 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2/18/c/solution

For this odd function, the Parseval identity from part (a) gives

\frac{1}{π} \int_{- π}^{π} (x^{3} - π^{2} x)^{2} d x = 144 \sum_{n = 1}^{\infty} \frac{1}{n ^{6}} .

The integrand is even. Expanding and integrating it gives

\frac{2}{π} \int_{0}^{π} (x^{6} - 2 π^{2} x^{4} + π^{4} x^{2}) d x = 2 π^{6} (\frac{1}{7} - \frac{2}{5} + \frac{1}{3}) = \frac{16 π ^{6}}{105} .

Dividing by

144

therefore proves

n = 1 \sum \infty \frac{1}{n ^{6}} = \frac{π ^{6}}{945} .

This evaluates the Riemann zeta function at six by using the norm of a simple polynomial.

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