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i want to get a 3D Plot in Octave of the function

$$f(r\cdot e^{i\theta})=\int_{-\pi}^\pi \frac{(1-r^2)\cdot\cos(x)}{1-2\cdot r\cdot\cos(\theta-x)+r^2}dx $$

what is called the Poisson-Integral for the function $\cos(x)$ for $0\leq r<1$. I really don't know how to get it and didn't find any answers in the internet. so maybe someone can help me.

Until now i got the following code but the output is far away from what i need:

r = linspace(0,0.999,10)
theta = pi*linspace(-1,1,10)
z = r.*exp(1i*theta)
[Y1,Y2]=meshgrid(sqrt(real(z).^2+imag(z).^2),imag(z))
for i=1:10
  for j=1:10
    n=Y1(i,j)
    m=Y2(i,j)
    y=@(x) (1-n.^2).*cos(x)/(1-2.*n.*cos(m-x)-n.^2)
    Z(i,j)= quad(y,-pi,pi)
  end
end

plot3(real(z),imag(z),Z)

I'm thankful for every answer or try to help me.

user904986
  • 21

1 Answers1

1

Let $t= x -\theta$ \begin{align} \int_{-\pi}^\pi \frac{(1-r^2)\cos x}{1-2r\cos(\theta-x)+r^2}dx =&(1-r^2) \int_{-\pi-\theta}^{\pi -\theta}\frac{\cos t\cos\theta- \sin t \sin\theta}{1-2r\cos t+r^2}dt\\ =& 2\cos\theta (1-r^2)\int_{0}^{\pi}\frac{\cos t}{1-2r\cos t+r^2}dt\\=& 2\pi r\cos\theta \end{align}

where, with $y=\tan\frac t2$

\begin{align} &\int_{0}^{\pi}\frac{\cos t}{1-2r\cos t+r^2}dt =\frac{1}{2r}\left(-\pi+ \frac{2(1+r^2)}{(1+r)^2} \int_{0}^{\infty}\frac{dy}{y^2+ \left(\frac{1-r}{1+r}\right)^2}\right) =\frac{\pi r}{1-r^2} \end{align}

Quanto
  • 120,125