# Integrating $\int_{a}^{+\infty} y \exp(-by)/(1-\exp(-cy)) \ dy$

Can any one help me calculate this integral :

$$\int_{a}^{+\infty} \frac{y\ \exp{(-by)}} {1-\exp{(-cy)}} \ dy$$

a, b & c are real constant numbers, b & c > 0

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Jack, two things will help: 1. What are $a,b,c$? (Or can they be any real number? 2. What have you tried to solve the integral in question? – Pedro Tamaroff May 21 '12 at 23:37
Peter thank you for trying to help me : 1. a, b, c are reels constants numbers b>0 & c>0. 2. I have tried to solve the integrale in expressing the denominator as an infinite serie and then permit the infinite somme and the integral but this lead to calaculate a non trivial infinite serie. – jack May 22 '12 at 0:09
Are you italian? I can help you with that infinite series. Maybe this interests you and this too. – Pedro Tamaroff May 22 '12 at 0:18

Assuming $a>0$, you can write the integrand as $\sum_{n=0}^\infty y \exp((-b-nc)y)$ and the integral becomes the convergent series $$\sum_{n=0}^\infty e^{-ab-acn} \frac{ab+acn+1}{(b+nc)^2}$$
According to Maple this can be written using a hypergeometric function: $$\frac{ab+1}{b^2 e^{ab}}\ {\mbox{_4F_3}(1,{\frac {b}{c}},{\frac {b}{c}},{\frac {1+ \left( c+b \right) a}{ac}};\,{\frac {c+b}{c}},{\frac {c+b}{c}},{\frac {ab+1}{ac}};\, {{\rm e}^{-ac}} )}$$ However, we can do somewhat better: first write $\dfrac{ab+acn+1}{(b+nc)^2} = \dfrac{a}{b+nc}+\dfrac{1}{(b+nc)^2}$. Now $$\sum_{n=0}^\infty e^{-ab-acn} \left( \dfrac{a}{b+nc} + \dfrac{1}{(b+nc)^2}\right) = {\rm e}^{-ab} \left( \dfrac{a}{c} {\rm LerchPhi} \left( {{\rm e}^{-ac}},1,{\frac {b}{c}} \right) + \dfrac{1}{c^2} {\rm LerchPhi} \left( {{\rm e}^{-ac}},2,{\frac {b}{c}} \right)\right)$$ where ${\rm LerchPhi}(t,m,v) = \sum_{n=0}^\infty \dfrac{t^n}{(v+n)^m}$.
In the special case $a=0$ the series becomes $$\sum_{n=0}^\infty \frac{1}{(b+nc)^2} = \Psi \left( 1,{\frac {b}{c}} \right) {c}^{-2}$$ where $\Psi(1,t) = \dfrac{d}{dt} \Psi(t)$ and $\Psi(t) = \dfrac{d}{dt} \ln \Gamma(t)$.