A Lebesgue integral into a sum

Question

Show that $\forall a,b \in ]0,+\infty[$

$\int_{]0,+\infty[ }\frac{te^{-at}}{1-e^{-bt}} d\lambda (t) = \sum_{n\geq 0} \frac{1}{(a+nb)^2}$

Where $\lambda$ is the measure of Lebesgue.

Answer 1

$$\int_0^{\infty} \frac {te^{-at}} {1-e^{-bt}}d\lambda (t)=\int_0^{\infty} te^{-at} (1+e^{-bt}+e^{-2bt}+e^{-3bt}+...)d\lambda (t).$$ This is equal to $$\int_0^{\infty} te^{-at} d\lambda (t)+\int_0^{\infty} te^{-(a+b)t} d\lambda (t)+\int_0^{\infty} te^{-(a+2b)t} d\lambda (t)+...$$ Now evaluate $\int_0^{\infty} te^{-(a+nn)t} d\lambda (t)$ using integration by parts.