Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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
add comment

1 Answer

up vote 2 down vote accepted

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: $$ \frac{ab+1}{b^2 e^{ab}}\ {\mbox{$_4$F$_3$}(1,{\frac {b}{c}},{\frac {b}{c}},{\frac {1+ \left( c+b \right) a}{ac}};\,{\frac {c+b}{c}},{\frac {c+b}{c}},{\frac {ab+1}{ac}};\, {{\rm e}^{-ac}} )} $$ 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 $\Psi(t) = \dfrac{d}{dt} \ln \Gamma(t)$.

share|improve this answer
    
Thank you very mauch Robert. –  jack May 22 '12 at 1:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.