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Sorry if this doesn't make any sense or if I did something obviously wrong, I was just playing around with taylor series' and then I got stuck.

I know from the geometric series that:

$$\large{\frac{1}{n}\sum_{k=1}^ne^{2\pi i \frac{k}{n}m}}=1_{n\mid m}$$

And letting $n$ get big I got:

$$\large{\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^ne^{2\pi i \frac{k}{n}m}}=1_{m=0}$$

And then I rewrote that as a Reimann sum:

$$\large{\int_{0}^1e^{2\pi i t m} dt}=1_{m=0}$$

Now I just supposed that some function $f$ could have a taylor series that represents it at all values $a$:

$$f(x+a)=\sum_{n=0}^\infty\frac{f^{n}(a)x^n}{n!}$$

Which means from the first integral that:

$$\int_{0}^1e^{-2\pi i t n}f(e^{2\pi i t}+a) dt=\frac{f^{n}(a)}{n!}$$

Cause all the other terms go away except for the $n^{th}$ one.

Then I tried to do integration by substitution and got:

$$z=e^{2\pi i t}+a$$ $$\frac{dz}{dt}=2\pi i e^{2\pi i t}=2\pi i(z-a)$$ $$\frac{1}{2\pi i}\frac{1}{(z-a)^{n+1}}\frac{dz}{dt}=\frac{1}{(z-a)^n}=e^{-2\pi i t n}$$

Which means that: $$\int_{t=0}^{t=1}\frac{n!}{2\pi i}\frac{f(z)}{(z-a)^{n+1}}dz=f^{n}(a)$$

But I don't know how to adjust the upper and lower bounds with out getting an integral that doesn't make any sense.

Cause at $t=0$ and at $t=1$ the function $e^{2\pi i t}$ has the same value.

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1 Answer 1

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What you have reached at is the Cauchy’s integral formula.

When doing integrals in the complex plane, it’s necessary to specify the whole path or curve along which the integral is evaluated, not merely the endpoints; otherwise the integral may be multi-valued.

We can let the path $C$ start from $1$, going along the unit circle counterclockwise, and ending in $1$. Then the formula can be written $$\int_C \frac{n!}{2\pi i}\frac{f(z)}{(z-a)^{n+1}}\,\mathrm{d}z = f^{(n)}(a).$$

Actually, since the path is closed, the exact endpoints does not matter. It suffices to say that $C$ is the unit circle counterclockwise, and the integration symbol can be replaced by $\displaystyle\oint_C$.

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