Re: MX2 2016 Integration Marathon
Okay, I think it's been long enough so I will post my solution.
If you feel you are making progress then keep trying yourself before reading the below!!
For starters, let's assume a and b are positive, as changing the sign of one of these guys literally just changes the sign of the integral, as sine is odd.
by basic trigonometric manipulations (specifically, product to differences and the half angle formula.) The motivation for the second manipulation is that the denominator tends to zero as x -> 0. This means that if we want to split the integral up as a sum or difference, the parts of the numerator should each tend to zero as well. so sines are good but cos's are bad.
So we have now reduced it to studying integrals of the form
where the third inequality came from integrating by parts, noting that the boundary term decays as 1/R and hence vanishes in the limit.
Now it is tempting to conclude that this final quantity is nothing more than the principal value integral of sin(x)/x over the real line. HOWEVER, there is a subtlety caused by c's sign! Note that if c is negative, then this final integral is negatively oriented! This means that in this case we will actually get the negative of the principal value integral. Hence we have
where the sgn function is defined to be 1 on the positive reals, -1 on the negative reals, and 0 at zero.
Plugging this back into our most recent expression for I, we get
The surprising upshot: the integral does not vary if you change the larger of the two parameters, but it does if you change the smaller one!