MHB Why Does Theorem 1.9 Use f(t) Instead of f(γ(t))?

[Math Amateur]

I am reading John B. Conway's book, "Functions of a Complex Variable I" (Second Edition) ...

I am currently focussed on Chapter IV: Complex Integration ... Section 1: Riemann-Stieljes Integral ... ...

I need help in fully understanding the first example on page 63 ... ...

The the first example on page 63 read as follows:

https://www.physicsforums.com/attachments/7447My question is as follows:

In the example we are supposed to be following Theorem 1.9 (see below) where$$\int_a^b f \ d \gamma = \int_a^b f(t) \gamma' (t) \ dt$$... BUT ... (my confusion ...) ... ... Conway writes ...$$\int_{ \gamma } \frac{1}{z} \ dz = \int_0^{ 2 \pi } e^{ -it } ( i e^{ it} ) \ dt$$ but ... ? ... apart from the fact that $$dz$$ does not appear in Theorem 1.9 ... ( I know it appears in the notation for a line integral ... but ...? )... it seems to me that $$e^{ -it} = f( \gamma (t) )$$ ... ... but $$f( \gamma (t) )$$ does not appear in Theorem 1.9 ... ?... but instead $$f(t)$$ appears in Theorem 1.9 ...
Can someone clarify the above by explaining in some detail what is actually going on in the above example from Conway ...Help will be much appreciated ...

Peter=========================================================================

The above post mentions Theorem 1.9 (Ch.4) and the definition of a line integral ... so I am providing MHB readers with access to both ... as follows:View attachment 7448

View attachment 7449Peter

 

[HallsofIvy]

I am reading John B. Conway's book, "Functions of a Complex Variable I" (Second Edition) ...

I am currently focussed on Chapter IV: Complex Integration ... Section 1: Riemann-Stieljes Integral ... ...

I need help in fully understanding the first example on page 63 ... ...

The the first example on page 63 read as follows:

My question is as follows:

In the example we are supposed to be following Theorem 1.9 (see below) where$$\int_a^b f \ d \gamma = \int_a^b f(t) \gamma' (t) \ dt$$... BUT ... (my confusion ...) ... ... Conway writes ...$$\int_{ \gamma } \frac{1}{z} \ dz = \int_0^{ 2 \pi } e^{ -it } ( i e^{ it} ) \ dt$$ but ... ? ... apart from the fact that $$dz$$ does not appear in Theorem 1.9 ... ( I know it appears in the notation for a line integral ... but ...? )

The "dz" is the differential of z over the path [math]\gamma[/math] so [math]dz= d\gamma[/math]. Here, the path of integration is given as [math]z= \gamma(t)= e^{it}[/math]. Every complex number, z, can be written [math]z= re^{i\theta}[/math] with modulus r and argument [math]\theta[/math]. On this path, the unit circle, r= 1 and [math]z= e^{it}[/math]. [math]dz= d\gamma= ie^{it}dt[/math]. Of course, since [math]z= e^{it}[/math], [math]\frac{1}{z}= \frac{1}{e^{it}}= e^{-it}[/math] so the integral becomes [math]\oint_0^{2\pi}\frac{1}{z}dz= \int_0^{2\pi}e^{-it}e^{it}dt= \int_0^{2\pi} dt= 2\pi[/math][/quote]... it seems to me that $$e^{ -it} = f( \gamma (t) )$$ ... ... but $$f( \gamma (t) )$$ does not appear in Theorem 1.9 ... ?... but instead $$f(t)$$ appears in Theorem 1.9 ...[/quote]
"f" is the function to be integrated, here 1/z. The function to be integrated is completely independent of the path of integration.

Can someone clarify the above by explaining in some detail what is actually going on in the above example from Conway ...Help will be much appreciated ...

Peter=========================================================================

The above post mentions Theorem 1.9 (Ch.4) and the definition of a line integral ... so I am providing MHB readers with access to both ... as follows:

Peter

 

[Math Amateur]

The "dz" is the differential of z over the path [math]\gamma[/math] so [math]dz= d\gamma[/math]. Here, the path of integration is given as [math]z= \gamma(t)= e^{it}[/math]. Every complex number, z, can be written [math]z= re^{i\theta}[/math] with modulus r and argument [math]\theta[/math]. On this path, the unit circle, r= 1 and [math]z= e^{it}[/math]. [math]dz= d\gamma= ie^{it}dt[/math]. Of course, since [math]z= e^{it}[/math], [math]\frac{1}{z}= \frac{1}{e^{it}}= e^{-it}[/math] so the integral becomes [math]\oint_0^{2\pi}\frac{1}{z}dz= \int_0^{2\pi}e^{-it}e^{it}dt= \int_0^{2\pi} dt= 2\pi[/math]

... it seems to me that $$e^{ -it} = f( \gamma (t) )$$ ... ... but $$f( \gamma (t) )$$ does not appear in Theorem 1.9 ... ?... but instead $$f(t)$$ appears in Theorem 1.9 ...[/quote]
"f" is the function to be integrated, here 1/z. The function to be integrated is completely independent of the path of integration.[/QUOTE]
Thanks for the help, HallsofIvy ...

Appreciate your assistance ...

Peter