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gerryliyana

  • 3 years ago

Using residu theorem, calculate integral

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  1. gerryliyana
    • 3 years ago
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    Using residu theorem, calculate: \[\int\limits_{0}^{\infty} \frac{ \cos 2x }{ 9 x^{2} + 4} dx \]

  2. anonymous
    • 3 years ago
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    i vaguely remember it is \(2\pi \sum\text{residues}\)

  3. anonymous
    • 3 years ago
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    actually \(2\pi i\)

  4. anonymous
    • 3 years ago
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    now how to compute the residues... zeros of the denominator are \(\frac{2}{3}i\) and \(-\frac{2}{3}i\)

  5. oldrin.bataku
    • 3 years ago
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    The residue theorem allows us to compute a contour integral using only the values of teh residues, more precisely:$$\oint_\gamma f(z)\mathrm{d}z=2\pi i+\sum_k\operatorname{Res}(f,a_k)$$ where \(a_k\) is the \(k\)-th residue.

  6. oldrin.bataku
    • 3 years ago
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    @satellite73 is correct so far; our denominator factors like \(9x^2+4=(3x+2i)(3x-2i)\) so we have poles at \(x=\pm\frac23i\).

  7. anonymous
    • 3 years ago
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    correct me if i am wrong but to find the residue of a simple pole, is it not the derivative evaluated at the singularity? it has been a while

  8. oldrin.bataku
    • 3 years ago
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    OOPS I meant \(2\pi i\sum_k\operatorname{Res}(f,a_k)\) -- a product, not a sum. Note we're dealing with an analytic continuation of our integrand in the complex plane, \(f(z)=\frac{\cos2z}{9z^2+4}\). Both of these poles are *simple* (order 1), so we may compute their residues as follows.$$\operatorname{Res}\left(f,\frac23i\right)=\lim_{z\to\frac23i}\left(z-\frac23i\right)f(z)=\lim_{z\to\frac23i}\frac{\cos2z}{3(3z+2i)}=\frac{\cos\frac23i}{12i}$$

  9. oldrin.bataku
    • 3 years ago
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    $$\operatorname{Res}\left(f,-\frac23i\right)=\frac{\cos\left(-\frac43i\right)}{-12i}=\frac{i}{12}\cos\frac43i$$Our other residue simplifies similarly$$\operatorname{Res}\left(f,\frac23i\right)=\frac{\cos\frac43i}{12i}=-\frac{i}{12}\cos\frac43i$$

  10. oldrin.bataku
    • 3 years ago
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    Okay, sorry. So recognize that our function is even, and thus \(\int_0^\infty f(x)\mathrm{d}x=\frac12\int_{-\infty}^\infty f(x)\mathrm{d}x\). The reasoning for this is that it is for our makes for a simple contour. We know our (new) integral can then be rewritten as follows.$$\int_{-\infty}^\infty f(x)\mathrm{d}x=\lim_{a\to\infty}\int_{-a}^af(x)\mathrm{d}x$$ Our definite integral inside our limit is equivalent to a contour integral of our analytic continuation \(f(z)\) along the real line from \(-a\) to \(a\).|dw:1365731595204:dw|Notice that we may draw a counterclockwise arc from \(-a\) to \(a\) so that we have a semicircle:|dw:1365731697019:dw|

  11. gerryliyana
    • 3 years ago
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    and then ?

  12. oldrin.bataku
    • 3 years ago
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    We're going to let \(a>\frac23\) so that our semicircle (which I will denote \(\gamma\)) encloses one of our simple poles in its interior, \(\frac23i\). Thus we can determine the contour integral along \(\gamma\) using Cauchy's residue theorem:$$\oint_\gamma f(z)\,\mathrm{d}z=2\pi i\left(-\frac{i}{12}\cos\left(\frac43i\right)\right)=\frac\pi6\cos\left(\frac43i\right)$$Now recognize that our semicircle \(\gamma\) may be broken into \(\gamma_1\) and \(\gamma_2\), our segment of the real line and our arc, respectively. It then follows that$$\oint_\gamma f(z)\,\mathrm{d}z=\oint_{\gamma_1} f(z)\,\mathrm{d}z+\oint_{\gamma_2} f(z)\,\mathrm{d}z$$Recalling that \(\oint_{\gamma_1}f(z)\,\mathrm{d}z=\int_{-a}^af(x)\mathrm{d}x\) and using our result using residues for \(\oint_\gamma f(z)\,\mathrm{d}z\), we have$$\int_{-a}^af(x)\,\mathrm{d}x=\frac\pi6\cos\left(\frac43i\right)-\oint_{\gamma_2}f(z)\,\mathrm{d}z$$

  13. oldrin.bataku
    • 3 years ago
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    Now let us consider \(\oint_{\gamma_2}f(z)\,\mathrm{d}z\) and look to the estimation lemma ( http://en.wikipedia.org/wiki/Estimation_lemma) for some help. Intuitively, we seek an upper-bound \(M\) on \(|f(z)|\) for all \(z\) along \(\gamma_2\) so that we may conclude \(|\oint_{\gamma_2}f(z)\,\mathrm{d}z|\le M\pi a \) where \(\pi a\) is te length of our arc, one-half the circumference of a circle with radius \(a\).

  14. oldrin.bataku
    • 3 years ago
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    hm, I'll admit I'm having a bit of trouble determining \(M\)... but once you do, you show that \(\lim_{a\to\infty}\oint_{\gamma_2}f(z)\mathrm{d}z\) converges to some number so that we may get an expression for \(\int_{-\infty}^{\infty}f(x)\,\mathrm{d}x\) and thus for \(\int_0^\infty f(x)\mathrm{d}x\).

  15. oldrin.bataku
    • 3 years ago
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    OOPS! I did my analytic continuation incorrect. Recognize \(\Re\{e^{2ix}\}=\cos 2x\), which means our extension should have been \(f(z)=\frac{e^{2iz}}{9z^2+4}\), so our residue is actually$$\frac{e^{-\frac43}}{12i}=-\frac{i}{12}e^{-\frac43}$$and thus$$\oint_\gamma f(z)\,\mathrm{d}z=2\pi i\times-\frac{i}{12}e^{-\frac43}=\frac\pi6e^{-\frac43}$$ Our bound is also now trivial, since \(|e^{2iz}|\le1,\,|z|=a\) and thus $$\left|\frac{e^{2ix}}{9z^2+4}\right|\le\frac1{9|z|^2-4}=\frac1{9a-4}$$ so we may let \(M=\frac1{9a-4}\) along \(\gamma_2\). The estimation lemma then tells us$$\left|\oint_{\gamma_2}f(z)\,\mathrm{d}z\right|\le\frac{\pi a}{9a^2-4}$$ We notice that as \(a\to\infty\), \(\frac{\pi a}{9a^2-4}\to0\) and therefore \(\oint_{\gamma_2}f(z)\,\mathrm{d}z\to0\). Thus we have $$\int_{-\infty}^\infty f(x)\,\mathrm{d}x=\frac\pi6e^{-\frac43}$$and therefore$$\int_0^\infty f(x)\,\mathrm{d}x=\frac\pi{12}e^{-\frac43}$$

  16. oldrin.bataku
    • 3 years ago
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    Oops, I meant \(1/(9a^2-4)\) not \(1/(9a-4)\).

  17. oldrin.bataku
    • 3 years ago
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    sorry about that all, I've never taken a complex analysis course since I'm still in high school so I had to try to piece together what I've learned from the Internet over the years. I can try to rewrite my answer in a more clear form if you'd like!

  18. gerryliyana
    • 3 years ago
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    why the residu is \(\frac{ e^{-4/3} }{ 12i } \) ???

  19. gerryliyana
    • 3 years ago
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    ok thank you so much @oldrin.bataku :)

  20. oldrin.bataku
    • 3 years ago
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    @gerryliyana our residue is \(\frac1{12i}e^{-\frac43}\) and determined as follows:$$\begin{align*}\operatorname{Res}\left(f,\frac23i\right)&=\lim_{z\to\frac23i}\frac{e^{2iz}}{9z^2+4}\cdot\left(z-\frac23i\right)\\&=\lim_{z\to\frac23i}\frac{e^{2iz}}{(3z+2i)(3z-2i)}\cdot\left(z-\frac23i\right)\\&=\lim_{z\to\frac23i}\frac{e^{2iz}}{3(z-\frac23i)(3z-2i)}\cdot\left(z-\frac23i\right)\\&=\lim_{z\to\frac23i}\frac{e^{2iz}}{3(3z-2i)}\\&=\frac{e^{2i\cdot\frac23i}}{3(3\cdot\frac23i-2i)}\\&=\frac{e^{\frac43i^2}}{3(-4i)}\\&=\frac{e^{-\frac43}}{12i}\end{align*}$$

  21. oldrin.bataku
    • 3 years ago
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    Oops, I broke that derivation. It's supposed to end up cancelling \(z-\frac23i\) to yield \(3(3z+2i)\) in the denominator, which then reduces to \(3(3\cdot\frac23i+2i)=3(4i)=12i\) -- no negative there.

  22. gerryliyana
    • 3 years ago
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    ok., thank you so much @oldrin.bataku :)...,

  23. gerryliyana
    • 3 years ago
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    @oldrin.bataku can you help me again on http://openstudy.com/study#/updates/5166a039e4b066fca6617d93 ???

  24. gerryliyana
    • 3 years ago
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    @oldrin.bataku , i've tried... \[\int\limits \frac{ e^{i2z} }{ 9z^{2}+4 } =-2 \pi i \alpha_{-1}\] \[f(z) \frac{ e^{i2z} }{ (z+i \frac{ 2 }{ 3 })(z-i \frac{ 2 }{ 3 }) } \] where \((z+i \frac{ 2 }{ 3 })(z-i \frac{ 2 }{ 3 }) \) is simply pole of singularity., then the residu is \[\alpha_{-1} = \lim_{z \rightarrow i \frac{ 2 }{ 3 }} (z - i \frac{ 2 }{ 3 }) f(z)\] \[\alpha_{-1} = \frac{ 1 }{ (m-1) !} \frac{ d^{m-1} }{ dz^{m-1} }\left[ (z-z_{o})^{m} f(z) \right]_{z=z_{o}} ::::>> m = 1\] \[\alpha_{-1} = \left[ (z-z_{o})f(z) \right]_{z=z_{o}} \] \[\alpha_{-1} = \lim_{z \rightarrow i \frac{ 2 }{ 3 }} (z-i \frac{ 2 }{ 3 }) \frac{ e^{i2z} }{ (z+i \frac{ 2 }{ 3 } )(z-i \frac{ 2 }{ 3 }) }\] \[\alpha_{-1} \lim_{z \rightarrow i \frac{ 2 }{ 3 }} \frac{ e^{i2z} }{ z+i \frac{ 2 }{ 3 } } = \frac{ e^{-4/3} }{ i \frac{ 4 }{ 3 } } \frac{ i }{ i} = -i \frac{ 3 }{ 4 } e^{-4/3}\] then \[\int\limits_{C_{4}} \frac{ e^{i2z} }{ 9z^{2}+4 } dz = -2 \pi i (-1) \frac{ 3 }{ 4 } e^{-4/3} = -\frac{ 3 }{ 2 } \pi e^{-4/3} \] \[\int\limits_{-\infty}^{\infty} \frac{ (\cos 2x + i \sin 2x)}{ 9x^{2}+4 } dx = \frac{ 3 }{ 2} \pi e^{-4/3}\] \[2 \int\limits_{0}^{\infty} \frac{ \cos 2x }{ 9x^{2}+4 } dx = \frac{ 3 }{ 2 } \pi e^{-4/3}\] \[\int\limits_{0}^{\infty} \frac{ \cos 2x }{ 9x^{2} + 4 } dx = \frac{ 3 }{ 4 } \pi e^{-4/3} \]

  25. oldrin.bataku
    • 3 years ago
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    Your denominator for your residue calculations is wrong. $$9z^2+4=9z^2-(-4)=(3z+2i)(3z-2i)=9\left(z+\frac23i\right)\left(z-\frac23i\right)$$Don't forget the \(9\) mate.

  26. oldrin.bataku
    • 3 years ago
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    though +1 for avoiding to explicitly need any estimates... I really should've remembered that \(\sin z\) is even... :-)

  27. gerryliyana
    • 3 years ago
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    how if i |dw:1366116750127:dw|

  28. oldrin.bataku
    • 3 years ago
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    You only want either semicircle.

  29. gerryliyana
    • 3 years ago
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    |dw:1366118390489:dw| Yes.., if i take countour an upper half plane to the singularity at z - i 2/3 \[\int\limits_{C_{1}+C_{2}+C_{3}+C_{4}} f(z) dz = 0\]

  30. gerryliyana
    • 3 years ago
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    \[I_{C_{4}} I (R,\theta) \] \[\lim_{R \rightarrow \infty} I (R,\theta) = 0 ==> Jordan's lemma\] \[\int\limits_{C_{1}+C_{2}+C_{3}+C_{4}} f(z) dz = 0\] \[f(z) = \frac{ e^{2iz} }{ z-i \frac{ 2 }{ 3 } }\] \[\int\limits_{C_{1}} + \int\limits_{C_{2}} + \int\limits_{C_{3}} + \int\limits_{C_{4}} \frac{ e^{2iz} }{ (9z^{2} +4)} dz = 0\] \[\int\limits_{-\infty}^{0} \frac{ f(z) }{ (....) } dz + \int\limits_{0}^{\infty} \frac{ f(z) }{ (....) } dz + \int\limits_{C_{2}} \frac{ f(z) }{ (.....) } dz = 0\] then \[\int\limits_{-\infty}^{\infty} \frac{ e^{i2z} }{ 9z^{2}+4 } dz = -\int\limits_{C_{1}} \frac{ e^{2iz} }{ 9z^{2} +4} dz = - \int\limits \frac{ e^{2iz} }{ 9z^{2}+4 } dz\]

  31. gerryliyana
    • 3 years ago
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    @oldrin.bataku correct me if i wrong..., i just wanna try with another way

  32. gerryliyana
    • 3 years ago
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    @oldrin.bataku still here ??

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