + \frac{i}{2} \right) + 6\log A + \frac{5}{6} \log 2 + \log \pi \end{align}$$

which according to Wolfram Alpha is approximately $0.4277662568$