this integral:
∫+∞0sinxxdx=π2
is very famous and had been discussed in the past days in this forum. and I have learned some elegant way to computer it. for example: using the identity:
∫+∞0e−xysinxdx=11+y2 and ∫∞0∫∞0e−xysinxdydx and Fubini theorem. the link is here:Post concern with sine integral
In this post, I want to discuss another way to computer it. since∫+∞0sinxxdx=12i∫+∞−∞eix−1xdx
this fact inspire me to consider the complex integral:∫Γeiz−1zdz
and Γ is the red path in the above figure, with counter-clockwise orientation, by Cauchy's theorem, we have
∫Γeiz−1zdz=0 the above integral can be written as:∫−ϵ−Reix−1xdx+∫Γϵeiz−1zdz+∫Rϵeix−1xdx+∫ΓReiz−1zdz
Let R→+∞ and ϵ→0, we have:
∫−ϵ−Reix−1xdx+∫Rϵeix−1xdx→∫+∞−∞eix−1xdx=2i∫+∞0sinxxdx
and ∫Γϵeiz−1zdz=∫0πeiϵeiθ−1ϵeiθiϵeiθdθ=i∫0π(cos(ϵeiθ)+isin(ϵeiθ)−1)dθ→0 as ϵ→0
so I am expecting that:∫ΓReiz−1zdz=−iπ when R→+∞
but I can't find it. Could you help me? Thanks very much.
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