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Math Help - Complex integration - powers of 100 - getting stuck

  1. #1
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    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 = \{<br />
z \in \mathbb{C}| ~ z \leq R \land Im \geq 0<br />
\}

    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!
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  2. #2
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    Quote Originally Posted by liquidFuzz View Post
    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 = \{<br />
z \in \mathbb{C}| ~ z \leq R \land Im \geq 0<br />
\}

    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!
    .
    Follow Math Help Forum on Facebook and Google+

  3. #3
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    Sep 2010
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    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?
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