Calculate .

How should I start?

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- September 6th 2011, 05:55 AMmath2011An integration problem in Fourier transform inversion
Calculate .

How should I start? - September 6th 2011, 06:15 AMchisigmaRe: An integration problem in Fourier transform inversion
You should start considering that is...

(1)

... and then compute the integral (1) integrating the complex variable function...

(2)

... along the path in the figure...

http://digilander.libero.it/luposabatini/MHF58.bmp

... if R tends to infinity. For doing that the Cauchy integral formula may be useful...

Kind regards

- September 7th 2011, 07:13 AMmath2011Re: An integration problem in Fourier transform inversion
Thank you. I get it now. The contour integral over which does not enclose a pole will be 0 according to Cauchy integral theorem (Cauchy Integral Formula -- from Wolfram MathWorld). Hence the countour integral over is equal to the countour integral of the semicircle in red ( and ). Then we can use either the Cauchy integral formula or the Residue formulas to calculate the semicircle countour integral. The Residue formula could be more convenient sometimes as we won't have to do partial fractions/complete square that we sometimes need to do when using Cauchy integral formula approach.

Thanks very much for helping me. (I didn't realize this is a complex integration and was trying to solve it as a real integral lol.) - September 7th 2011, 12:18 PMchisigmaRe: An integration problem in Fourier transform inversion
The formula I spoke about gives the integral of an f(z) along a closed path ...

(1)

... where the are the residues of the poles of f(z) inside . In Your case the only pole of f(z) inside is at z=j [see figure]...

http://digilander.libero.it/luposabatini/MHF58.bmp

... and the residue is...

(2)

If R tends to infinity the integral along the 'bif half circle' tends to 0 so that the result is...

(3)

Kind regards

- September 8th 2011, 06:22 AMmath2011Re: An integration problem in Fourier transform inversion
Thanks for writing out the solutions. I calculated without changing the sign in the beginning, i.e. I evaluated the LHS of below equation that you wrote.

The solution I got for the last step is instead of .

Why does equation (1) hold? How can the LHS and RHS yield different results if they are equal? - September 8th 2011, 06:55 AMchisigmaRe: An integration problem in Fourier transform inversion
Because and have the same real part and the inaginary part of Your integral is 0 the substituion is perfectly possible... but why the substitution?... the reason is that using the value of the function on the 'upper big half circle' is and it tends to 0 if R tends to infinity, so that the Jordan's lemma is satisfied...

Kind regards

- September 10th 2011, 09:46 AMmath2011Re: An integration problem in Fourier transform inversion
Thank you for explaining. It all makes sense now.