Evaluating Real Integrals using Residue Calculus

[phagist_]

Homework Statement

$$I = \int_{-\infty}^{\infty} \frac{{e}^{ax}}{1+{e}^{x}} dx \; \; 0 < a < 1$$
a) Show that the improper real integral is absolutely convergent.
b) Integrating around the closed rectangle $$\boldsymbol{R}$$ with corners $$-R, R, R+2\pi\iota, -R+2\pi\iota$$ use residue calculus to evaluate the real improper integral $$I$$ for $$0<a<1$$
c) In completing part (b) show carefully that;
$$\lim_{R\rightarrow\infty} \int_{R}^{R+2\pi\iota} \frac{{e}^{az}}{1+{e}^{z}} dz = 0 \; \; \: {\; \; \; 0<a<1$$
$$\lim_{R\rightarrow\infty} \int_{-R}^{-R+2\pi\iota} \frac{{e}^{az}}{1+{e}^{z}} dz = 0 \; \; \: {\; \; \; 0<a<1$$

Homework Equations

for a) I know that if $$\lim_{t\rightarrow\infty}\int_{-t}^{t} |f(x)| dx$$ 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)$$\oint_{\Gamma }^{} f(z) dz = 2\pi\iota \sum Res(f,{a}_{k})$$

The Attempt at a Solution

b) I found the poles are when $$1+{e}^{x} = 0$$ which is when $$x = \pi\iota \pm 2n\pi\iota\; , n \epsilon \mathbb{Z}$$ 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 :)

 

[lanedance]

for a) consider breaking the integral into increments, and try to show |f(x)| < |g(x)| for some arbitrary g(x) which is absolutely convergent on that interval...

 

[lanedance]

for b) try writing out the integral for each side of the rectangles contribution to the contour & consider the contribution of each as R gets big:
- the base becomes the integral you're trying to evaluate
- the sides should tend to zero in the limit as pointed out in c)
- the top will need further evaulation, but usually the aim is to easily evaluate it (say as tending to zero in the limit) or write as a multiple of the integeral you're trying to find (variable change may helP)

once you've done that, you'll need to eveualte the residues & equate with the contour integral and you ahouls be there

 

[jackmell]

And for (c) on the right leg, let x=R+iy and just use the ML-inequality:

$\left|i\int_{0}^{2\pi}\frac{e^{aR}e^{aiy}}{1+e^R e^{iy}}dy\right|\leq\int_0^{2\pi}\frac{|e^{aR}|}{|1-e^R|}dy$

for large enough R. Now, what happens to that expression if a is less than one and R goes to infinity?

 

[Gib Z]

To do (a) an effective method is to apply the squeeze theorem to the integrand. Since 0<a<1, we can find lower and upper bounds for the integral. Since the integrand is strictly positive and decreasing approaching zero in both directions, the integral must converge.

This is because Sum (a_k) exists if a_k is positive and decreasing to zero (from the theorem that bounded monotone sequences converge). Interpreting the integral as a Riemann sum gives the result.

 

[phagist_]

ok thanks for the replies.
I'm having trouble thinking of a suitable function to apply the squeeze theorem with.

If I use $$\int_{-\infty}^{\infty} \frac{{e}^{ax}}{{e^{x}}} dx = \int_{-\infty}^{\infty} {e}^{x(a-1) dx$$.

The function works in the $$\lim_{x\rightarrow\infty}$$ but not as $$\lim_{x\rightarrow-\infty}$$.

I can't really think of any other similar functions which bound the function from above.

 

[Gib Z]

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

Since 0<a<1, $e^{ax} < e^x$ so you can bound your integral by using that.

 

[Count Iblis]

That inequality holds for x > 0. If you split the intergal in two parts, from minus infinity to zero and from zero to infinity, you can do a change of variables x = -u in the former part and then rewrite the new integrand so that it becomes of the same form as the present integrand.