Complex Analysis Questions: Singularities and Integrals

[latentcorpse]

two questions here:

(i) my notes say that $\frac{1}{e^{\frac{1}{z}}-1}$ has an isolated singularity at $z=\frac{1}{2 \pi i n}, n \in \mathbb{Z} \backslash \{0\}$

i can't see this though...

(ii) let $b \in \mathbb{R}$. show

$\int_{-\infty}^{\infty} e^{-x^2} \cos{(2bx)} dx = e^{-b^2} \sqrt{\pi}$

we take a rectangular contour $\Gamma$ with vertices (R,0) , (R,b) , (-R,b) , (-R,0)
this integrand is holomorphic on the entire complex plane so Cauchy's Theorem tells us that

[itex]\int_{\Gamma}e^{-z^2} \cos{(2bz)} dz=0[/latex

now we define
$\gamma_1$ between (-R,0) and (R,0)
$\gamma_2$ between (R,0) and (R,b)
$\gamma_3$ between (R,b) and (-R,b)
$\gamma_4$ between (-R,b) and (-R,0)
so that $\Gamma= \gamma_1 \cup \gamma_2 \cup \gamma_3 \cup \gamma_4$

now on $\gamma_2$ we use Jordan's lemma

the length of the curve is b

and we have $|e^{-z^2} \cos{(2bz)}| \leq |e^{-z^2}|$ as $|\cos{(2bz)}| \leq 1$

we parameterise $\gamma_2$ by $z=R+it, t \in [0,b]$

so $|e^{-z^2}| = |e^{-(R+it)^2}| \leq e^{t^2-R^2} \leq e^{b^2-R^2}$

so $\int_{\gamma_2} e^{-z^2} \cos{(2bz)} dz \leq be^{b^2-R^2} \rightarrow 0$ as $R \rightarrow \infty$

and $\gamma_4$ vanishes simlarly

my problem is how to treat $\gamma_3$. is it meant to be rearranged so that we get the integral over $\gamma_3$ is proportional to the integral over $\gamma_1$. i was going to say it ISN'T because we want it to be non-zero so that the integral over $\gamma_1$ doesn't end up as 0.

this is what I've been trying:

parameterise $\gamma_3$ by $z=z+ib , x \in [-R,R]$

$\int_{\gamma_3} f(z) dz = \int_R^{-R} e^{-(x+ib)^2} \cos{(2b(x+ib))} dx$

do i continue down this road or am i supposed to be using Jordan's lemma here:

$|e^{-z^2} \cos{(2bz)}| \leq |e^{-z^2}| = |e^{-x^2} e^{-2ixb} e^{b^2}| = |e^{b^2-x^2}| \leq e^{b^2}$ as $x \in [-R,R]$ and the length of this contour is b so

we'd end up with our desired integral being equal to $-be^{b^2}$ which is a bit off

any advice...

 

[tiny-tim]

(i) my notes say that $\frac{1}{e^{\frac{1}{z}}-1}$ has an isolated singularity at $z=\frac{1}{2 \pi i n}, n \in \mathbb{Z} \backslash \{0\}$

Hi latentcorpse!

When z = 1/2nπi, 1/z = 2nπ1, so e^1/z = 1.

(ii) let $b \in \mathbb{R}$. show

$\int_{-\infty}^{\infty} e^{-x^2} \cos{(2bx)} dx = e^{-b^2} \sqrt{\pi}$

we take a rectangular contour …

Wouldn't it be easier to write cos(2bx) = Re(e^2bxi), and then complete the square ?

 

[latentcorpse]

of course it would lol. I'm such a tool. i literally just finished reading that part of my notes before trying this question and completely ignored them!
oh well ... learn by doing examples eh?