# Thread: Complex integration - powers of 100 - getting stuck

1. ## Complex integration - powers of 100 - getting stuck

I'm getting stuck while I'm trying to solve a integral.
$$\int_0^\infty \frac{1}{x^{100} + 1}\,dx.$$

What I've got so far:
Since the integer is even I get:
$\displaystyle $\int_0^\infty \frac{1}{x^{100} + 1}\,dx$$ = $\displaystyle \frac{1}{2} $\int_{-\infty}^\infty \frac{1}{x^{100} + 1}\,dx$$

Using the complex plane. More precise, the area of a half circle in the upper part. of the plane.
$\displaystyle \frac{1}{2} $\int_{C_R} \frac{1}{z^{100} + 1}\,dz$$; $C_R = \{
z \in \mathbb{C}| ~ z \leq R \land Im \geq 0
\}$

So now I calculate the roots of the divisor.
$z^{100} + 1 = 0$
$z = e^{\frac{i\pi}{100}(1 + 2k)}$ ; $k = 0, 1, 2 ...$

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

Argument-variance to find roots.
$C_2$ = the arch going from +R to -R on the real axis.
$C_1$ is -R to +R on the real axis.

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

$C_2$ : R is chosen big enough to include any roots in the half circle. Along the rim?
$z = R e^{i\theta}~; 0 \leq \theta \leq \pi, R \rightarrow \infty$
$Arg (z^{100} + 1) = 100\pi$ - Wich means that there are $\frac{100\pi}{2\pi} = 50$ 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!

2. Originally Posted by liquidFuzz
I'm getting stuck while I'm trying to solve a integral.
$$\int_0^\infty \frac{1}{x^{100} + 1}\,dx.$$

What I've got so far:
Since the integer is even I get:
$\displaystyle $\int_0^\infty \frac{1}{x^{100} + 1}\,dx$$ = $\displaystyle \frac{1}{2} $\int_{-\infty}^\infty \frac{1}{x^{100} + 1}\,dx$$

Using the complex plane. More precise, the area of a half circle in the upper part. of the plane.
$\displaystyle \frac{1}{2} $\int_{C_R} \frac{1}{z^{100} + 1}\,dz$$; $C_R = \{
z \in \mathbb{C}| ~ z \leq R \land Im \geq 0
\}$

So now I calculate the roots of the divisor.
$z^{100} + 1 = 0$
$z = e^{\frac{i\pi}{100}(1 + 2k)}$ ; $k = 0, 1, 2 ...$

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

Tonio

Argument-variance to find roots.
$C_2$ = the arch going from +R to -R on the real axis.
$C_1$ is -R to +R on the real axis.

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

$C_2$ : R is chosen big enough to include any roots in the half circle. Along the rim?
$z = R e^{i\theta}~; 0 \leq \theta \leq \pi, R \rightarrow \infty$
$Arg (z^{100} + 1) = 100\pi$ - Wich means that there are $\frac{100\pi}{2\pi} = 50$ 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!
.

3. Mhmm... I don't quite follow you.

$\frac{(1+2k)\pi}{100} < \pi$ Do you mean that you use the fact $k = 0, 1, 2 ... 49$ in the expression you end up in, since there's 50 roots in the area?
$k = 0 : \frac{\pi}{100}$ all the way up to $k= 49 : \frac{99\pi}{100}$ equals 'not quite' $\pi$

How is that usable while solving this?