How to evaluate $ \int_{-\infty}^\infty {e^{ax} \over 1 +e^x } \; dx $

Question

Possible Duplicate:
Possibility to simplify $\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{\pi }{{\sin \pi a}}} $

Given that $0 < a < 1$ how to evaluate by the method of residues $$ \int_{-\infty}^\infty {e^{ax} \over 1 +e^x } \; dx $$

In particular, see Argon's answer to that question. – Pedro Sep 05 '12 at 19:30 — Pedro, Sep 05 '12 at 19:30
@PeterTamaroff how did you change the sum into integral? – hasExams Sep 06 '12 at 04:23 — hasExams, Sep 06 '12 at 04:23
I explain it in detail in the question. – Pedro Sep 06 '12 at 13:11 — Pedro, Sep 06 '12 at 13:11

score 4 · Accepted Answer · answered Sep 05 '12 at 16:16

4

Substituting $u=e^x$, so that $dx = \dfrac{du}{u}$, the integral becomes

$$\int_0^{\infty} \dfrac{u^{a-1}}{1+u}du$$

How might you solve this? You've tagged the question as homework, so I'll leave it here for now, but if you're still stuck, post in the comments.

answered Sep 05 '12 at 16:16

Clive Newstead

64,925

does this method work when $0<a<1$ ? this needs a little editing and picture of contour is missing – hasExams Sep 05 '12 at 16:18
@testuser: Yes. – Clive Newstead Sep 05 '12 at 16:19
I must be an idiot :((( – hasExams Sep 05 '12 at 16:20
@testuser: Possible, but unlikely! Just follow through that solution with $b=1$ all the way through. [Or work it out yourself $-$ it's not too hard using fairly standard complex methods.] – Clive Newstead Sep 05 '12 at 16:22
yeah the answer is $ \pi \over \sin (a \pi )$ – hasExams Sep 05 '12 at 16:24

score 3 · Answer 2 · answered Sep 05 '12 at 19:08

3

Recalling the $\beta$ function,

$$ \beta( n,m ) = \int_{0}^{1} u^{n-1}\,(1-u)^{m-1}\, du = \frac{\Gamma(n)\Gamma(m)}{\Gamma(m+n)} \,.$$

Using the substitution $ 1+u=\frac{1}{t} $ casts $ \int_0^{\infty} \dfrac{u^{a-1}}{1+u}du $ in terms of the $\beta$ function,

$$\int_0^{\infty} \frac{u^{a-1}}{1+u}du = \int _{0}^{1}\!{t}^{-a} \left( 1-t \right) ^{a-1}{dt} =\Gamma(a)\Gamma(1-a) = \frac{\pi}{\sin (a \pi )} \,. $$

answered Sep 05 '12 at 19:08

Mhenni Benghorbal

47,431

@Downvoter:I do not know what you are downvoting for. I gave a way to evaluate the integral in a way different from the complex techniques. – Mhenni Benghorbal Sep 05 '12 at 21:04

How to evaluate $ \int_{-\infty}^\infty {e^{ax} \over 1 +e^x } \; dx $

2 Answers2