# Having trouble evaluating this integral using the binomial theorem.

If I wanted to evaluate $\int_{C(0,1)}(z+\frac{1}{z})^{2n}\frac{1}{z}dz$ using the Binomial theorem what would my result be ?

So far I've rearranged the integral until we have $\int\frac{(z^2+1)^{2n}}{z^{2n+1}}dz$

Then by using the binomial theorem on the numerator we obtain

$(1+z^2)^{2n}=\sum_{k=0}^{2n} \binom{2n}{k}(z^2)^k=\binom{2n}{0}+\binom{2n}{1}z^2+...+ \binom{2n}{2n}(z^2)^{2n}$

Then I thought that I should decompose the original integral into a sum of new integrals and evaluate them using Cauchy's integral formula for derivatives, and it seemed like this would yield some fruitful result , but when i applied it I realised that if we take the derivative of $f(z_0)$ 2n times all but the final integral $\binom{2n}{2n}\int \frac{z^{2n+1}}{z^{2n+1}}dz$ will give zero as the result, using this approach.

This last integral then would give $\int dz=\int_{0}^{2\pi} ie^{it}dt$ after parametrisation which also yields zero ?

I don't feel like this is right at all, would anyone have any advice to guide me through where I'm going wrong, your help is much appreciated in advance.

• Think about residues. Note that all but one of the powers integrates to zero. – Cameron Williams Apr 18 '18 at 13:31
• @CameronWilliams we actually are'nt doing residues this year, My lecturer said that won't Come up until Complex analysis 2 – excalibirr Apr 18 '18 at 13:33
• What can you say about the integrals of $z^k$ for integer $k$ ? – Yves Daoust Apr 18 '18 at 13:34
• @CameronWilliams is there a way to do it without residues, We've been working onCauchy's formula and theorem , Taylor's theorem and the binomial theorem, so I think she wants us to just use those. – excalibirr Apr 18 '18 at 13:34
• @YvesDaoust that they can't be differentiated 2n times for any value except for k=2n ? – excalibirr Apr 18 '18 at 13:38

Hint: for $n\in\Bbb Z$, $n\ne -1$ the function $z\longmapsto z^n$ has primitive; for $n = -1$, calculate yourself $$\int_{|z| = 1}\frac1z\,dz$$ with the obvious parametrization $z = e^{i\theta}$ (also works for $n\ne -1$).