Complex Analysis again

[AKG]

Homework Statement

1. Evaluate the following integrals using residues:
a)

$$\int _0 ^{\infty} \frac{x^{1/4}}{1 + x^3}dx$$

b)

$$\int _{-\infty} ^{\infty} \frac{\cos (x)}{1 + x^4}dx$$

c)

$$\int _0 ^{\infty} \frac{dx}{p(x)}$$

where p(x) is a poly. with no zeros on {x > 0}

d)

$$\int _{-\infty} ^{\infty}\frac{\sin ^2(x)}{x^2}dx$$

2. Let A be a complex constant lying outside the real interval [-1,1]. Using residues, prove that:

$$\int _{-1} ^1 \frac{dx}{(x-A)\sqrt{1-x^2}} = \frac{\pi }{\sqrt{A^2 - 1}}$$,

with the appropriate determination of $\sqrt{A^2 - 1}$.

Homework Equations

Let f(z) be analytic except for isolated singularities a_j in a region $\Omega$. Then

$$\frac{1}{2\pi i}\int _{\gamma }f(z)dz = \sum _j n(\gamma , a_j)\mbox{Res} _{z=a_j}f(z)$$

for any cycle $\gamma$ which is homologous to zero in $\Omega$ and does not pass through any of the points a_j.

The Attempt at a Solution

1.a) I made the substitution z = x^1/4, giving:

$$\int _0 _{\infty} \frac{x^{1/4}}{1 + x^3}dx$$

$$= 4\int _0 ^{\infty} \frac{z^4}{1 + z^{12}}dz$$

$$= 2\int _{-\infty} ^{\infty} \frac{z^4}{1 + z^{12}}dz$$

$$= 4\pi i\sum _{\mbox{Im} (z) > 0}\mbox{Res}f(z)$$

I know how to give expressions for these residues, but I don't know a good way to compute this thing. I've used rotationaly symmetry to express this as (a sum of 6 things) times (one of the residues) but it's still ugly.

b)

$$\int _{-\infty} ^{\infty} \frac{\cos x}{1 + x^4}dx$$

$$= \mbox{Re}\left (\int _{-\infty} ^{\infty} \frac{e^{ix}}{1 + x^4}dx \right )$$

$$= \mbox{Re}\left (2\pi i \sum _{\mbox{Im} (z) > 0} \mbox{Res} \frac{e^{iz}}{1 + z^4} \right )$$

I know the relevant poles are $e^{3i\pi /4}$ and $e^{i\pi /4}$, so I know how to find expressions for the residues at these poles, but again I don't have a neat way to compute this.

c) If p is constant or linear, the integral doesn't exist. Otherwise, the integral does exist, but I have no clue really how to compute it for arbitrary p.

d) Again, not much clue.

2. Well I can compute that the residue at A is (1 - A²)^-1/2. It's a matter of making a clever choice of arc over which to integrate, or possibly a parametrized family of arcs and then taking the limits as the parameters of the family tend to desired limits, but I can't see what this clever choice would be. Any hints?

 

[d_leet]

I would think that for c the integral would evaluate to zero, since there are no residues in the interval you're integrating over, but I'm not absolutely certain that that is correct.

EDIT: On second thought that probably isn't correct because to me the question seems to imply that the polynomial only has no zeros only on the positive real line since they specify x>=0 and you can't give complex numbers that order. So if the polynomial has any complex zeros the best answer I think you could give would be 2*pi*i*Sum(x_i) where x_i are all the complex zeroes of the polynomial... But then if the polynomial is real, we should be expecting a real answer so you woudl probably want only the really part of that which again I have a feeling should turn out to e zero.

I'm sorry if this doesn't help you very much, I can't remember very much from my complex analysis course, but that problem interested me, and maybe some of this might help.

 

[mjsd]

I'll do one for you in full...(PF willing)... the rest should be similar... these are tedious but method is generally clear.
$$\displaystyle{\int_{-\infty}^\infty \frac{\cos (az)}{z^4+1}\; dz}$$
$$\displaystyle{=\lim_{R\rightarrow \infty}\int_\text{loop} - \int_\text{upper arc}, a>0}$$
it can be easily shown that
$$\displaystyle{\int_\text{upper arc}\rightarrow 0}$$ as $$R\rightarrow \infty$$ since
$$\displaystyle{\left|\frac{e^{iz}}{z^4+1}\right| \leq \left|\frac{1}{z^4+1}\right| \leq \left|\frac{1}{R^4-1}\right| \sim \frac{1}{R^4}}$$ and so ML-estimate gives function to go like
$$\displaystyle{\frac{1}{R^4}.\pi R \sim \frac{1}{R^3}}$$ and integral vanish as R goes to infinity.

So, we want Real part of
$$\displaystyle{\int_\text{loop} \frac{e^{iaz}}{z^4+1}\; dz=2\pi i \left( \text{Res}\left[\frac{e^{iaz}}{z^4+1}, e^{i\pi/4}\right]+\text{Res}\left[\frac{e^{iaz}}{z^4+1}, e^{i3\pi/4}\right]\right)}$$
Residues can be evaluated using
$$\displaystyle{\text{Res}\left[f(z)/g(z), z_0\right]=\frac{f(z_0)}{g'(z_0)}}$$ where at $$z_0, g(z)$$ has a simple zero, $$g'(z)$$ is the derivative.

therefore, we have
$$\displaystyle{\int_\text{loop} \frac{e^{iaz}}{z^4+1}\; dz= 2\pi i \left[\frac{e^{ia e^{i\pi/4}}}{4(e^{i 3\pi/4})}+ \frac{e^{ia e^{i 3\pi/4}}}{4(e^{i 9\pi/4})}\right]}$$

after some quick manipulation using Euler formula:
$$\displaystyle{= \frac{\pi i}{2} e^{-a/\sqrt{2}}\left( e^{i(a/\sqrt{2}-3\pi/4)}+ e^{-i(a/\sqrt{2}+\pi/4)} \right)}$$

convert $$i \rightarrow e^{i\pi/2}$$ and multiply into bracket, it becomes
$$\displaystyle{= \frac{\pi}{2} e^{-a/\sqrt{2}}\left( e^{i(a/\sqrt{2}-\pi/4)}+ e^{-i(a/\sqrt{2}-\pi/4)} \right)}$$
$$\displaystyle{= \pi e^{-a/\sqrt{2}} \cos (a/\sqrt{2}-\pi/4)}$$
$$\displaystyle{= \pi e^{-a/\sqrt{2}} (\cos (a/\sqrt{2})\cos (\pi/4) +\sin (a/\sqrt{2})\sin (\pi/4))}$$
so altogether you get
$$\displaystyle{\int_{\infty}^{\infty} \frac{\cos(az)}{z^4+1}\; dz = \frac{\pi}{\sqrt{2}}e^{-a/\sqrt{2}}\left[\cos(a/\sqrt{2})+ \sin(a/\sqrt{2})\right], a>0}$$

NB: this is a standard textbook problem, so don't charge me for posting full solutions to Homework Forum