Answer: Integral

The definite integral to evaluate was:

\int_0^{2\pi} \log\left(1 + x^2 - 2x\cos\theta\right) d\theta

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

\int \log(1+x\theta)d\theta

maybe we’d have a better shot.

You might be used to pulling this sort of trick:

\log(24)=\log(2*2*2*3) = 3\log(2) + \log(3)

We can do the same sort of thing by factoring that quadratic inside the logarithm. So we want to factor

x^2 - 2x\cos\theta + 1

Set it equal to zero and use the quadratic formula to find the roots.

x^2 - 2x\cos\theta + 1 = 0
x = \frac{2\cos\theta \pm \sqrt{4\cos^2\theta - 4}}{2} = \cos\theta \pm \sqrt{-(1-\cos^2\theta)} = \cos\theta \pm i\sin\theta = e^{\pm i\theta}

therefore

\int_0^{2\pi} \log\left(1 + x^2 - 2x\cos\theta\right) d\theta
= \int_0^{2\pi} \log\left((x-e^{i\theta})(x-e^{-i\theta}) \right) d\theta
= \int_0^{2\pi} \log\left((1-xe^{i\theta})(1-xe^{-i\theta}) \right) d\theta
= \int_0^{2\pi} \left(\log(1-xe^{i\theta}) + \log(1-xe^{-i\theta})\right) d\theta

now, so long as \|t\|<1, we can expand \log(1-t) as a Taylor series:
\log(1-t) =-t - \frac{t^2}{2} - \frac{t^3}{3} - \ldots = \sum_{n=1}^{\infty}-\frac{t^n}{n}

applying to the integral:
\int_0^{2\pi} \left(\log(1-xe^{i\theta}) + \log(1-xe^{-i\theta})\right) d\theta
=\int_0^{2\pi} \sum_{n=1}^\infty \frac{-x^ne^{in\theta}}{n} + \sum_{n=1}^\infty \frac{-x^ne^{-in\theta}}{n} d\theta
= -\sum_{n=1}^\infty \frac{x^n}{n} \int_0^{2*\pi}e^{in\theta}+e^{-in\theta} d\theta
= -\sum_{n=1}^\infty \frac{x^n}{n} \int_0^{2*\pi}2*\cos(n\theta) d\theta = 0

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2 Responses to “Answer: Integral”

  1. kiwi Says:

    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.

  2. meichenl Says:

    cool, thanks kiwi!

    Working through it as you suggest, I at first got 4 \pi \ln x + 4 \pi^2 i, so it looks like I need to be careful with the branch of the natural logarithm I’m choosing

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