# Monthly Archives: May 2017

Prove that  $$\int^1_0 \frac{\cos(\log x)}{x^2+1}\,dx = \frac{\pi}{4}\mathrm{sech}\left( \frac{\pi}{2}\right)$$ First note that $$2 \int^1_0 \frac{\cos(\log x) }{x^2+1}\,dx = \int^\infty_0 \frac{\cos(\log x)}{x^2+1}\,dx$$ Integrate the following function $$f(z) = \frac{e^{i\log(z)}}{z^2+1}$$ Around a semi-circle in the upper half place. Where we avoid the branch … Continue reading
Theorem Let $f$ be analytic function in the unit circle $|z|\leq 1$  such that $f\neq 0$ . Then $$\int^{2\pi}_0f(e^{it})\,dt =2\pi \, f(0)$$ $$\textit{proof}$$ Since the function $f$ is analytic in and on the … Continue reading