# Tag Archives: semi

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

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## Integrating a function around three branches using a semi-circle contour

[Ex] Watson’s complex integration $$\int^{\pi/2}_{0}\cos(nt)\cos^m(t)\,dt=\frac{\pi \Gamma(m+1)}{2^{m+1}\Gamma\left(\frac{n+m+2}{2}\right)\Gamma\left(\frac{2-n+m}{2}\right)}$$ $$\textit {solution}$$ Let us integrate the following function $$f(z) = z^{n-m-1}\left(1+z^2\right)^m$$ We choose the principle logarithm where $$\log(z) = \log|z|+\mathrm{Arg}(z)$$ Note that the function $z^{n-m-1} = e^{(n-m-1)\log(z)}$ will have a branch cut on the … Continue reading