(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[tex]I = \int_{-\infty}^{\infty} \frac{{e}^{ax}}{1+{e}^{x}} dx \; \; 0 < a < 1[/tex]

a) Show that the improper real integral is absolutely convergent.

b) Integrating around the closed rectangle [tex]\boldsymbol{R}[/tex] with corners [tex]-R, R, R+2\pi\iota, -R+2\pi\iota[/tex] use residue calculus to evaluate the real improper integral [tex]I[/tex] for [tex]0<a<1[/tex]

c) In completing part (b) show carefully that;

[tex]\lim_{R\rightarrow\infty} \int_{R}^{R+2\pi\iota} \frac{{e}^{az}}{1+{e}^{z}} dz = 0 \; \; \: {\; \; \; 0<a<1[/tex]

[tex]\lim_{R\rightarrow\infty} \int_{-R}^{-R+2\pi\iota} \frac{{e}^{az}}{1+{e}^{z}} dz = 0 \; \; \: {\; \; \; 0<a<1[/tex]

2. Relevant equations

for a) I know that if [tex]\lim_{t\rightarrow\infty}\int_{-t}^{t} |f(x)| dx[/tex] then the integral is absolutely convergent, is there any other way to do this without evaluating the integral? (since I am not sure how to evaluate using either Residue Calculus or Real Analysis, hence this post)

b)[tex]\oint_{\Gamma }^{} f(z) dz = 2\pi\iota \sum Res(f,{a}_{k})[/tex]

3. The attempt at a solution

b) I found the poles are when [tex] 1+{e}^{x} = 0[/tex] which is when [tex]x = \pi\iota \pm 2n\pi\iota\; , n \epsilon \mathbb{Z}[/tex] but am not sure how to continue.

c) I'm thinking I'm going to have apply a similar argument as we did when we showed the contribution to the integrals from the arc of a semicircle who's radius tends to infinity is 0.

I have evaluated real integrals using Residue Calculus before, but that is when the region of integration was a semi circle who's radius went to infinity and didn't contribute to the integral. Also I have tackled cases which have had polynomials (i.e a finite number of zeros) in the denominator, but never with an exponential.

Any help would be greatly appreciated :)

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# Homework Help: Evaluating Real Integrals using Residue Calculus

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