# area defined as variable resedue question..

• Jan 16th 2010, 05:13 AM
transgalactic
area defined as variable resedue question..
i need to calculate this integral from plus to minus infinity$\displaystyle f(z)=\frac{z}{e^{2\pi iz^2}-1}\\$

in this area

$\displaystyle \gamma _r=\left \{ |z|=r \right \},n<r^2<n+1$

i need to find the points which turn to zero in the denominator
and non zero in the numerator.
i got two such points
$\displaystyle z=\pm \sqrt{n}$
by using this formula
$\displaystyle res(\sqrt{a})=\frac{p(a)}{q(a)'}$
$\displaystyle res(\sqrt{n})=\frac{1}{4\pi i}$
$\displaystyle res(-\sqrt{n})=\frac{1}{4\pi i}$

the third point is z=0 but for it we have both numerator and denominator 0
i calculated the residium for it by $\displaystyle res(f(x),a)=\lim_{x->a}(f(x)(x-a))$ formula
but then
my prof says some stuff that involves the area
he says that my points are 0 +1 -1 +2^(0.5) -2^(0.5) etc.. because the denominator goes to zero
for each point have a residiu and i need to sum the residiums inside.
but here the area is not defined
its not like (by radius 3)

i dont know what point are inside the area
• Jan 17th 2010, 03:07 AM
HallsofIvy
Quote:

Originally Posted by transgalactic
i need to calculate this integral from plus to minus infinity$\displaystyle f(z)=\frac{z}{e^{2\pi iz^2}-1}\\$

in this area

$\displaystyle \gamma _r=\left \{ |z|=r \right \},n<r^2<n+1$

This simply makes no sense. You say you want to integrate "from plus to minus infinity", but then you say that |z|= r. They can't both be true.

Perhaps you mean that you want to integrate around the circle z= |r|
And saying that "$\displaystyle n< r^2< n+1$" simply says that $\displaystyle \sqrt{n}< r< \sqrt{n+1}$

Quote:

i need to find the points which turn to zero in the denominator
and non zero in the numerator.
i got two such points
$\displaystyle z=\pm \sqrt{n}$
by using this formula
$\displaystyle res(\sqrt{a})=\frac{p(a)}{q(a)'}$
$\displaystyle res(\sqrt{n})=\frac{1}{4\pi i}$
$\displaystyle res(-\sqrt{n})=\frac{1}{4\pi i}$
$\displaystyle e^{2\pi i z^2}- 1$ is 0 only when $\displaystyle e^{2\pi iz^2}= 1$ which happens only when $\displaystyle 2\pi i z^2$ is a multiple of $\displaystyle 2\pi i$: $\displaystyle z^2= m$ for some integer m so the singularities are at $\displaystyle -\sqrt{m}$ and $\displaystyle \sqrt{m}$. If you are integrating around the path |z|= r, with $\displaystyle \sqrt{n}< r< \sqrt{n+1}$, then there are two poles for every positive integer $\displaystyle m\le n$

Quote:

the third point is z=0 but for it we have both numerator and denominator 0
i calculated the residium for it by $\displaystyle res(f(x),a)=\lim_{x->a}(f(x)(x-a))$ formula
but then
my prof says some stuff that involves the area
he says that my points are 0 +1 -1 +2^(0.5) -2^(0.5) etc.. because the denominator goes to zero
for each point have a residiu and i need to sum the residiums inside.
but here the area is not defined
its not like (by radius 3)

i dont know what point are inside the area