Upper Bound for Integral of e^(iz)/z over a semicircle

Today, we show that the integral of e^(iz)/z over a semicircular arc goes to 0 as the radius goes to infinity.
Original video (sin(x)/x using complex analysis):

[Claudiostuff]

I was stuck in this problem. So glad I found your video! Very clear explanation, thanks

[devinbalian]

Using the symmetry of the function to go from [0, pi] to [0, pi/2] was so helpful, thanks!

[angelmendez-rivera]

To do the transformation from the integral on [0, π] to [0, π/2] rigorously, you can add the integral on [0, π/2] to the integral on [π/2, π]. To demonstrate that the integral on [π/2, π] is equal to the integral on [0, π/2], notice that exp[–R·sin(θ)] = exp[–R·sin(θ – π/2 + π/2)]. Let θ – π/2 |—> θ, hence [π/2, π] |—> [0, π/2], and exp[–R·sin(θ)] |—> exp[–R·sin(θ + π/2)]. Since sin(θ + π/2) = sin(π/2 – θ), exp[–R·sin(θ + π/2)] = exp[–R·sin(π/2 – θ)], and by the King's property of the integral, the integral of exp[–R·sin(π/2 – θ)] on [0, π/2] is equal to the integral of exp[–R·sin(θ)] on [0, π/2]. Therefore, the integral of exp[–R·sin(θ)] on [0, π] is equal to the integral of 2·exp[–R·sin(θ)] on [0, π/2]. Q. E. D.

In fact, a more general theorem can be proven with this method. Given some Riemann integrable f : [–1, 1] —> R, let ° denote functional composition, and so, the integral of f°sin on [0, π] is equal to the integral of 2·f°sin on [0, π/2].