anonymous
  • anonymous
Using residu theorem, calculate integral
Geometry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Using residu theorem, calculate: \[\int\limits_{0}^{\infty} \frac{ \cos 2x }{ 9 x^{2} + 4} dx \]
anonymous
  • anonymous
i vaguely remember it is \(2\pi \sum\text{residues}\)
anonymous
  • anonymous
actually \(2\pi i\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
now how to compute the residues... zeros of the denominator are \(\frac{2}{3}i\) and \(-\frac{2}{3}i\)
anonymous
  • anonymous
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.
anonymous
  • anonymous
@satellite73 is correct so far; our denominator factors like \(9x^2+4=(3x+2i)(3x-2i)\) so we have poles at \(x=\pm\frac23i\).
anonymous
  • anonymous
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
anonymous
  • anonymous
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.$$\operatorname{Res}\left(f,\frac23i\right)=\lim_{z\to\frac23i}\left(z-\frac23i\right)f(z)=\lim_{z\to\frac23i}\frac{\cos2z}{3(3z+2i)}=\frac{\cos\frac23i}{12i}$$
anonymous
  • anonymous
$$\operatorname{Res}\left(f,-\frac23i\right)=\frac{\cos\left(-\frac43i\right)}{-12i}=\frac{i}{12}\cos\frac43i$$Our other residue simplifies similarly$$\operatorname{Res}\left(f,\frac23i\right)=\frac{\cos\frac43i}{12i}=-\frac{i}{12}\cos\frac43i$$
anonymous
  • anonymous
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|
anonymous
  • anonymous
and then ?
anonymous
  • anonymous
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$$\int_{-a}^af(x)\,\mathrm{d}x=\frac\pi6\cos\left(\frac43i\right)-\oint_{\gamma_2}f(z)\,\mathrm{d}z$$
anonymous
  • anonymous
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 \(a\).
anonymous
  • anonymous
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 \(\int_0^\infty f(x)\mathrm{d}x\).
anonymous
  • anonymous
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$$\int_0^\infty f(x)\,\mathrm{d}x=\frac\pi{12}e^{-\frac43}$$
anonymous
  • anonymous
Oops, I meant \(1/(9a^2-4)\) not \(1/(9a-4)\).
anonymous
  • anonymous
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!
anonymous
  • anonymous
why the residu is \(\frac{ e^{-4/3} }{ 12i } \) ???
anonymous
  • anonymous
ok thank you so much @oldrin.bataku :)
anonymous
  • anonymous
@gerryliyana our residue is \(\frac1{12i}e^{-\frac43}\) and determined as follows:$$\begin{align*}\operatorname{Res}\left(f,\frac23i\right)&=\lim_{z\to\frac23i}\frac{e^{2iz}}{9z^2+4}\cdot\left(z-\frac23i\right)\\&=\lim_{z\to\frac23i}\frac{e^{2iz}}{(3z+2i)(3z-2i)}\cdot\left(z-\frac23i\right)\\&=\lim_{z\to\frac23i}\frac{e^{2iz}}{3(z-\frac23i)(3z-2i)}\cdot\left(z-\frac23i\right)\\&=\lim_{z\to\frac23i}\frac{e^{2iz}}{3(3z-2i)}\\&=\frac{e^{2i\cdot\frac23i}}{3(3\cdot\frac23i-2i)}\\&=\frac{e^{\frac43i^2}}{3(-4i)}\\&=\frac{e^{-\frac43}}{12i}\end{align*}$$
anonymous
  • anonymous
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.
anonymous
  • anonymous
ok., thank you so much @oldrin.bataku :)...,
anonymous
  • anonymous
@oldrin.bataku can you help me again on http://openstudy.com/study#/updates/5166a039e4b066fca6617d93 ???
anonymous
  • anonymous
@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 \[\int\limits_{C_{4}} \frac{ e^{i2z} }{ 9z^{2}+4 } dz = -2 \pi i (-1) \frac{ 3 }{ 4 } e^{-4/3} = -\frac{ 3 }{ 2 } \pi e^{-4/3} \] \[\int\limits_{-\infty}^{\infty} \frac{ (\cos 2x + i \sin 2x)}{ 9x^{2}+4 } dx = \frac{ 3 }{ 2} \pi e^{-4/3}\] \[2 \int\limits_{0}^{\infty} \frac{ \cos 2x }{ 9x^{2}+4 } dx = \frac{ 3 }{ 2 } \pi e^{-4/3}\] \[\int\limits_{0}^{\infty} \frac{ \cos 2x }{ 9x^{2} + 4 } dx = \frac{ 3 }{ 4 } \pi e^{-4/3} \]
anonymous
  • anonymous
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.
anonymous
  • anonymous
though +1 for avoiding to explicitly need any estimates... I really should've remembered that \(\sin z\) is even... :-)
anonymous
  • anonymous
how if i |dw:1366116750127:dw|
anonymous
  • anonymous
You only want either semicircle.
anonymous
  • anonymous
|dw:1366118390489:dw| Yes.., if i take countour an upper half plane to the singularity at z - i 2/3 \[\int\limits_{C_{1}+C_{2}+C_{3}+C_{4}} f(z) dz = 0\]
anonymous
  • anonymous
\[I_{C_{4}} I (R,\theta) \] \[\lim_{R \rightarrow \infty} I (R,\theta) = 0 ==> Jordan's lemma\] \[\int\limits_{C_{1}+C_{2}+C_{3}+C_{4}} f(z) dz = 0\] \[f(z) = \frac{ e^{2iz} }{ z-i \frac{ 2 }{ 3 } }\] \[\int\limits_{C_{1}} + \int\limits_{C_{2}} + \int\limits_{C_{3}} + \int\limits_{C_{4}} \frac{ e^{2iz} }{ (9z^{2} +4)} dz = 0\] \[\int\limits_{-\infty}^{0} \frac{ f(z) }{ (....) } dz + \int\limits_{0}^{\infty} \frac{ f(z) }{ (....) } dz + \int\limits_{C_{2}} \frac{ f(z) }{ (.....) } dz = 0\] then \[\int\limits_{-\infty}^{\infty} \frac{ e^{i2z} }{ 9z^{2}+4 } dz = -\int\limits_{C_{1}} \frac{ e^{2iz} }{ 9z^{2} +4} dz = - \int\limits \frac{ e^{2iz} }{ 9z^{2}+4 } dz\]
anonymous
  • anonymous
@oldrin.bataku correct me if i wrong..., i just wanna try with another way
anonymous
  • anonymous
@oldrin.bataku still here ??

Looking for something else?

Not the answer you are looking for? Search for more explanations.