$\int^{2 \pi}_0 \frac{1}{ \sqrt{5}+\cos t}dt$, $\int^{2 \pi}_0 \frac{\cos^2t}{ 5-3\cos t}dt$ - Cauchy integral?

Question

Compute the integrals $$\int^{2 \pi}_0 \frac{1}{ \sqrt{5}+\cos t}dt$$ and $$\int^{2 \pi}_0 \frac{\cos^2t}{ 5-3\cos t}dt$$

I am stucked on these problems since a good while. Is there someone is able to compute these integrals?

Answer 1

There are several methods that work. For the first, we can use the eccentric anomaly $$\sin E=\frac{\sqrt{1-e^2}\sin t}{1+e\cos t}, \cos E=\frac{\cos t+e}{1+e\cos t}, dE=\frac{\sqrt{1-e^2}}{1+e\cos t}dE$$ So $$\int_9^{2\pi}\frac1{\sqrt5+\cos t}dt=\frac1{\sqrt5}\int_0^{2\pi}\frac{dE}{\sqrt{1-\frac15}}=\pi$$ For the other, $$\int_0^{2\pi}\frac{\cos^2t}{5-3\cos t}dt=\int_0^{2\pi}\left(-\frac13\cos t-\frac59+\frac{\frac{5}9}{1-\frac35\cos t}\right)dt$$ And wrap that third term up with the eccentric anomaly $E$ again.

Of course, since this is http://math.stackexchange.com, I should be doing the integrals via folding techniques, but I am too tired at this hour to present such a solution.

Answer 2

Like Cameron Williams says, you can perform these by contour integration.

Write $$ \int^{2 \pi}_0 \frac{1}{ \sqrt{5}+\cos t}\mathrm{d}t = \int^{2\pi}_0 \frac{\mathrm{d}t}{\sqrt{5} + \frac{1}{2} e^{it} + \frac{1}{2}e^{-it}} = \oint_{|z|=1} \frac{\mathrm{d}z}{iz \left (\sqrt{5} + \frac{z}{2} + \frac{1}{2z} \right )}, $$ where we have substituted $z = e^{it}$ in the last step, and the contour $|z|=1$ is the unit circle, which must be traversed counterclockwise once. Simplifying the denominator gives us $$ = -2i\oint_{|z|=1} \frac{\mathrm{d}z}{\left ( z^2 + 2 \sqrt{5}\; z + 1 \right )}. $$ We find that the quadratic $z^2 + 2\sqrt{5}\;z + 1$ has roots $-\sqrt{5} \pm 2$, and the root $z_+=-\sqrt{5}+2$ has $|z_+|< 1$. So the integral picks up the residue at that root, $$ =(-2i)(2\pi i)\frac{1}{-\sqrt{5}+2-(-\sqrt{5}-2)}=\frac{4\pi}{4}=\pi. $$

For the second one, once again you could use contour integration, $$ \int^{2 \pi}_0 \frac{\cos^2t}{ 5-3\cos t}\mathrm{d}t = \oint_{|z|=1} \frac{\mathrm{d}z}{iz} \frac{z^4 + 2z^2+1}{2(10z^2-3z^3-3z)} = \oint_{|z|=1} \frac{\mathrm{d}z}{-2iz^2} \frac{(z^2+1)^2}{(z-3)(3z-1)}. $$ It's important to note that since $(z^2+1)=(z+i)(z-i)$, no cancellations ensue. The poles in the unit disk are a double pole at $z=0$, and a pole at $z = 1/3$. $$ = \frac{2\pi i}{-2i} \left (10 -\frac{25}{54} \right )=-\frac{515}{54}\pi $$ PS: I might have made algebra / arithmetic mistakes. Note that for the double pole at zero, you need the derivative at $z = 0$ of the non-singular term, which may be found by using the binomial approximation and keeping terms to first order in $z$. See Cauchy's Integral Formula for more.

Answer 3

I'm nearly certain that there exists a simpler way, but a messy / brute-force way to solve the first integral might be as follows. Write, $$ \frac{1}{\sqrt{5} + \cos t} = \frac{1}{\sqrt{5}} \sum_{k = 0}^\infty \left (\frac{-1}{\sqrt{5}} \right )^k \cos^k t $$ The series should converge for real $t$ owing to $\cos t$ being bounded. Then, using this result, we may write $$ \frac{1}{\sqrt{5} + \cos t} = \frac{1}{\sqrt{5}} \sum_{k = 0}^\infty \left (\frac{-1}{\sqrt{5}} \right )^k \frac{1}{ 2^{k} } \sum_{n=0}^k \binom{k}{n} \cos ( (2n-k) t). $$ Now, it looks like we've created a mess, but observing that $$ \int_0^{2\pi} \cos( m x ) \;\mathrm{d}x = \cases{2\pi \qquad m=0\\0 \qquad m\neq 0}, $$ we see that only the $n = k/2$ terms can contribute, and these exist only if $k$ is even. Thus, the integral $$ \int_0^{2\pi} \frac{1}{\sqrt{5} + \cos t} = 2\pi \left [ 1 + \frac{1}{\sqrt{5}} \sum_{j = 0}^\infty \left ( \frac{-1}{2 \sqrt{5}} \right)^{2j} \binom{2j}{j} \right ] = 2\pi \left [ 1 + \frac{1}{\sqrt{5}} \sum_{j = 0}^\infty \left ( \frac{1}{20} \right)^{j} \binom{2j}{j} \right ] $$ I'm not sure where to go from here.