Tuesday, 6 August 2013

complex analysis - Proving intmathbbRfracsin(x)xdx=pi using the residue theorem




I've been searching the web for a way to prove that sin(x)/x=π with complex analysis, because I have a problem of consistency.




I found two, carried in the following link : Computing sinxxdx with residue calculus. But then I wanted to find it by using the fact that :



sin(x)/x=Im(eix/x)



To do so, I shifted the integration by i in the complex plane, showing that the integral on [,] is equal to the one on [i,i], since the integrand vanish on the vertical borders of the rectangle when we tend to infinity. I did this to get rid of the pole on the contour of integration.



Then, by using the residues theorem on the contour C1:z=yi,y[R,R] and C2:z=Reiti,t[0,π]. Given the integral on C2 vanishes, we have then that :



sin(x)/x=Im(eix/x)=Im(2πiRes(eiz/z,0))=2π




Which gives me the answer with a factor of 2. I don't understand were did I go wrong ? I think it has something to do with the fact that I take the imaginary part of the integral, but I don't really know... Can someone spot my mistake ?



If needed, I can provide further detail on my calculations.


Answer



The problem with your computation is that
iisin(x)xdxIm(iieixxdx)

since the integrand on the left has no singularities and the integrand on the right has one at 0. One way to get around this is to use Im(eix1x) which also has no singularities. Another is to use the whole of the sine function as I did in this answer to your cited question.






Using the contours
γ1=[R,R]
and
γ2=Rei[0,π]
Consider the integral
sin(x)xdx=Im(eix1xdx)=Im(γ1γ2eiz1zdxγ2eiz1zdz)

Since there are no singularities of eix1x, the integral over γ1γ2 is 0. However, the integral over γ2 is a bit more interesting.



Note that
|γ2eizzdz|π0eRsin(θ)dθ=2π/20eRsin(θ)dθ2π/20e2Rθ/πdθ=πR(1eR)πR0
and
γ21zdz=π01ReiθiReiθdθ=π0idθ=iπ
Putting together (3), (4), and (5), we get
sin(x)xdx=Im(iπ)=π


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find lim without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...