The antiderivative of $\sum_{n\gt 0}\frac{x}{n(x+n)}$

Question

I tried to calculate $\int\sum_{n\gt 0}\frac{x}{n(x+n)}\, \mathrm dx$: $$\begin{align}\int\sum_{n\gt 0}\frac{x}{n(x+n)}&=\sum_{n\gt 0}\frac{1}{n}\int\left(1-\frac{n}{x+n}\right)\, \mathrm dx \\&=\sum_{n\gt 0}\left(\frac{1}{n}\int\mathrm dx -\int \frac{\mathrm dx}{x+n}\right)\\&\overset{x+n\to x}{=}\sum_{n\gt 0}\left(\frac{x}{n}-\ln (x+n)\right)+C.\end{align}$$ It seems however, that the 'result' fails to converge for any $x$ (whereas $\sum_{n\gt 0}\frac{x}{n(x+n)}$ converges everywhere except for negative integers). How could a correct antiderivative be obtained?

Answer 1

You need to choose the constants of integration wisely:

\begin{align*} \int_{0}^{x} \sum_{n=1}^{\infty} \frac{t}{n(t+n)} \, \mathrm{d}t &= \sum_{n=1}^{\infty} \int_{0}^{x} \frac{t}{n(t+n)} \, \mathrm{d}t \\ &= \sum_{n=1}^{\infty} \left[ \frac{t}{n} - \log(t+n) \right]_{t=0}^{t=x} \\ &= \sum_{n=1}^{\infty} \left( \frac{x}{n} - \log\left(1+\frac{x}{n}\right) \right). \end{align*}

So in general,

$$ \int \sum_{n=1}^{\infty} \frac{x}{n(x+n)} \, \mathrm{d}x = \sum_{n=1}^{\infty} \left( \frac{x}{n} - \log\left(1+\frac{x}{n}\right) \right) + C. $$

Answer 2

Using partial fractions, we have $$ \sum_{n=1}^{\infty} \frac{x}{n(x+n)} = \sum_{n=1}^{\infty} \frac{1}{n}-\frac{1}{n+x} $$ This series is well-known; see, for instance, the link: $$ = - \gamma + \psi_0(1+x) = -\gamma + \Gamma'(1+x)/\Gamma(1+x) $$Integration is now easy. $$ \int -\gamma + \Gamma'(1+x)/\Gamma(1+x)\,dx = -\gamma x + \log|\Gamma(1+x)|+ C $$

• https://en.wikipedia.org/wiki/Digamma_function
• https://dlmf.nist.gov/5.7#E6