Fourier Transformation integral

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SUMMARY

The discussion centers on the integration of the function F(u) defined as F(u) = ∫ f(x)e^{-2πixu} dx, with limits extending to positive and negative infinity. The participant initially attempts integration by parts, leading to an incorrect expression due to the unspecified nature of f(x). The imaginary unit i is confirmed as i = √-1, but the integration approach requires clarification of f(x) to ensure accuracy in the evaluation of the integral.

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rugapark
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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
 
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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!
 

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