How Can You Solve an Integral with a 2nd Order Pole on the Real Axis?

[gysush]

I'm working on a complex analysis problem from Arfken.

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...

 

[fzero]

Try integrating by parts once to get an integrand with only a simple pole.

 

[gysush]

Thank you!

 

[jackmell]

I'll give you a quick sketch. Perhaps you can go through it and fill in the details. Also, that's a removable singularity in the original problem. But how about considering the integral:

\int_C \frac{e^{iaz}-e^{ibz}}{z^2}dz

where C is the half-disc in the upper half-plane with an indentation around the now simple pole. We would then have:

\int_{-\infty}^0+\int_{\Gamma}+\int_{0}^{\infty}+\int_U=0 so that (after some justification):

\int_{-\infty}^{\infty}+\int_{\Gamma}=0

where the integral over U is the half-circle arc and gamma is the indentation at the origin and that is just -i\pi r

where:

r=\mathop\text{Res}_{z=0}\left\{\frac{e^{iaz}-e^{ibz}}{z^2}\right\}=i(a-b)

I'm assuming the sine terms drop out because they are odd but I've not gone through this carefully and the integral over the large semi-circle tends to zero but I've not checked that.

So assuming all that is ok, then we have:

\int_{-\infty}^{\infty}=-\pi(a-b)=\pi(b-a)

Note I'm getting a minus sign different from yours. You can look into that. I left out details that you can go through.

 

[gysush]

Yes, thank you as well. But the answer fell out almost trivially (since I already know Integral[sinx/x] once I did the integration by parts. Just had to make sure I took out the b or the a once I diff cos(ax)