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gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
Using residu theorem, calculate: \[\int\limits_{0}^{\infty} \frac{ \cos 2x }{ 9 x^{2} + 4} dx \]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i vaguely remember it is \(2\pi \sum\text{residues}\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
actually \(2\pi i\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
now how to compute the residues... zeros of the denominator are \(\frac{2}{3}i\) and \(\frac{2}{3}i\)
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
@satellite73 is correct so far; our denominator factors like \(9x^2+4=(3x+2i)(3x2i)\) so we have poles at \(x=\pm\frac23i\).
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
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.$$\operatorname{Res}\left(f,\frac23i\right)=\lim_{z\to\frac23i}\left(z\frac23i\right)f(z)=\lim_{z\to\frac23i}\frac{\cos2z}{3(3z+2i)}=\frac{\cos\frac23i}{12i}$$
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
$$\operatorname{Res}\left(f,\frac23i\right)=\frac{\cos\left(\frac43i\right)}{12i}=\frac{i}{12}\cos\frac43i$$Our other residue simplifies similarly$$\operatorname{Res}\left(f,\frac23i\right)=\frac{\cos\frac43i}{12i}=\frac{i}{12}\cos\frac43i$$
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
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:dwNotice that we may draw a counterclockwise arc from \(a\) to \(a\) so that we have a semicircle:dw:1365731697019:dw
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
and then ?
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
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$$\int_{a}^af(x)\,\mathrm{d}x=\frac\pi6\cos\left(\frac43i\right)\oint_{\gamma_2}f(z)\,\mathrm{d}z$$
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
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 upperbound \(M\) 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, onehalf the circumference of a circle with radius \(a\).
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
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 \(\int_0^\infty f(x)\mathrm{d}x\).
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
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 $$\left\frac{e^{2ix}}{9z^2+4}\right\le\frac1{9z^24}=\frac1{9a4}$$ so we may let \(M=\frac1{9a4}\) along \(\gamma_2\). The estimation lemma then tells us$$\left\oint_{\gamma_2}f(z)\,\mathrm{d}z\right\le\frac{\pi a}{9a^24}$$ We notice that as \(a\to\infty\), \(\frac{\pi a}{9a^24}\to0\) 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$$\int_0^\infty f(x)\,\mathrm{d}x=\frac\pi{12}e^{\frac43}$$
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
Oops, I meant \(1/(9a^24)\) not \(1/(9a4)\).
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
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!
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
why the residu is \(\frac{ e^{4/3} }{ 12i } \) ???
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
ok thank you so much @oldrin.bataku :)
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
@gerryliyana our residue is \(\frac1{12i}e^{\frac43}\) and determined as follows:$$\begin{align*}\operatorname{Res}\left(f,\frac23i\right)&=\lim_{z\to\frac23i}\frac{e^{2iz}}{9z^2+4}\cdot\left(z\frac23i\right)\\&=\lim_{z\to\frac23i}\frac{e^{2iz}}{(3z+2i)(3z2i)}\cdot\left(z\frac23i\right)\\&=\lim_{z\to\frac23i}\frac{e^{2iz}}{3(z\frac23i)(3z2i)}\cdot\left(z\frac23i\right)\\&=\lim_{z\to\frac23i}\frac{e^{2iz}}{3(3z2i)}\\&=\frac{e^{2i\cdot\frac23i}}{3(3\cdot\frac23i2i)}\\&=\frac{e^{\frac43i^2}}{3(4i)}\\&=\frac{e^{\frac43}}{12i}\end{align*}$$
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
ok., thank you so much @oldrin.bataku :)...,
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
@oldrin.bataku can you help me again on http://openstudy.com/study#/updates/5166a039e4b066fca6617d93 ???
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
@oldrin.bataku , i've tried... \[\int\limits \frac{ e^{i2z} }{ 9z^{2}+4 } =2 \pi i \alpha_{1}\] \[f(z) \frac{ e^{i2z} }{ (z+i \frac{ 2 }{ 3 })(zi \frac{ 2 }{ 3 }) } \] where \((z+i \frac{ 2 }{ 3 })(zi \frac{ 2 }{ 3 }) \) is simply pole of singularity., then the residu is \[\alpha_{1} = \lim_{z \rightarrow i \frac{ 2 }{ 3 }} (z  i \frac{ 2 }{ 3 }) f(z)\] \[\alpha_{1} = \frac{ 1 }{ (m1) !} \frac{ d^{m1} }{ dz^{m1} }\left[ (zz_{o})^{m} f(z) \right]_{z=z_{o}} ::::>> m = 1\] \[\alpha_{1} = \left[ (zz_{o})f(z) \right]_{z=z_{o}} \] \[\alpha_{1} = \lim_{z \rightarrow i \frac{ 2 }{ 3 }} (zi \frac{ 2 }{ 3 }) \frac{ e^{i2z} }{ (z+i \frac{ 2 }{ 3 } )(zi \frac{ 2 }{ 3 }) }\] \[\alpha_{1} \lim_{z \rightarrow i \frac{ 2 }{ 3 }} \frac{ e^{i2z} }{ z+i \frac{ 2 }{ 3 } } = \frac{ e^{4/3} }{ i \frac{ 4 }{ 3 } } \frac{ i }{ i} = i \frac{ 3 }{ 4 } e^{4/3}\] then \[\int\limits_{C_{4}} \frac{ e^{i2z} }{ 9z^{2}+4 } dz = 2 \pi i (1) \frac{ 3 }{ 4 } e^{4/3} = \frac{ 3 }{ 2 } \pi e^{4/3} \] \[\int\limits_{\infty}^{\infty} \frac{ (\cos 2x + i \sin 2x)}{ 9x^{2}+4 } dx = \frac{ 3 }{ 2} \pi e^{4/3}\] \[2 \int\limits_{0}^{\infty} \frac{ \cos 2x }{ 9x^{2}+4 } dx = \frac{ 3 }{ 2 } \pi e^{4/3}\] \[\int\limits_{0}^{\infty} \frac{ \cos 2x }{ 9x^{2} + 4 } dx = \frac{ 3 }{ 4 } \pi e^{4/3} \]
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
Your denominator for your residue calculations is wrong. $$9z^2+4=9z^2(4)=(3z+2i)(3z2i)=9\left(z+\frac23i\right)\left(z\frac23i\right)$$Don't forget the \(9\) mate.
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
though +1 for avoiding to explicitly need any estimates... I really should've remembered that \(\sin z\) is even... :)
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
how if i dw:1366116750127:dw
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.2
You only want either semicircle.
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
dw:1366118390489:dw Yes.., if i take countour an upper half plane to the singularity at z  i 2/3 \[\int\limits_{C_{1}+C_{2}+C_{3}+C_{4}} f(z) dz = 0\]
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
\[I_{C_{4}} I (R,\theta) \] \[\lim_{R \rightarrow \infty} I (R,\theta) = 0 ==> Jordan's lemma\] \[\int\limits_{C_{1}+C_{2}+C_{3}+C_{4}} f(z) dz = 0\] \[f(z) = \frac{ e^{2iz} }{ zi \frac{ 2 }{ 3 } }\] \[\int\limits_{C_{1}} + \int\limits_{C_{2}} + \int\limits_{C_{3}} + \int\limits_{C_{4}} \frac{ e^{2iz} }{ (9z^{2} +4)} dz = 0\] \[\int\limits_{\infty}^{0} \frac{ f(z) }{ (....) } dz + \int\limits_{0}^{\infty} \frac{ f(z) }{ (....) } dz + \int\limits_{C_{2}} \frac{ f(z) }{ (.....) } dz = 0\] then \[\int\limits_{\infty}^{\infty} \frac{ e^{i2z} }{ 9z^{2}+4 } dz = \int\limits_{C_{1}} \frac{ e^{2iz} }{ 9z^{2} +4} dz =  \int\limits \frac{ e^{2iz} }{ 9z^{2}+4 } dz\]
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
@oldrin.bataku correct me if i wrong..., i just wanna try with another way
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.1
@oldrin.bataku still here ??
 one year ago
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