0

I am finding difficulty with the following couple of problems

a) show that $$\int_{-\infty}^\infty \frac{\cos(\pi x)}{2x-1} \ne\frac{\pi}{2}$$

I've been trying to do this over a semi-circle, but keep getting the answer -pi instead of -pi/2. It is correct that the arc integrates to 0 right?

Any help would be greatly appreciated. Thank you!

Jan Eerland
  • 28,671
J. Bant
  • 561
  • Is that really the correct integrand? – mrf Nov 25 '15 at 14:19
  • should be (2x-1) integrand. Fixed now – J. Bant Nov 25 '15 at 14:19
  • Ok then, what function are you integrating, and over which curve? You probably need an indentation around $x=\frac12$ (which will only pick up half the residue in the limit). – mrf Nov 25 '15 at 14:21
  • Singularity at x = 1/2, simple pole there. So by residue thm, the integral is 2pii lim z->1/2 of (z-1/2)f(z) which is piie^i*pi/2 if im not mistaken – J. Bant Nov 25 '15 at 14:23
  • You still haven't said which function $f(z)$ you are integrating, or which curve you are integrating over (all needed for using the residue theorem). If you don't provide those, we can only guess where your error is. – mrf Nov 25 '15 at 14:24
  • f(z) = (e^ipiz)/ 2z-1. Integrating over semi-circle from -R to R in upper plane. Thank you for the help! – J. Bant Nov 25 '15 at 14:27
  • Well, then as I said already a number of comments ago, the integral along the indentation around $x=\frac12$ will only pick up half the residue at $\frac12$. See for example http://math.stackexchange.com/questions/319959 – mrf Nov 25 '15 at 14:30
  • Why do I need an indentation here? – J. Bant Nov 25 '15 at 14:36
  • $$\lim_{a\to\infty}\int_{-a}^{a}\frac{\cos(\pi x)}{2x-1}\space\text{d}x=\lim_{a\to\infty}\left[\frac{1}{2}\text{Si}\left(\frac{\pi}{2}-\pi x\right)\right]{-a}^{a}=\frac{1}{2}\lim{a\to\infty}\left[\text{Si}\left(\frac{\pi}{2}-\pi x\right)\right]_{-a}^{a}=-\frac{\pi}{2}$$ – Jan Eerland Nov 25 '15 at 18:13

0 Answers0