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=y−i,y∈[−R,R] and C2:z=Reit−i,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
∫∞−i−∞−isin(x)xdx≠Im(∫∞−i−∞−ieixxdx)
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(eix−1x) 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(∫∞−∞eix−1xdx)=Im(∫γ1∪γ2eiz−1zdx−∫γ2eiz−1zdz)
Since there are no singularities of eix−1x, the integral over γ1∪γ2 is 0. However, the integral over γ2 is a bit more interesting.
Note that
|∫γ2eizzdz|≤∫π0e−Rsin(θ)dθ=2∫π/20e−Rsin(θ)dθ≤2∫π/20e−2Rθ/πdθ=πR(1−e−R)≤πR→0
and
∫γ21zdz=∫π01ReiθiReiθdθ=∫π0idθ=iπ
Putting together (3), (4), and (5), we get
∫∞−∞sin(x)xdx=Im(iπ)=π
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