Calculating Fourier Integral with a Change of Variable | Homework Help

[mhill]

Homework Statement

calculate the Fourier transform $$\int_{-\infty}^{\infty}dx e^{iux} (a+x)^{-1/2}$$ and $$\int_{-\infty}^{\infty}dx e^{iux} (a-x)^{-1/2}$$

Homework Equations

see statement above

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

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

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

$$\int_{0}^{\infty}dt sin(ut) (t)^{-1/2}$$ 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

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

 

[mighty2000]

Your approach is correct. By making the change of variables, you have transformed the integrals into a more manageable form. However, you may need to be careful with the limits of integration when making the substitution. In the first integral, the limits should be from 0 to infinity and in the second integral, the limits should be from -infinity to 0. This is because when you substitute t = a + x, you are essentially shifting the integration limits as well. As for the constant, it will depend on the value of a. If a is positive, then the constant will be 1, but if a is negative, the constant will be -1. This can be seen by plugging in a = 1 into the original integrals and comparing it to the result you obtained using distribution theory. I hope this helps. Good luck with your calculations!