Question

Using residu theorem, calculate integral

Answer 1

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

Answer 2

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

Answer 3

actually \(2\pi i\)

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

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

Answer 8

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.$$\operatorname{Res}\left(f,\frac23i\right)=\lim_{z\to\frac23i}\left(z-\frac23i\right)f(z)=\lim_{z\to\frac23i}\frac{\cos2z}{3(3z+2i)}=\frac{\cos\frac23i}{12i}$$

Answer 9

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

Answer 10

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 \(a\).Notice that we may draw a counterclockwise arc from \(-a\) to \(a\) so that we have a semicircle: