Integral
Let $rho in RR^+$, calculate
$int_0^(2pi) log(1-2rhocost+rho^2) dt$
Have fun!
$int_0^(2pi) log(1-2rhocost+rho^2) dt$
Have fun!
Risposte
Considering $0<=\rho<1$ we can write $f(x) = \log(1-2\rho\cosx+\rho^2) = -\sum_{n=1}^{\infty} {2\rho^n}/n \cos(nx)$.
Setting $f_n(x) = - {2\rho^n}/n \cos(nx)$ we have $|f_n(x)| <= {2\rho^n}/n$.
Also $\sum_{n=1}^{\infty} {2\rho^n}/n = 2\log(1-\rho)$, so $f(x)$ converges uniformly by Weierstrass M-test.
This gives $\int_0^{2\pi}f(x)dx = \sum_{n=1}^{\infty}\int_0^{2\pi} f_n(x)dx = -\sum_{n=1}^{\infty} {2\rho^n}/{n^2}\sin(2\pi n) = 0$.
I'm going to think about $\rho>=1$ later.
Setting $f_n(x) = - {2\rho^n}/n \cos(nx)$ we have $|f_n(x)| <= {2\rho^n}/n$.
Also $\sum_{n=1}^{\infty} {2\rho^n}/n = 2\log(1-\rho)$, so $f(x)$ converges uniformly by Weierstrass M-test.
This gives $\int_0^{2\pi}f(x)dx = \sum_{n=1}^{\infty}\int_0^{2\pi} f_n(x)dx = -\sum_{n=1}^{\infty} {2\rho^n}/{n^2}\sin(2\pi n) = 0$.
I'm going to think about $\rho>=1$ later.
I can't give you a hint about the road you suggested (the standard way by residue theorem)... I can only say that I know different ways to solve that integral.
I great help can be found here.
I great help can be found here.
The standard way should be to put $z=e^{it}$ so that the integral becomes $\int_{|z|=1}1/{iz}\log[1-\rho(z+1/z)+\rho^2]dz$ and use the residue theorem. However I'm not able to compute the residue of this function, any hints on that?
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