- #1
mahler1
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Homework Statement .
Let ##\gamma_r:[0,\pi] \to \mathbb C## be given by ##\gamma_r(t)=re^{it}##. Prove that ##\lim_{r \to \infty} \int_{\gamma_r} \dfrac{e^{iz}}{z}dz=0##.
The attempt at a solution.
The only thing I could do was:
## \int_{\gamma_r} \dfrac{e^{iz}}{z}dz=\int_0^{\pi} \dfrac{e^{ire^{it}}ire^{it}}{re^{it}}dt##
This last expression equals to ##\int_0^{\pi} \dfrac{e^{ire^{it}}ire^{it}}{re^{it}}dt=\int_0^{\pi} e^{ire^{it}}idt##. At this point I got stuck, this expression ## e^{ire^{it}}## looks horrible to me and I don't know what to do with it.
The curve ##\gamma_r## is the upper half of the circle of radius ##r##, according to the statement as the radius of the circle tends to infinity, the value of the integral goes to zero. I would like to understand intuitively why this is the case and how the function ##\dfrac{e^{iz}}{z}## behaves.
I would appreciate some help and suggestions. Thanks in advance.
Let ##\gamma_r:[0,\pi] \to \mathbb C## be given by ##\gamma_r(t)=re^{it}##. Prove that ##\lim_{r \to \infty} \int_{\gamma_r} \dfrac{e^{iz}}{z}dz=0##.
The attempt at a solution.
The only thing I could do was:
## \int_{\gamma_r} \dfrac{e^{iz}}{z}dz=\int_0^{\pi} \dfrac{e^{ire^{it}}ire^{it}}{re^{it}}dt##
This last expression equals to ##\int_0^{\pi} \dfrac{e^{ire^{it}}ire^{it}}{re^{it}}dt=\int_0^{\pi} e^{ire^{it}}idt##. At this point I got stuck, this expression ## e^{ire^{it}}## looks horrible to me and I don't know what to do with it.
The curve ##\gamma_r## is the upper half of the circle of radius ##r##, according to the statement as the radius of the circle tends to infinity, the value of the integral goes to zero. I would like to understand intuitively why this is the case and how the function ##\dfrac{e^{iz}}{z}## behaves.
I would appreciate some help and suggestions. Thanks in advance.