- #1

gysush

- 26

- 0

Integrate[ (cos(b*x)-cos(a*x))/(x^2), {x, -Infinity, Infinity}]

and show that it is equal to Pi*(a-b)

Attempt:

I first look at a related problem (or one that I think is related).

=> f(x) = sin(x)/x

=> f(z) = sin(z)/z

For this problem I can look at P[Integral(f(z)dz)]

then...sin(z)/z = [exp(i*z) - exp(-i*z)]/2*i*z = Integral 1 - Integral 2

where Integral 1 is the +i and Integral 2 is -i => this determines if we should close our contour in the UHP or LHP

Then Integral 1 = P[exp(i*z)/2*i*z] = Pi*i*Res(f(0))

Res(f(0)) is simple to calculate => 1/2*i

Likewise, Integral 2 = P[exp(-i*z)/2*i*z] = -Pi*i*Res(f(0))

Then Integral 1 - 2 = Pi*i/2*i - (-Pi*i/2*i) = pi

Upon inspection of my original integral I realize I have a 2nd order pole on the axis instead of a simple pole. Prof said the above method is only valid for simple poles. I then examine my notes/boas/arfken/leia for an example of 2nd order pole on real axis as in this problem...but have not been successful.

One example that I see is of form f(cos,sin)/(x^2 - a^2) which has two simple poles at a and -a

Another is of the form f(cos,sin)/(x^2 + a^2) which has two simple poles at +/- ia

For this integral, do not need principal value. We can consider, i.e. sin/(x^2 + a^2)...and from there calculate Integral[exp(iz)/z)...and set it equal to sum of residues etc...then separate out im/real parts and math them on LHS and RHS.

I'm kinda lost on the problem that I originally posted...