Complex integral, 2 problems

**arbolis** · August 3rd 2010, 07:15 PM

1)I must calculate $\int _{|z|=1} \frac{\cos (e^{-z})}{z^2} dz$ . I'm not sure if I should see it as the real part of $\int _{|z|=1} \frac{e^ {i (e^{-z})}}{z^2} dz$ .
Anyway the problem is obviously when $z=0$ .
I get an infinite residue: $Res(f,z=0)=\lim _{z \to 0} \frac{\cos (e^{-z})}{z}=+\infty$ .

2)I'd like to get an example of a real integral where when I use complex analysis methods and I choose a contour of integration, I fall over a singularity on the real axis.

Thanks in advance.

**TheEmptySet** · August 3rd 2010, 08:17 PM

Notice that

$f(z)=\cos(e^z) \implies f(0)=\cos(0)$

so f does not have a zero at zero.

So to calculate the residue at 0 we have

$g(z)=\frac{\cos(e^z) }{z^2}$

So the residue is

$\displaystyle \lim_{z \to 0}\frac{d}{dz}z^2g(z)=\lim_{z\to 0}\frac{d}{dz}\cos(e^{z})=-\sin(1)$

**chiph588@** · August 3rd 2010, 08:21 PM

Originally Posted by arbolis

2)I'd like to get an example of a real integral where when I use complex analysis methods and I choose a contour of integration, I fall over a singularity on the real axis.

Thanks in advance.

Try $\displaystyle \int_{-\infty}^\infty \frac{\sin x}{x}dx$ by integrating the complex function $\displaystyle f(z)=\frac{e^{iz}}{z}$ .

**TheEmptySet** · August 3rd 2010, 08:47 PM

I have noticed that you have been calculating a lot of residues this help.

Suppose you have a complex function $f(z)$ that is holomorphic except for a finite number of poles. Now if $a$ is a pole of order k then $f(z)$ has a Laurent expansion at a that looks like

$\displaystyle f(z)=\sum_{n=-k}^{\infty}a_{n}(z-a)^{n}$
Then

$g(z)=(z-a)^{k}f(z)=\sum_{n=-k}^{\infty}a_{n}(z-a)^{n+k}$
The residue is the coefficient at $a_{-1}$ to recover this from $g(z)$ we take take $k-1$ derivatives to get

$\displaystyle \frac{d^{k-1}}{dz^{k-1}}g(z)=(k-1)!\sum_{n=-1}^{\infty}a_{n}(z-a)^{n+1}$

This gives that

$\displaystyle \frac{d^{k-1}}{dz^{k-1}}g(a)=\lim_{z \to a}(k-1)!\sum_{n=-1}^{\infty}a_{n}(a-a)^{n+1}=(k-1)!a_{-1}$

This gives that

$\text{Res}(f,z=a)=\lim_{z\to a}\frac{1}{(k-1)!}\frac{d^{k-1}}{dz^{k-1}}(z-a)^{k}f(z)$

**arbolis** · August 4th 2010, 08:49 AM

Thanks chip, I'll try it.
Thanks TheEmptySet, that was very useful. However the function f(z) should be $\cos (e^{-z})$ but it doens't change the fact that $f(0) \neq 0$ . Also, how did you know that g(z) had a pole of order 2? Is it because it gave me infinity when I thought it was of order 1?
Correct me if I'm wrong in the following: If a function f has a singularity at $z_0$ and if $\lim _{z \to z_0} (z-z_0) f(z) =\pm \infty$ then $z_0$ is either a pole of order greater than 1 or an essential singularity. On the other hand, if $\lim _{z \to z_0} (z-z_0) f(z) =0$ then f has a removable singularity at $z_0$ . While if $\lim _{z \to z_0} (z-z_0) f(z) =k$ with $k \in \mathbb{C}$ and $k \neq 0$ , then f has a pole of order 1 at $z_0$ .

**TheEmptySet** · August 4th 2010, 12:33 PM

For this problem we had a function of the form

$\displaystyle \frac{\cos(e^{-z})}{z^2}$ the numerator does not have a zero at z=0 and is analytic so the constant term of its power series must be non zero and the denominator has a zero or order 2. The first term in its Laurent series must look like $\frac{a_0}{z^2}$ So it has a pole of order 2.

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