I'm getting stuck while I'm trying to solve a integral.
![\[ \int_0^\infty \frac{1}{x^{100} + 1}\,dx.\]](http://latex.codecogs.com/png.latex?\[ \int_0^\infty \frac{1}{x^{100} + 1}\,dx.\])
What I've got so far:
Since the integer is even I get:
![\displaystyle \[ \int_0^\infty \frac{1}{x^{100} + 1}\,dx\]](http://latex.codecogs.com/png.latex?\displaystyle \[ \int_0^\infty \frac{1}{x^{100} + 1}\,dx\])
=
![\displaystyle \frac{1}{2} \[ \int_{-\infty}^\infty \frac{1}{x^{100} + 1}\,dx\]](http://latex.codecogs.com/png.latex?\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} \[ \int_{C_R} \frac{1}{z^{100} + 1}\,dz\]](http://latex.codecogs.com/png.latex?\displaystyle \frac{1}{2} \[ \int_{C_R} \frac{1}{z^{100} + 1}\,dz\] )
;
So now I calculate the roots of the divisor.
})
;
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, }\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.
= the arch going from +R to -R on the real axis.
is -R to +R on the real axis.
:
No roots on the real axis.
: R is chosen big enough to include any roots in the half circle. Along the rim?
 = 100\pi)
- Wich means that there are
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!
LinkBack URL
About LinkBacks
Originally Posted by liquidFuzz 
