2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Expand $\frac 1 {1-e^{-bt}} $ as $\sum e^{-bnt}$. $\endgroup$ Commented Jan 19, 2020 at 0:19
  • $\begingroup$ @KaviRamaMurthy Can you expand on your hint? I still don't see it. $\endgroup$
    – Math1000
    Commented Jan 22, 2020 at 1:28

1 Answer 1

0
$\begingroup$

$$\int_0^{\infty} \frac {te^{-at}} {1-e^{-bt}}d\lambda (t)=\int_0^{\infty} te^{-at} (1+e^{-bt}+e^{-2bt}+e^{-3bt}+...)d\lambda (t).$$ 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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .