# Integration Error

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.

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 $$\displaystyle\oint_C$$.