Evaluate this $\int_{0}^{\infty}{x^s-1 \over x-1}\cdot{1-e^{-ax} \over 1-e^{ax}}dx$

Question

$$\int_{0}^{\infty}{x^s-1 \over x-1}\cdot{1-e^{-ax} \over 1-e^{ax}}dx$$

$$\int_{0}^{\infty}{x^s-1 \over x-1}\cdot{1-e^{-ax} \over 1-e^{ax}}\cdot{e^{ax}\over e^{ax}}dx$$

$$\int_{0}^{\infty}-{x^s-1 \over x-1}\cdot{e^{-ax}}dx$$

Suppose $s=2$

$$\int_{0}^{\infty}-{x^2-1 \over x-1}\cdot{e^{-ax}}dx$$

$(x^2-1)=(x+1)(x-1)$

$$-\int_{0}^{\infty}{e^{-ax}}(1+x)dx$$

Integrating using by parts

$u=1+x$ $\rightarrow$ $du=dv$

$dv=e^{-ax}$ $\rightarrow$ $v=-{{e^{-ax}} \over {a}}$

$=\left.{-(1+x)e^{-ax}\over a}\right|_{0}^{\infty}-{1\over a}\int_{0}^{\infty}e^{-ax}dx$

$=\left.{-(1+x)e^{-ax}\over a}\right|_{0}^{\infty}+{1\over a^2}\left.e^{-ax}\right|_{0}^{\infty}$

$=-{1\over a}-{1\over a^2}$

Let $s=k$ for k is an integer $k\ge1$ ; restriction to integers for sake of easiness

The factorisation of $x^k-1$ is quite lengthy, so integrating via this method seem tedious. How can one evaluate this integral using another less effort strategy?

Claude Leibovici · Accepted Answer · 2016-05-31T14:21:44.320

4

Admitting that $s$ is a positive integer, $$\frac{x^s-1}{x-1}=\sum_{k=0}^{s-1}x^k$$ So, $$\int_{0}^{\infty}{x^s-1 \over x-1}\cdot{1-e^{-ax} \over 1-e^{ax}}dx=\sum_{k=0}^{s-1}\int_{0}^{\infty}x^k{1-e^{-ax} \over 1-e^{ax}}dx$$ Now, assuming $a>0$, $$\int x^k{1-e^{-ax} \over 1-e^{ax}}dx=-\int x^k e^{-ax}dx=\frac{1}{a^{k+1}}\Gamma (k+1,a x)$$ where appears the incomplete gamma function.

Using bounds $$\int_{0}^{\infty} x^k{1-e^{-ax} \over 1-e^{ax}}dx=-\frac{1}{a^{k+1}} \Gamma (k+1)=-\frac{k!}{a^{k+1}} $$

edited May 31 '16 at 14:21

answered May 31 '16 at 11:55

Claude Leibovici

260,315

Imcomplete gamma function that is new to me, when I read about I might ask you some questions. – gymbvghjkgkjkhgfkl May 31 '16 at 12:46
@Chinacat. For sure ! Any time you want. Cheers :-) – Claude Leibovici May 31 '16 at 12:48
1

I forgot my manner, thank you for the answer – gymbvghjkgkjkhgfkl May 31 '16 at 12:52
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@Chinacat. Thank you for your good manner !! – Claude Leibovici May 31 '16 at 12:53

score 3 · Answer 2 · answered May 31 '16 at 12:04

3

Getting rid of the syntactic sugar, $$ I(s,a)=\int_{0}^{+\infty}\frac{x^s-1}{x-1}\cdot\frac{1-e^{-ax}}{1-e^{ax}}\,dx = -\int_{0}^{+\infty}\frac{x^s-1}{x-1}e^{-ax}\,dx$$ hence the problem is equivalent to finding a Laplace transform. If $s\in\mathbb{N}^+$, since $$ \frac{x^s-1}{x-1} = \sum_{k=0}^{s-1} x^k $$ and $\int_{0}^{+\infty}x^k e^{-ax}\,dx = \frac{k!}{a^{k+1}}$, it follows that: $$ I(s,a) = -\sum_{k=0}^{s-1}\frac{k!}{a^{k+1}}.$$

answered May 31 '16 at 12:04

Jack D'Aurizio

353,855

I would like to see the full steps, please try not to skip steps. I know your answer is nice and tidy but for the sake of the questioner I need to fully understand the result also. Sorry if I have been too harassed on you. – gymbvghjkgkjkhgfkl May 31 '16 at 12:45
@Chinacat: what steps do you think are missing? – Jack D'Aurizio May 31 '16 at 13:00
Did you took the sum out of the integral? – gymbvghjkgkjkhgfkl May 31 '16 at 13:04
@Chinacat: yes I did. With our assumptions, it is a finite sum. – Jack D'Aurizio May 31 '16 at 13:05
I like the use of phrase "getting rid of syntactic sugar" +1. – Paramanand Singh Jun 02 '16 at 10:44

Evaluate this $\int_{0}^{\infty}{x^s-1 \over x-1}\cdot{1-e^{-ax} \over 1-e^{ax}}dx$

2 Answers2