# Monthly Archives: April 2017

Prove for $a,b > 0$ $$\int^{2\pi}_0\frac{dt}{a^2\cos^2 t +b^2\sin^2 t} = \frac{2\pi}{ab}$$ $$\textit{solution}$$ Let us integrate the following function $$f(z) = \frac{1}{z}$$ Around the ellipse $$\oint_{\gamma}f(z)\,dz =2\pi i\,\mathrm{Res}(f,0)$$ The parametrization of the ellipse $\gamma(t) = a\cos(t)+ib\sin(t)$ $$\oint_{\gamma}f(z)\,dz=\int^{2\pi}_0 \frac{-a\sin t+ib\cos t}{a\cos t+ib\sin t} … Continue reading Posted in Contour Integration | | Leave a comment ## Integrating a fraction of exponential and trignometric using rectangular contour [Ex 41 ] Watson’s complex integration$$\int^\infty_0 \frac{\sin(ax)}{e^{2\pi x}-1}\,dx = \frac{1}{4}\coth\left(\frac{a}{2} \right)-\frac{1}{2a}\textit{solution}$$By integrating the following function$$f(z) = \frac{e^{iaz}}{e^{2\pi z}-1}$$The function is analytic in and on the contour, indented at the poles of the function Hence by … Continue reading Posted in Contour Integration | | Leave a comment ## Integrating around a triangular contour for Fresnel integral$$\int^\infty_0\frac{\cos(x)}{\sqrt{x}}\,dx =\int^\infty_0\frac{\sin(x)}{\sqrt{x}}\,dx =\sqrt{\frac{\pi}{2}}\textit{solution}$$Consider the following function$$f(z)=z^{-1/2}\,e^{iz}$$Where we choose the principle root for $z^{-1/2}=e^{-1/2\log(z)}$. By integrating around the following contour$$\int_{C_r}f(z)\,dz+\int_{r}^R f(x)\,dx+\int_{\gamma}f(z)\,dz+\int^{iR}_{ir}f(x)\,dx = 0$$Taking the integral around the small quarter circle with r\to 0$$\left| … Continue reading

[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