The definite integral to evaluate was:
I didn’t know how to do it, so I posted on the physics forum, and somebody named ‘adriank’ helped me out.
Here’s how it goes:
It would be hard to find a substitution to make this work directly. But if the logarithm were something more congenial, like
maybe we’d have a better shot.
You might be used to pulling this sort of trick:
We can do the same sort of thing by factoring that quadratic inside the logarithm. So we want to factor
Set it equal to zero and use the quadratic formula to find the roots.
now, so long as , we can expand as a Taylor series:
applying to the integral: