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.

therefore

now, so long as , we can expand as a Taylor series:

applying to the integral:

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Tags: calculus, definite integral, integrals, logarithm, power series, Taylor series

This entry was posted on December 10, 2008 at 6:43 pm and is filed under problems and solutions. You can follow any responses to this entry through the RSS 2.0 feed.
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December 12, 2008 at 1:59 am

If x > 1 then the value of the integral is 4 pi log(x) – you can show this by writing

log(1 – xe^ith) = log(-xe^ith) + log(1 – 1/xe^ith) and using the power series for the last term which now converges for x > 1.

December 12, 2008 at 2:26 am

cool, thanks kiwi!

Working through it as you suggest, I at first got , so it looks like I need to be careful with the branch of the natural logarithm I’m choosing