Friday, November 1, 2019

integration - A Complex approach to sine integral

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:
+0exysinxdx=11+y2 and 00exysinxdydx 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+eix1xdx
this fact inspire me to consider the complex integral:Γeiz1zdz
Integral path

and Γ is the red path in the above figure, with counter-clockwise orientation, by Cauchy's theorem, we have
Γeiz1zdz=0 the above integral can be written as:ϵReix1xdx+Γϵeiz1zdz+Rϵeix1xdx+ΓReiz1zdz
Let R+ and ϵ0, we have:
ϵReix1xdx+Rϵeix1xdx+eix1xdx=2i+0sinxxdx
and Γϵeiz1zdz=0πeiϵeiθ1ϵeiθiϵeiθdθ=i0π(cos(ϵeiθ)+isin(ϵeiθ)1)dθ0 as ϵ0

so I am expecting that:ΓReiz1zdz=iπ when R+
but I can't find it. Could you help me? Thanks very much.

No comments:

Post a Comment

analysis - Injection, making bijection

I have injection f:AB and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...