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I am going through some worksheets to study for an exam and one question is to use the mean value theorem:

$$u(a)=\frac{1}{V_r}\int_{B_r(a)}u$$

with a suitable harmonic function to show that:

$$ \int_0^{2\pi} \frac{2+\cos\phi}{5+4\cos\phi}\,\mathrm{d}\phi=\pi $$

How do I begin?

Fei Cao
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dan
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  • I haven't been able to find a harmonic function $u$ that works, but if you can such a $u$ so that $u(0,0)=\frac{1}{2}$ and $u\Big(\cos\theta,\sin\theta\Big)=\frac{2+\cos\theta}{5+4\cos\theta}$ then we can say $$\frac{1}{2}=u(0,0)=\frac{\int_{\gamma}u\mathrm{d}s}{\int_{\gamma}\mathrm{d}s}=\frac{1}{2\pi}\int_0^{2 \pi}u\big(\cos\theta,\sin \theta\big)\mathrm{d}\theta=\frac{1}{2\pi}\int_0^{2 \pi}\frac{2+\cos\theta}{5+4\sin\theta}\mathrm{d}\theta$$ Here, $\gamma$ is your standard parameterization of the unit circle $\big(\cos t,\sin t\big):t\in [0,2\pi)$. – user429040 Aug 16 '21 at 21:34
  • The function is equivalent to $\frac{2 \cos^2 \phi/2 + 1}{8 \cos^2 \phi/2 + 1}$. – Toby Mak Aug 18 '21 at 01:25

2 Answers2

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We can try Poisson kernel.

$$ P_r(\theta)=\frac{1-r^2}{1-2r\cos\theta+r^2} = \Re \left(\frac{1+re^{i\theta}}{1-re^{i\theta}}\right). $$

To have $\cos \theta$ on the numerator, we use $1+P_r(\theta)$. Then $$ 1+P_r(\theta) = \frac{2-2r\cos\theta}{1-2r\cos\theta+r^2} = 1+\Re \left(\frac{1+re^{i\theta}}{1-re^{i\theta}}\right). $$

Put $r=1/2$, we have $$ 1+P_{1/2}(\theta) = \frac{8-4\cos\theta}{5-4\cos\theta}.$$ Then we use $\theta+\pi$ in place of $\theta$. $$ 1+P_{1/2}(\theta+\pi)=\frac{8+4\cos\theta}{5+4\cos\theta} = 1+\Re \left(\frac{1-\frac12 e^{i\theta}}{1+\frac 12e^{i\theta}}\right). $$ Use $f(z) = 1+\frac{1-z/2}{1+z/2}$ on the neighborhood of the closed unit disk $\overline{\mathbb{D}}$, then we have by the Mean Value Theorem for $u(z)= \Re(f(z))$, $$ \frac1{2\pi} \int_0^{2\pi} \frac{8+4\cos\theta}{5+4\cos\theta} d\theta =u(0) = 2.$$ The result follows by dividing both sides by $4$.

Sungjin Kim
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2

The Poisson kernels or harmonic functions are certainly useful, but here is a direct way to solve the problem: Letting $z=e^{i\phi},$ one has $$dz=iz~d\phi,\cos\phi=\frac{e^{i\phi}+e^{-i\phi}}2=\frac {z+z^{-1}}2.$$ By substitution, the original integral becomes $$\int_{|z|=1}\frac {2+\frac{z+z^{-1}}2}{5+4\cdot \frac{z+z^{-1}}2}\cdot \frac 1{iz}dz$$ $$=\int_{|z|=1}\frac {z^2+4z+1}{4iz(z+\frac 1 2)(z+2)}~dz=2\pi i\left(\frac 1 {4i}+\frac 1{4i}\right)=\pi$$ by the residue thereom, where the residues are evaluated at the poles $z=0$ and $z=-\frac 1 2.$

Pythagoras
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