Integral sum thing 600

$\displaystyle \int_{0}^{1} \sum_{n=1}^{\infty} \frac{cos(\frac{\pi n}{5})}{n} dx = \sum_{n=1}^{\infty} \frac{cos(\frac{\pi n}{5})}{n}$ because there are no $x$ terms in the summation, so to the integral it is just a constant.

Note that $\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln(1-x)$ .

$\displaystyle \sum _{n=1}^{\infty } \frac{\cos(\frac{\pi n}{5})}{n}$ is the real part of $\displaystyle\sum _{n=1}^{\infty } \frac{e^{\frac{\pi ni}{5}}}{n}$ .

$\displaystyle\sum _{n=1}^{\infty } \frac{e^{\frac{\pi ni}{5}}}{n} = -\ln(1-e^{\frac{\pi i}{5}})$ .

To obtain the real part of a complex logarithm, we rewrite the complex number into polar form. Its modulus will be its real part and its argument will be its imaginary part.

$-\ln(1-e^{\frac{\pi i}{5}}) = -\ln(2\sin(\frac{\pi}{10})e^{\frac{9i\pi}{10}})$

The real part is $-\ln(2\sin(\frac{\pi}{10}))$ . We take out a factor of $\frac{1}{2}$ to get $-\frac{1}{2}\ln(4\sin^2(\frac{\pi}{10}))$ . Using double angle formulae and the fact that $\cos(\frac{\pi}{5}) = \frac{1+\sqrt{5}}{4}$ , we arrive at $\frac{1}{2}\ln(\frac{3+\sqrt{5}}{2})$ .