Drew Williamson

2022-09-06

How can we evaluate

Elisa Spears

The $\sqrt{2}$ is pretty much irrelevant. Notice
$I={\int }_{0}^{\pi }\frac{dx}{1+\mathrm{tan}\left(x{\right)}^{\sqrt{2}}}={\int }_{\pi }^{0}\frac{d\left(\frac{\pi }{2}-u\right)}{1+\left(\mathrm{tan}\left(\frac{\pi }{2}-u\right){\right)}^{\sqrt{2}}}={\int }_{0}^{\pi }\frac{du}{1+\mathrm{cot}\left(u{\right)}^{\sqrt{2}}}$
and
$\frac{1}{1+\mathrm{tan}\left(x{\right)}^{\sqrt{2}}}=\frac{\mathrm{cos}\left(x{\right)}^{\sqrt{2}}}{\mathrm{cos}\left(x{\right)}^{\sqrt{2}}+\mathrm{sin}\left(x{\right)}^{\sqrt{2}}},\phantom{\rule{2em}{0ex}}\frac{1}{1+\mathrm{cot}\left(u{\right)}^{\sqrt{2}}}=\frac{\mathrm{sin}\left(u{\right)}^{\sqrt{2}}}{\mathrm{sin}\left(u{\right)}^{\sqrt{2}}+\mathrm{cos}\left(u{\right)}^{\sqrt{2}}}.$
so
$2I={\int }_{0}^{\pi }\frac{dv}{1+\mathrm{tan}\left(v{\right)}^{\sqrt{2}}}+{\int }_{0}^{\pi }\frac{dv}{1+\mathrm{cot}\left(u{\right)}^{\sqrt{2}}}$
$={\int }_{0}^{\pi }\left[\frac{\mathrm{cos}\left(v{\right)}^{\sqrt{2}}}{\mathrm{cos}\left(v{\right)}^{\sqrt{2}}+\mathrm{sin}\left(v{\right)}^{\sqrt{2}}}+\frac{\mathrm{sin}\left(v{\right)}^{\sqrt{2}}}{\mathrm{sin}\left(v{\right)}^{\sqrt{2}}+\mathrm{cos}\left(v{\right)}^{\sqrt{2}}}\right]dv={\int }_{0}^{\pi }\frac{\mathrm{cos}\left(v{\right)}^{\sqrt{2}}+\mathrm{sin}\left(v{\right)}^{\sqrt{2}}}{\mathrm{cos}\left(v{\right)}^{\sqrt{2}}+\mathrm{sin}\left(v{\right)}^{\sqrt{2}}}dv$
which is ${\int }_{0}^{\pi }1dv=\pi$. Ultimately, this is a symmetry argument.

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