Power series involving the Von Mangoldt-Function

Question

I've been studying a proof of the Prime Number Theorem, given by D. V. Widder, where in one part he uses the identity $$\sum_{n=1}^{\infty}\frac{(\Lambda(n)-1)}{1-e^{-nx}}e^{-nx} = \sum_{n=1}^{\infty}(\log(n)-\tau(n))e^{-nx},$$ where $\Lambda(n)$ is the Von Mangoldt-Function and $\tau(n)$ is the number of divisors of n.

Can anybody help or give a hint how to see this identity?

Answer 1

Expand in the LHS $\frac{1}{1-e^{-nx}}$ as $\sum_{m=1}^{\infty}{e^{-mnx}}$.

Then the coefficient in front of $e^{-nx}$, where $n \geq 1$, is $\sum_{d|n}{\Lambda(d)-1}$.

But it’s easy to see $\sum_{d|n}{\Lambda(d)}=\log(n)$, which concludes.