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I need to compute the Cauchy Integral Formula $${1\over{2{\pi}i}}\int {f(z)\over(z-a)^{k+1}}$$

$$Where\, k=3\,and\,z=e^{sin(z)}\,at\, z=i$$

I need help figuring out how to do this exactly, I get to this point: $${1\over{2{\pi}i}}\int {e^{sin(z)}\over(z-i)^{4}}$$ From here I am unsure of what I am suppose to do to evaluate this integral. I have done the differentiation side and gotten this ugly thing: $$e^{sin(i)}(cos^3(i)-3sin(i)cos(i)-cos(i))\over 6$$ Basically I want to show that my integral equals the $3^{rd}$ Taylor coefficient but am unsure of how.

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    you mean $z-i$ not $i-a$. I have fixed this for you. – 2017-02-15

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To see this write out the Taylor series

$$f(z) = \sum_{n=0}^\infty {f^{(n)}(a)(z-a)^n\over n!}$$

Now divide and integrate term-by-term (you should have already proven this is a legal maneuver in class)

$${1\over 2\pi i}\oint {f(z)\over (z-a)^{k+1}}\,dz = \sum_{n=0}^\infty {1\over 2\pi i}\oint {f^{(n)}(a)(z-a)^{n-k-1}\over n!}$$

Since all terms except the $k=n$ term have analytic primitives on the entire curve, we see their integrals are zero, and we are left only with

$${1\over 2\pi i}\oint {f^{(k)}(a)\over k!(z-a)}={f^{(k)}(a)\over k!}$$

Now in your case $k=3$ and the result follows.

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    So I am plugging in my 3rd derivative into the integral, doing it over 6z-6i and integrating it? – 2017-02-15
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    @Heavenly96 you plug nothing in, you don't actually compute the integral, you use the theorem to allow you to just compute the right derivative, then divide by $3!$. – 2017-02-15
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    I have already computed the right derivative though, I was asked to use the actual integral formula this time, or am I overthinking what I need to do? If I've already solved the right side then I have it finished? As for proving the maneuver in class we haven't actually solved any Cauchy Integral problems, we were told to work on it outside of class and talk about it again on Thursday. – 2017-02-15
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    @heavenly96 with a function as ugly as the one you have, I cannot imagine they actually intend for you to go through a parameterization to actually compute this integral. If they are telling you to use the integral formula, then that's exactly what you need to do. I would say you are already finished. – 2017-02-15