## gerryliyana Group Title Using residu theorem, calculate integral one year ago one year ago

1. gerryliyana Group Title

Using residu theorem, calculate: $\int\limits_{0}^{\infty} \frac{ \cos 2x }{ 9 x^{2} + 4} dx$

2. satellite73 Group Title

i vaguely remember it is $$2\pi \sum\text{residues}$$

3. satellite73 Group Title

actually $$2\pi i$$

4. satellite73 Group Title

now how to compute the residues... zeros of the denominator are $$\frac{2}{3}i$$ and $$-\frac{2}{3}i$$

5. oldrin.bataku Group Title

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 Group Title

@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. satellite73 Group Title

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 Group Title

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 Group Title

$$\operatorname{Res}\left(f,-\frac23i\right)=\frac{\cos\left(-\frac43i\right)}{-12i}=\frac{i}{12}\cos\frac43i$$Our other residue simplifies similarly$$\operatorname{Res}\left(f,\frac23i\right)=\frac{\cos\frac43i}{12i}=-\frac{i}{12}\cos\frac43i$$

10. oldrin.bataku Group Title

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 Group Title

and then ?

12. oldrin.bataku Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

Oops, I meant $$1/(9a^2-4)$$ not $$1/(9a-4)$$.

17. oldrin.bataku Group Title

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 Group Title

why the residu is $$\frac{ e^{-4/3} }{ 12i }$$ ???

19. gerryliyana Group Title

ok thank you so much @oldrin.bataku :)

20. oldrin.bataku Group Title

@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 Group Title

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 Group Title

ok., thank you so much @oldrin.bataku :)...,

23. gerryliyana Group Title

@oldrin.bataku can you help me again on http://openstudy.com/study#/updates/5166a039e4b066fca6617d93 ???

24. gerryliyana Group Title

@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 Group Title

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 Group Title

though +1 for avoiding to explicitly need any estimates... I really should've remembered that $$\sin z$$ is even... :-)

27. gerryliyana Group Title

how if i |dw:1366116750127:dw|

28. oldrin.bataku Group Title

You only want either semicircle.

29. gerryliyana Group Title

|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 Group Title

$I_{C_{4}} I (R,\theta)$ $\lim_{R \rightarrow \infty} I (R,\theta) = 0 ==> Jordan's lemma$ $\int\limits_{C_{1}+C_{2}+C_{3}+C_{4}} f(z) dz = 0$ $f(z) = \frac{ e^{2iz} }{ z-i \frac{ 2 }{ 3 } }$ $\int\limits_{C_{1}} + \int\limits_{C_{2}} + \int\limits_{C_{3}} + \int\limits_{C_{4}} \frac{ e^{2iz} }{ (9z^{2} +4)} dz = 0$ $\int\limits_{-\infty}^{0} \frac{ f(z) }{ (....) } dz + \int\limits_{0}^{\infty} \frac{ f(z) }{ (....) } dz + \int\limits_{C_{2}} \frac{ f(z) }{ (.....) } dz = 0$ 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 Group Title

@oldrin.bataku correct me if i wrong..., i just wanna try with another way

32. gerryliyana Group Title

@oldrin.bataku still here ??