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- #1

- Jan 17, 2013

- 1,667

\(\displaystyle \frac{1}{2}\log^2\left(\phi\right)-\frac{1}{4}\log^2\left( \frac{1+\phi }{4}\right)-\arctan^2\left(\sqrt{\phi}\right)\)

where $\phi$ is the golden ratio .

The numeric value proved an equivalence to the value of the integral .

Can anybody simplify it a little bit , or should I leave it like this ?