Integral of $\int_0^{2\pi} \frac{e^{-it }dt}{e^{it}-z}$

Question

Sorry if this question seems stupid, but I am confused here: Does it follow that $$ I(z)=\int_0^{2\pi} \frac{e^{-it}dt}{e^{it}-z} = 0 $$ For every $z$ with $|z|<1$? I think this is true.

I converted this integral to a line integral around the unit circle $S$: $$ I(z)=\int_S \frac{d\xi}{i\xi^2(\xi-z)} $$ I wanted to say the integrand has a primitive, but it seems I need logarithm to express it, which I am not quite sure can be holomorphic on $S$.

Answer 1

Using partial fractions we get: $\dfrac{1}{\xi^2(\xi-z)} = -\dfrac{1/z^2}{\xi}-\dfrac{1/z}{\xi^2}+\dfrac{1/z^2}{\xi-z}$

Therefore, $I(z) = \displaystyle\int_{S}\dfrac{\,d\xi}{i\xi^2(\xi-z)} = -\dfrac{1}{iz^2}\int_S\dfrac{\,d\xi}{\xi} - \dfrac{1}{iz}\int_S\dfrac{\,d\xi}{\xi^2}+\dfrac{1}{iz^2}\int_S\dfrac{\,d\xi}{\xi-z}$.

The first two of these three integrals are very straight forward to evaluate.

The third one is a direct application of Cauchy's Integral Formula. (Actually, all three are).

After evaluating these integrals, you do indeed get $0$ as the final answer.

Answer 2

Another take on it:

We have:

$I(z) = \int_0^{2\pi} \dfrac{e^{-it}dt}{e^{it} - z} = \int_0^{2\pi} \dfrac{e^{-2it}dt}{1 - e^{-it}z}, \tag{1}$

and since

$\vert e^{-it}z \vert = \vert e^{-it} \vert \vert z \vert = \vert z \vert < 1, \tag{2}$

we also have

$\dfrac{1}{1 - e^{-it}z} = \sum_0^\infty (e^{-it}z)^n = \sum_0^\infty e^{-int}z^n; \tag{3}$

the series on the right-hand side converges absolutely and uniformly by virtue of (2); thus we may exploit term-by-term integration in the evaluation of

$I(z) = \int_0^{2\pi} \dfrac{e^{-it}dt}{e^{it} - z} = \int_0^{2\pi} \dfrac{e^{-2it}dt}{1 - e^{-it}z} = \int_0^{2\pi} e^{-2it}(\sum_0^\infty e^{-int}z^n)dt$ $= \int_0^{2\pi} (\sum_0^\infty e^{-i(n + 2)t}z^n)dt = \sum_0^\infty \int_0^{2\pi}(e^{-i(n + 2)t}z^n)dt , \tag{4}$

and since

$\int_0^{2\pi}(e^{-i(n + 2)t}z^n)dt = z^n \int_0^{2\pi} e^{-i(n + 2)t}dt =  \dfrac{z^n}{-i(n + 2)}(e^{-i(n + 2)t} \mid_0^{2\pi}$ $ = \dfrac{z^n}{-i(n + 2)}(e^{-2(n + 2)\pi i} - e^0) = \dfrac{z^n}{-i(n + 2)}(1 - 1) = 0; \tag{5}$

it follows that $I(z) = 0$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!