# Urgent, residue and winding numbers

• Nov 11th 2008, 07:26 PM
jbpellerin
Urgent, residue and winding numbers
1. Evaluate the integrals

$\int ^{2\pi} _0 \frac{dt}{2+sint}$

and

$\int ^{\infty} _0 \frac{x^2}{x^4 +x}dx$

2. Let $z_1,z_2,z_3,z_4 \epsilon \mathbb{C}$ be the vertices of a closed quadrilateral $\Box$ such that $[z_1,z_2] \oplus [z_2,z_3]\oplus [z_3,z_4]\oplus [z_4,z_1]$ traverses $\partial \Box$ in counterclockwise direction (abusing notation, we shall write $\partial \Box$ for that curve) show that $v(\partial \Box,z)$ is 0 for $z\epsilon \mathbb{C}  \Box$ (C-minus Box) and 1 if $z$ is an interior point of $\Box$ (Hint: do not attempt to parametrize $\partial \Box$..)
• Nov 11th 2008, 07:40 PM
Mathstud28
Quote:

Originally Posted by jbpellerin
1. Evaluate the integrals

$\int ^{2\pi} _0 \frac{dt}{2+sint}$

and

$\int ^{\infty} _0 \frac{x^2}{x^4 +x}dx$

2. Let $z_1,z_2,z_3,z_4 \epsilon \mathbb{C}$ be the vertices of a closed quadrilateral $\Box$ such that $[z_1,z_2] \oplus [z_2,z_3]\oplus [z_3,z_4]\oplus [z_4,z_1]$ traverses $\partial \Box$ in counterclockwise direction (abusing notation, we shall write $\partial \Box$ for that curve) show that $v(\partial \Box,z)$ is 0 for $z\epsilon \mathbb{C}  \Box$ (C-minus Box) and 1 if $z$ is an interior point of $\Box$ (Hint: do not attempt to parametrize $\partial \Box$..)

For the first ones do you need to do it by CA?
• Nov 11th 2008, 07:42 PM
jbpellerin
CA? I'm pretty sure the prof wants us to use the residue theorem
theres an example in the book with cos instead of sin and I'm working on getting it to work for this question
• Nov 11th 2008, 08:44 PM
ThePerfectHacker
Quote:

Originally Posted by jbpellerin
1. Evaluate the integrals

$\int ^{2\pi} _0 \frac{dt}{2+sint}$

Notice that,
$\int_0^{2\pi} \frac{dt}{2+\sin t} = \oint \limits_{|z|=1} \frac{1}{2+\tfrac{1}{2i}(z - z^{-1})} \cdot \frac{1}{iz} dz$

Thus, we get,
$\oint_{|z|=1} \frac{2dz}{z^2+4iz - 1}$

Now apply the residue theorem.
• Nov 11th 2008, 09:05 PM
ThePerfectHacker
Quote:

Originally Posted by jbpellerin
$\int ^{\infty} _0 \frac{x^2}{x^4 +x}dx$

By integrating over partial circular contour of angle $2\pi/n$ we get $\int_0^{\infty} \frac{dx}{x^n+1} = \frac{(\pi/n)}{\sin (\pi/n)}$ for $n\geq 2$.

$\int_0^{\infty} \frac{x^2}{x^4+x} dx=\int_0^{\infty} \frac{x}{x^3+1} = \int_0^{\infty} \frac{dx}{x^2-x+1} - \int_0^{\infty} \frac{dx}{x^3+1}$

Now,
$\int_0^{\infty} \frac{dx}{x^2-x+1} = \int_0^{\infty} \frac{dx}{\left( x - \frac{1}{2} \right)^2 + \frac{3}{4} } = \int_{-1/2}^{\infty} \frac{dt}{t^2 + \frac{3}{4}}$
That computes to,
$\frac{2}{\sqrt{3}}\tan^{-1} \left( \frac{2t}{\sqrt{3}} \right) \bigg|_{-1/2}^{\infty} = \frac{\pi}{\sqrt{3}} + \frac{\pi}{3\sqrt{3}} = \frac{4\pi}{3\sqrt{3}}$

And by above result,
$\int_0^{\infty}\frac{dx}{x^3+1} = \frac{(\pi/3)}{\sin (\pi/3)} = \frac{2\pi}{3\sqrt{3}}$

Thus, the final answer is, $\frac{4\pi}{3\sqrt{3}}-\frac{2\pi}{3\sqrt{3}} = \frac{2\pi}{3\sqrt{3}}$
• Nov 11th 2008, 09:12 PM
ThePerfectHacker
Quote:

Originally Posted by jbpellerin
2. Let $z_1,z_2,z_3,z_4 \epsilon \mathbb{C}$ be the vertices of a closed quadrilateral $\Box$ such that $[z_1,z_2] \oplus [z_2,z_3]\oplus [z_3,z_4]\oplus [z_4,z_1]$ traverses $\partial \Box$ in counterclockwise direction (abusing notation, we shall write $\partial \Box$ for that curve) show that $v(\partial \Box,z)$ is 0 for $z\epsilon \mathbb{C} ` \Box$ (C-minus Box) and 1 if $z$ is an interior point of $\Box$ (Hint: do not attempt to parametrize $\partial \Box$..)

If $z$ is outside the rectange then computing the winding number is to evaluate $\oint_{\partial \Box}\frac{dw}{w-z}=0$ by Cauchy's theorem. If $z$ is inside then it can be shown, here (post #8 with long proof), that if $C$ is a tiny circle around $z$ still inside $\partial \Box$ then $\oint_{\partial \Box} \frac{dw}{w-z} = \oint_{C} \frac{dw}{w-z} = 2\pi i$. And so the winding number is $1$.
• Nov 11th 2008, 09:51 PM
jbpellerin
Whenever I make a thread on here I always hope you're online because you always have excellent answers :)
thank you again for the help
• Nov 12th 2008, 05:10 AM
ThePerfectHacker
Quote:

Originally Posted by jbpellerin
Whenever I make a thread on here I always hope you're online because you always have excellent answers :)

(Happy) You should not say that because I have a terrible ego problem.
.....Many members here compain about my addiction of myself to myself.
• Nov 12th 2008, 12:04 PM
Mathstud28
Quote:

Originally Posted by ThePerfectHacker
.....Many members here compain about my addiction of myself to myself.

Here here! (Wink)...jk