Originally Posted by

**liquidFuzz** I'm getting stuck while I'm trying to solve a integral.

$\displaystyle \[ \int_0^\infty \frac{1}{x^{100} + 1}\,dx.\]$

What I've got so far:

Since the integer is even I get:

$\displaystyle \displaystyle \[ \int_0^\infty \frac{1}{x^{100} + 1}\,dx\]$ = $\displaystyle \displaystyle \frac{1}{2} \[ \int_{-\infty}^\infty \frac{1}{x^{100} + 1}\,dx\] $

Using the complex plane. More precise, the area of a half circle in the upper part. of the plane.

$\displaystyle \displaystyle \frac{1}{2} \[ \int_{C_R} \frac{1}{z^{100} + 1}\,dz\] $; $\displaystyle C_R = \{

z \in \mathbb{C}| ~ z \leq R \land Im \geq 0

\}$

So now I calculate the roots of the divisor.

$\displaystyle z^{100} + 1 = 0$

$\displaystyle z = e^{\frac{i\pi}{100}(1 + 2k)}$ ; $\displaystyle k = 0, 1, 2 ...$

This is where I get stuck. I can find out how many of the roots lies in the upper part of the complex plane.

As many as their imaginary part is positive:

So, $\displaystyle Im\left(e^{\frac{i\pi}{100}(1 + 2k)}\right)>0\Longleftrightarrow \sin\frac{(1+2k)\pi}{100}>0\Longleftrightarrow 0<\frac{(1+2k)\pi}{100}<\pi$ ...and we get an easy inequality.

Tonio

Argument-variance to find roots.

$\displaystyle C_2$ = the arch going from +R to -R on the real axis.

$\displaystyle C_1$ is -R to +R on the real axis.

$\displaystyle C_1$ : $\displaystyle z^{100} = -1 \neq 0$ No roots on the real axis.

$\displaystyle C_2$ : R is chosen big enough to include any roots in the half circle. Along the rim?

$\displaystyle z = R e^{i\theta}~; 0 \leq \theta \leq \pi, R \rightarrow \infty$

$\displaystyle Arg (z^{100} + 1) = 100\pi$ - Wich means that there are$\displaystyle \frac{100\pi}{2\pi} = 50 $ roots in my area.

This is about it, I don't know how to get further on this one. If anyone could wave their magic wand over it. That would really make my weekend.

Cheers!