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Fourier integral help

  • Thread starter mhill
  • Start date
188
1
1. Homework Statement

calculate the fourier transform [tex] \int_{-\infty}^{\infty}dx e^{iux} (a+x)^{-1/2} [/tex] and [tex] \int_{-\infty}^{\infty}dx e^{iux} (a-x)^{-1/2} [/tex]

2. Homework Equations

see statement above

3. The Attempt at a Solution

i believe that making a change of variable (a+x)=t or (a-x)=t depending of the problem i can convert the initial problem into this one

[tex] \int_{-\infty}^{\infty}dx e^{iut} (t)^{-1/2} [/tex]

and using the property of the square root [tex] (-x)^{1/2}=ix^{1/2} [/tex] with "i" the root of -1 then this integral is equal to the real valued integral

[tex] \int_{0}^{\infty}dt sin(ut) (t)^{-1/2} [/tex] which is for me easier to calculate, my question is if the reasoning made is correct or if i forgot a constant in the results.

also using distribution theory and tables i reachet to the result

[tex] \int_{-\infty}^{\infty}dx e^{iux} (x)^{-1/2}=C(-iu)^{-1/2}sgn(u) [/tex] with 'C' a constant relating [tex] \Gamma (-1/2) [/tex] and [tex] -i\pi [/tex] sgn(x) is the sign function that values -1 for negative values and 1 for positive ones.
 
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