A logarithmic poisson like integral
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A logarithmic poisson like integral
For the values of $a$ for which the integral converges , find the general form of:
$$\int_{0}^{1}\frac{\ln \left ( x^2-2x \cos a +1 \right )}{x}\, {\rm d}x$$
$$\int_{0}^{1}\frac{\ln \left ( x^2-2x \cos a +1 \right )}{x}\, {\rm d}x$$
Imagination is much more important than knowledge.
Re: A logarithmic poisson like integral
By using the series for -ln(1-x), we can establish the equivalency of
$$-\sum_{n=1}^{\infty}\frac{e^{ina}}{n}x^{n-1}=\frac{\ln|1-xe^{ia}|}{x}=\frac{\ln(x^{2}-2x\cos(a)+1)}{2x}$$
Take the real part and we find that:
$$-\sum_{n=1}^{\infty}\frac{\cos(na)}{n}x^{n-1}=\frac{\ln|1-xe^{ia}|}{x}=\frac{\ln(x^{2}-2x\cos(a)+1)}{2x}$$
Integrate from $[0,1]$:
$$-\sum_{n=1}^{\infty}\frac{\cos(na)}{n^{2}}=\int_{0}^{1}\frac{\ln(x^{2}-2x\cos(a)+1)}{x}dx$$
We can also note that $$\sum_{n=1}^{\infty}\frac{\sin(na)}{n}=\frac{\pi -a}{2}$$
Integrate this w.r.t 'a' to obtain sum needed:
$$\sum_{n=1}^{\infty}\frac{\cos(na)}{n^{2}}=\frac{a^{2}}{4}-\frac{\pi a}{2}+C\tag{1}$$
What is C?.
The left side of the cos series above we can call $f(a)$. Notice that $f(0)=\zeta(2)$
$$f(0)=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+.....$$
$$f(\pi) = -1+\frac{1}{2^{2}}-\frac{1}{3^{2}}+.....$$
Notice that $$\frac{1}{2^{2}}+\frac{1}{4^{2}}+\frac{1}{6^{2}}+....$$ is equal to
$\frac{1}{4}f(0)$ and $\frac{1}{2}(f(0)+f(\pi))$
So, we have:
$$\frac{f(0)}{4}=\frac{f(0)+f(\pi)}{2}$$
$$-1/2f(0)=f(\pi)$$
From (1), the constant of integration, C, becomes:
$$-1/2C=\frac{\pi^{2}-2\pi^{2}}{4}+C$$
$$C=\frac{\pi^{2}}{6}=\zeta(2)$$
Thus, therefore, and whence, inserting into (1), our final solution to said integral is:
$$\boxed{-2\left(\zeta(2)-\frac{\pi}{2}a+\frac{a^{2}}{4}\right)}$$
$$-\sum_{n=1}^{\infty}\frac{e^{ina}}{n}x^{n-1}=\frac{\ln|1-xe^{ia}|}{x}=\frac{\ln(x^{2}-2x\cos(a)+1)}{2x}$$
Take the real part and we find that:
$$-\sum_{n=1}^{\infty}\frac{\cos(na)}{n}x^{n-1}=\frac{\ln|1-xe^{ia}|}{x}=\frac{\ln(x^{2}-2x\cos(a)+1)}{2x}$$
Integrate from $[0,1]$:
$$-\sum_{n=1}^{\infty}\frac{\cos(na)}{n^{2}}=\int_{0}^{1}\frac{\ln(x^{2}-2x\cos(a)+1)}{x}dx$$
We can also note that $$\sum_{n=1}^{\infty}\frac{\sin(na)}{n}=\frac{\pi -a}{2}$$
Integrate this w.r.t 'a' to obtain sum needed:
$$\sum_{n=1}^{\infty}\frac{\cos(na)}{n^{2}}=\frac{a^{2}}{4}-\frac{\pi a}{2}+C\tag{1}$$
What is C?.
The left side of the cos series above we can call $f(a)$. Notice that $f(0)=\zeta(2)$
$$f(0)=1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+.....$$
$$f(\pi) = -1+\frac{1}{2^{2}}-\frac{1}{3^{2}}+.....$$
Notice that $$\frac{1}{2^{2}}+\frac{1}{4^{2}}+\frac{1}{6^{2}}+....$$ is equal to
$\frac{1}{4}f(0)$ and $\frac{1}{2}(f(0)+f(\pi))$
So, we have:
$$\frac{f(0)}{4}=\frac{f(0)+f(\pi)}{2}$$
$$-1/2f(0)=f(\pi)$$
From (1), the constant of integration, C, becomes:
$$-1/2C=\frac{\pi^{2}-2\pi^{2}}{4}+C$$
$$C=\frac{\pi^{2}}{6}=\zeta(2)$$
Thus, therefore, and whence, inserting into (1), our final solution to said integral is:
$$\boxed{-2\left(\zeta(2)-\frac{\pi}{2}a+\frac{a^{2}}{4}\right)}$$
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