Evaluate the following integral along circle with radius $1$ and center at $0$

Question

I know how to manipulate the top integral to the bottom, and from the ($n!$) term, I think we need to use the Cauchy Integral Formula, but I'm not sure how. I have an exam tomorrow so any help would be appreciated :)

Answer 1

I think you are on the right track. Indeed, we can use the Cauchy's integral formula to compute the above integral. By Cauchy's integral formula, if $f$ is holomorphic (analytic) inside and on a positively-oriented simple closed contour $C$ and $z_0$ is inside $C$, then $$f^{(2n)}(z_0) = \frac{(2n)!}{2 \pi i} \int_C \frac{f(z)}{(z - z_0)^{2n + 1}}$$

I will leave you to check $$\frac{\left(z + \frac{1}{z}\right)^{2n}}{z} = \frac{(1 + z^2)^{2n}}{z^{2n + 1}}$$

Hence after letting $f = (1 + z^2)^{2n}$, to compute the above integral, it suffices to compute $f^{(2n)}(0)$. Now the trick is to do a binomial expansion. By binomial theorem, $$(1 + z^2)^{2n} = {{2n}\choose{0}} z^{2(2n)} + {{2n}\choose{1}} z^{2(2n - 1)} + \dots + {{2n}\choose{n}} z^{2n} + \dots + {{2n}\choose{2n}}$$

Next, consider each term $z^k$. If $k < 2n$, then $(z^k)^{(2n)} = 0$. If $k > 2n$, then $(z^k)^{(2n)} = A z^{2n - k}$ for some $A \in \mathbb{N}$ where $2n - k > 0$. So $(z^k)^{(2n)}(0) = 0$. Hence after substituting $0$, the only non-vanishing term is ${{2n}\choose{n}} z^{2n}$.

I will leave you to check $$\left({{2n}\choose{n}} z^{2n}\right)^{(2n)} = \frac{(2n)!}{n! \cdot n!}(2n!) = \frac{((2n)!)^2}{(n!)^2}$$ which should give you the integral after substitution.