# Characteristic function in prbability

1. Mar 11, 2009

### O.J.

http://mathworld.wolfram.com/CharacteristicFunction.html
very neat. But then I tried out of boredom integrating the expression by parts where u = the exponential term and v = f (x) (or P(x)). The integral came out nicely as I got a term similar to the left hand side except with a different coefficient. Anyway, the evaluated integral was infinite, which contradicts with this link. Is there something wrong with my logic? Can any of you try evaluating the expression by integration by parts and show ur results? thank u

2. Mar 11, 2009

### John Creighto

I'm not sure what integral you are evaluating. From what I see a Taylor series expansion is done on the exponential function and the moments come out by definition. That said. I didn't really read the link.

3. Mar 11, 2009

### O.J.

The integral im talking about is the integral that defines the characteristic function. the definition.

4. Mar 11, 2009

### John Creighto

Show your steps. I think my previous comment still applies.

5. Mar 11, 2009

### O.J.

I don't know how to use that Latex math language :(....

6. Mar 11, 2009

### John Creighto

You should learn, it is easy.