# Proving a conjecture in complex analysis

1. Jun 12, 2013

### Mandelbroth

How would I go about proving that, for a curve in the complex plane $\alpha$ and a real number $\beta$,

$$\exists\alpha,\beta: \frac{x}{2\pi i}\int\limits_\alpha \frac{\Gamma(z+\frac{1}{2})\Gamma(-z)x^{\beta z}}{\Gamma(\frac{3}{2}-z)}\, dz = \arctan{x}?$$

The poles of the integrand are $z_0\in\left\{-(n+\frac{1}{2}), n, n+\frac{3}{2}\right\}$ for a non-negative integer n. This is as far as I've gotten in terms of work. I can't think where to go from here. I don't know if it's true, but either way I intend to prove or disprove it.

And then, if I can prove the existence of a curve and a real number that satisfy the relation, how might I find those?

Thanks for any help in advance.

2. Jun 14, 2013

### Mandelbroth

Sorry. I'm bumping the thread. I still have no idea where to go. Any help is greatly appreciated.

3. Jun 14, 2013

### Office_Shredder

Staff Emeritus
It seems like the only way you are going to prove existence is to find a specific example. If you can calculate the residue at each pole as a function of x and beta, hope there's an obvious subset which when summed over yields arctan(x) (having a list of different infinite sums which yields arctan(x) would help for this part).

Thinking about it some more you might be able to get away with existence without construction if the sum of the residues diverges. For example if the nth pole had a residue of 1/n, and you wanted the integral to equal 2, then there is some combination of poles that you could use to get the residues to sum to 2 without explicitly knowing what they are. But in the complex plane this isn't always going to work - for example if every residue has a positive real and complex part, you can't add them up in a way to make the imaginary part go to zero to get to arctan(x). But if there's some nice sign switching going on it might be do-able

Last edited: Jun 14, 2013
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