Fourier Transformation integral

rugapark
Messages
22
Reaction score
0
I'm trying to integrate a function which is given as

F(u)= \int f(x)e^{-2}^{\pi} ^{i} ^{x} ^{u} dx

with limits of +ve and -ve infinity

integrating by parts gives me

\frac{1}{2} f(x)^{2}e^{-2}^{\pi}^{i}^{x}^{u}-\frac{1}{2} \int f(x)^{2}xe^{-2}^{\pi}^{i}^{x}^{u}dx

fisrt off, is the i in the equation an imaginary number ( I am pretty sure it is)
and after integration how do I apply the infinite limits, also is my integration correct?

cheers
 
Physics news on Phys.org
Hi Ruga,

rugapark said:
fisrt off, is the i in the equation an imaginary number ( I am pretty sure it is)
Yes, i = \sqrt{-1}.

rugapark said:
and after integration how do I apply the infinite limits, also is my integration correct?
Without knowing what f(x) is, your integral is not correct. In general, \int{f(x)} dx will not resemble \frac{1}{2}f(x)^2 (or do you know that f(x) = x?), so choosing v' = f(x) in the integration by parts method does not yield v = \frac{1}{2}f(x)^2.

Could you provide some context (e.g., the complete problem statement) as we may find the requisite expression for f(x) there. If you're just trying to evaluate the integral in full generality, then we're out of luck!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top