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Math Help - Integral

  1. #1
    Super Member Showcase_22's Avatar
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    Integral

    How do I evaluate \int_0^{2 \pi} \frac{(2e^{it})^5+3}{2e^{it}-i}~dt?

    I've tried using substitution and splitting it into partial fractions, but they end up taking many sides of paper and leading to dead ends!
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    Quote Originally Posted by Showcase_22 View Post
    How do I evaluate \int_0^{2 \pi} \frac{(2e^{it})^5+3}{2e^{it}-i}~dt?

    I've tried using substitution and splitting it into partial fractions, but they end up taking many sides of paper and leading to dead ends!
    This looks like one of those integrals that can only be evaluated by complex variable methods. If you make the substitution z = e^{it} then it becomes \oint_C\frac{(2z)^5+3}{2z-i}\,\frac{dz}{iz} (C = unit circle). You can evaluate this by finding the residues at z=0 and z=i/2.
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  3. #3
    Super Member Showcase_22's Avatar
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    residues at z=0 and z=i/2.
    I'm afraid I haven't studied residues yet and, upon checking my notes, there doesn't appear to be anything on them at all!

    I have <br /> <br />
\oint_C\frac{(2z)^5+3}{2z-i}\,\frac{dz}{iz}<br />
and i need to calculate the residues at z=0 and z=\frac{i}{2}. I found this method on wikipedia and i'll try and follow through that (http://en.wikipedia.org/wiki/Residue_(complex_analysis).

    Parameterisation of the unit circle r(t)=\cos t+ i \sin t where t \in[0, 2 \pi].

    This means I have to calculate \oint_0^{2 \pi} \frac{(2e^{it})^5+3}{2e^{it}-i}~\frac{dt}{ie^{it}}

    \oint_0^{2 \pi} \frac{32e^{5it}+3}{2e^{it}-i} \frac{dt}{ie^{it}}

    \oint_0^{2 \pi} \frac{32 \sum_{j=0}^{\infty} \frac{t^j}{5(j!)}+3}{2 \sum_{j=0}^{\infty}\frac{t^j}{j!}-i} \frac{dt}{i \sum_{j=0}^{\infty} \frac{t^j}{j!}}

    \oint_0^{2 \pi} \frac{32 \sum_{j=0}^{\infty} \frac{t^j}{5(j!)}+3}{2i \sum_{j=0}^{\infty}\left( \frac{t^j}{j!} \right)^2+\sum_{j=0}^{\infty} \frac{t^j}{j!}}~dt

    \oint_0^{2 \pi} \frac{32 \sum_{j=0}^{\infty} \frac{t^j}{5(j!)}+3}{(2i+1) \sum_{j=0}^{\infty} \left( \left( \frac{t^j}{j!} \right)^2+\frac{t^j}{j!}\right)}~dt

    \frac{32}{2i+1} \oint_0^{2 \pi} \frac{ \sum_{j=0}^{\infty} \frac{t^j}{5(j!)}}{ \sum_{j=0}^{\infty} \left( \left( \frac{t^j}{j!} \right)^2+\frac{t^j}{j!}\right)}~dt+\frac{3}{2i+1} \oint_0^{2 \pi}\frac{1}{ \sum_{j=0}^{\infty} \left( \left( \frac{t^j}{j!} \right)^2+\frac{t^j}{j!}\right)}~dt

    \frac{32}{2i+1} \oint_0^{2 \pi} \sum_{j=0}^{\infty} \frac{  \frac{t^j}{5(j!)}}{  \left( \left( \frac{t^j}{j!} \right)^2+\frac{t^j}{j!}\right)}~dt+\frac{3}{2i+1} \oint_0^{2 \pi}\frac{1}{ \sum_{j=0}^{\infty} \left( \left( \frac{t^j}{j!} \right)^2+\frac{t^j}{j!}\right)}~dt

    \frac{32}{2i+1} \oint_0^{2 \pi}\left( \frac{ \frac{1}{5}}{2}+\frac{ \frac{t}{5}}{t^2+t}+.... \right)~dt+\frac{3}{2i+1} \oint_0^{2 \pi}\sum_{j=0}^{\infty} \frac{j!^2}{t^{2j}+j! t^j}~dt

    \frac{32}{2i+1} \oint_0^{2 \pi}\left( \frac{ \frac{1}{5}}{2}+\frac{ 1}{5t(t+1)}+.... \right)~dt+\frac{3}{2i+1} \oint_0^{2 \pi}\left( \frac{1}{2}+\frac{1}{t^2+t}+\frac{4}{t^4+2t^2}+...  . \right)~dt

    \frac{32}{2i+1} \oint_0^{2 \pi}\left(\frac{1}{10}+\frac{ 1}{5t(t+1)}+.. \right)~dt+\frac{3}{2i+1} \oint_0^{2 \pi}\left( \frac{1}{2}+\frac{1}{t^2+t}+\frac{4}{t^4+2t^2}+.. \right)~dt

    However, \oint_\gamma \frac{1}{t^n}~dt=0 for n \in \mathbb{Z}, \ n \neq 1

    This means that the following terms are going to have some value:

    \frac{32}{2i+1} \oint_0^{2 \pi} \frac{1}{5t(t+1)}~dt+\frac{3}{2i+1} \oint_0^{2 \pi} \frac{1}{t(t+1)}~dt

    \frac{32}{5(2i+1)} \oint_0^{2 \pi} \frac{1+t-t}{t(t+1)}~dt+\frac{3}{2i+1} \oint_0^{2 \pi} \frac{1+t-t}{t(t+1)}~dt

    \frac{32}{5(2i+1)} \oint_0^{2 \pi} \frac{1}{t}-\frac{1}{t+1}~dt+\frac{3}{2i+1} \oint_0^{2 \pi} \frac{1}{t}-\frac{1}{t+1}~dt

    \frac{32}{5(2i+1)} \oint_0^{2 \pi} \frac{1}{t}~dt-\frac{32}{5(2i+1)} \oint_0^{2 \pi} \frac{1}{t+1}~dt+\frac{3}{2i+1} \oint_0^{2 \pi} \frac{1}{t}~dt-\frac{3}{2i+1}\oint_0^{2 \pi}\frac{1}{t+1}~dt

    \left( \frac{32}{5(2i+1)}+\frac{3}{2i+1} \right) \oint_0^{2 \pi} \frac{1}{t}~dt- \left(\frac{32}{5(2i+1)}+\frac{3}{2i+1} \right) \oint_0^{2 \pi} \frac{1}{t+1}~dt

    hmm, evaluating this will give me a \ln 0..... =S
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