Confirm that ∫∞0t−1sintdt=π/2.
The guide book I am using gives the following help:
Consider ∫γz−1eizdz, where for $0
Exercise IV.4.20. For r with $0
Using the hint, I know that ∫γz−1eizdz=0 for the Cauchy theorem, with which ∫[s,r]z−1eizdz+∫γrz−1eizdz+∫[−r,−s]z−1eizdz−∫γsz−1eizdz=0, but I do not know what else to do here, could someone help me please? Thank you very much.
Answer
Define a path in the Complex Plane:
Now consider ∫Ceizzdz=∫arceizzdz+∫Arceizzdz+∫−r−Reizzdz+∫Rreizzdz
By parametizing the integrals over the arcs, and letting r→0 for arc and R→∞ for Arc, we see that ∫arc→i∫0πdθ and ∫Arc→0.
So we have ∫Ceizzdz=PV∫∞−∞eizzdz−πi
Since the contour does not enclose any poles, the entire contour integral is 0.
So 0=PV∫∞−∞eizzdz−πi
And so ∫∞−∞sin(z)zdz=π
Lastly since sin(z)z is an even function:
∫∞0sin(z)zdz=π2
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