# Category Archives: Contour Integration

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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## Creating Difficult integrals by the residue theorem

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

## 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

[Ex 9] Watson’s complex integration $$\int^{\infty}_{-\infty}e^{-x^2}\,\cos(2ax)dx=e^{-a^2}\sqrt{\pi}$$ $$proof$$ Integrate the following function $$f(z) = e^{-z^2}$$ Use the following contour Note that the function is entire, hence $$\int^{L}_{-L} e^{-t^2}\,dt+\int^{L+ai}_{L} e^{-t^2}\,dt+\int^{-L}_{-L+ai} e^{-t^2}\,dt+\int^{-L+ai}_{L+ai}e^{-t^2}\,dt=0$$ For the forth integral use the substitution $x= t-ai$ $$\int^{-L+ai}_{L+ai}e^{-t^2}\,dt=\int^{-L}_{L}e^{-(x+ai)^2}\,dx=-e^{a^2}\int^{L}_{-L}e^{-x^2}\,e^{-2iax}dx$$ Take … Continue reading