Best of MathNerds - The Integral Challenge

Hint 2

Let I = \int_0^{\pi/2} \ln (\sin \theta) \, d\theta. Using Hint 1, we have

\begin{eqnarray} I &=& \int_0^{\pi/2} \ln \left( 2 \sin \left( \frac{\theta}{2} \right) \cos \left( \frac{\theta}{2} \right) \right) \, d\theta \\ &=& \int_0^{\pi/2} \ln 2 \, d\theta + \int_0^{\pi/2} \ln \sin \left( \frac{\theta}{2} \right) \, d\theta + \int_0^{\pi/2} \ln \cos \left( \frac{\theta}{2} \right) \, d\theta. \end{eqnarray}

Now make the substitution \phi = \theta/2.

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The Rest of the Solution

Make the substitution \alpha = \pi/2 - \phi in the integral containing \cos \phi. Then we obtain:

\int_0^{\pi/4} \ln (\cos \phi) \, d\phi = -\int_{\pi/2}^{\pi/4} \ln (\sin \alpha) \, d\alpha = \int_{\pi/4}^{\pi/2} \ln (\sin \alpha) \, d\alpha

Hence equation (*) above becomes

\begin{eqnarray} I &=& \frac{\pi}{2} \ln 2 + 2\int_0^{\pi/4} \ln (\sin \phi) \, d\phi + 2\int_{\pi/4}^{\pi/2} \ln (\sin \alpha) \, d\alpha \\ &=& \frac{\pi}{2} \ln 2 + 2\int_0^{\pi/2} \ln (\sin \alpha) \, d\alpha \\ &=& \frac{\pi}{2} \ln 2 + 2I. \end{eqnarray}

Solving for I, we get

I = -\frac{\pi}{2} \ln 2.

Self-Similarity

Notes by Allen Stenger

This method works because there is a recursive structure to the function. This is easier to see if you work on the interval [0, \pi]: because the function is symmetric about \theta = \pi/2, the integral on this larger interval is twice the integral on [0, \pi/2], so it's really the same problem as before. Using the half-angle formulas as above we can write

\ln (\sin \theta) = \ln 2 + \ln \left( \sin \left( \frac{\theta}{2} \right) \right) + \ln \left( \sin \left( \frac{\pi}{2} - \frac{\theta}{2} \right) \right) .

This expresses the given function in terms of two stretched and shifted copies of itself: \ln ( \sin (\theta/2) ) on the interval [0, \pi] is a stretched version of the original curve on the interval [0, \pi/2], and \ln ( \sin (\pi/2 - \theta/2) ) is a flipped version of this stretched curve. The graphs of the three functions are shown below.

graph of log sine part 1

graph of log sine part 2

In geometric terms the three curves are similar, and because one curve can be made up from the others (plus a constant factor) we might say it is self-similar.

By integrating this recursion over [0, \pi] we get an implicit formula for I, namely

2I = \pi \ln 2 + 2I + 2I

(the first 2I comes from the symmetry over the interval [0, \pi] and the second and third 2I come because the functions are stretched out twice as wide and therefore have twice the area under the curve). This implicit formula is easy to solve, giving us

I = -\frac{\pi}{2} \ln 2

as before.

The idea of recursion is very important in many areas of mathematics, and the particular form called self-similarity is important in fractals. You can read some more about integrals, fractals, and self-similarity in the Reference by Strichartz.

Reference: Robert S. Strichartz, "Evaluating Integrals Using Self-Similarity". American Mathematical Monthly, volume 107 number 4 (April 2000), pp. 316-326.

The Integral Challenge

Hint 1

Hint 2

Hint 3

The Rest of the Solution

Self-Similarity