Complex analysis - partial fraction expansion

In summary, you are trying to find a function which will match the summation at z=n, but you are not sure why the residue would show up in the other terms of the Laurent series. The series around ai is good only down to -ai, so the Laurent series might not converge to your desired function everywhere.
  • #1
yy205001
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Homework Statement


Show that:
Ʃ(-1)n/(n^2+a^2) (from n=0 to ∞) = pi/[asinh(pi*a)], a[itex]\neq in[/itex], n[itex]\in Z[/itex].


Homework Equations



f(z) = f(0) + Ʃbn(1/(z-an)+1/an) (from n=1 to ∞) , where bn is the residue of f(z) at an.

The Attempt at a Solution



The main problem is I don't how to pick the function f(z) to start with. I am thinking is pi*cot(pi*z) a good starting point? But in this case, we are dealing with the hyperbolic function, so coth will be a better choice??

Any help is appreciated!
 
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  • #2
First of all the sum is over [itex]n \in \mathbb{Z}[/itex], and the correct statement is
[tex]\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+a^2}=\frac{\pi}{a \sinh(a \pi)}.[/tex]
Now find a function with residua [itex](-1)^n[/itex] at [itex]z=n[/itex] for [itex]n \in \mathbb{Z}[/itex].

You are close with the choice of your function, but the one you gave has residua [itex]+1[/itex]. Multiply it with an entire function which takes the values [itex](-1)^n[/itex] at [itex]z=n \in \mathbb{Z}[/itex]. Then choose an appropriate contour to integrate over and finally deform it appropriately to use the residue theorem.
 
  • #3
From my notes, i found that Ʃ(-1)nf(n) (from -∞ to ∞) = - Ʃ Res(pi*csc(pi*z), zj) (poles zj of f(z)).

So, should I let f(n) = 1/(n2+a2) in this case?
 
  • #4
vanhees71 said:
First of all the sum is over [itex]n \in \mathbb{Z}[/itex], and the correct statement is
[tex]\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+a^2}=\frac{\pi}{a \sinh(a \pi)}.[/tex]
Now find a function with residua [itex](-1)^n[/itex] at [itex]z=n[/itex] for [itex]n \in \mathbb{Z}[/itex].

You are close with the choice of your function, but the one you gave has residua [itex]+1[/itex]. Multiply it with an entire function which takes the values [itex](-1)^n[/itex] at [itex]z=n \in \mathbb{Z}[/itex]. Then choose an appropriate contour to integrate over and finally deform it appropriately to use the residue theorem.

I've been trying myself to sort out this Laurent series stuff, but I'm pretty sure I'm not following this. Would you mind answering some questions?

Are we trying to find a function g(z) to expand in a Laurent series which will match the summation at z = n?

I understand that any appropriate g would have simple poles at ±ai, so would have just the ##a_{-1}## term in its Laurent series. But why would this residue necessarily show up in the other terms of the Laurent series?

Does this Laurent series converge to g everywhere? Somehow I'm thinking that the series around ai is good only down to -ai -- that is converges in a radius of 2a around the point +ai.

Or am I completely on the wrong track here?

If you can explain, I would very much appreciate it.
 

1. What is partial fraction expansion in complex analysis?

Partial fraction expansion in complex analysis is a method used to simplify rational functions. It involves decomposing a rational function into a sum of simpler functions, called partial fractions. This technique is commonly used in integration and solving differential equations.

2. How is partial fraction expansion different in complex analysis compared to real analysis?

In complex analysis, the partial fraction expansion involves complex numbers and variables, whereas in real analysis, it only involves real numbers and variables. This means that the partial fractions in complex analysis can have complex coefficients and the decomposition may also include complex roots.

3. What is the purpose of using partial fraction expansion?

The purpose of using partial fraction expansion is to simplify complex rational functions, making them easier to integrate or differentiate. It also helps in solving complex differential equations and evaluating complex integrals.

4. What are some common techniques used for partial fraction expansion?

Some common techniques used for partial fraction expansion include the Heaviside cover-up method, the method of undetermined coefficients, and the method of residues. Other techniques such as synthetic division and long division can also be used for partial fraction expansion.

5. Can partial fraction expansion be used for all rational functions in complex analysis?

Yes, partial fraction expansion can be used for all proper rational functions in complex analysis. However, some functions may require additional techniques or methods to fully decompose them into partial fractions.

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