A logarithmic poisson like integral

Calculus (Integrals, Series)
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Tolaso J Kos
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A logarithmic poisson like integral

#1

Post by Tolaso J Kos »

For the values of $a$ for which the integral converges , find the general form of:

$$\int_{0}^{1}\frac{\ln \left ( x^2-2x \cos a +1 \right )}{x}\, {\rm d}x$$
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galactus
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Re: A logarithmic poisson like integral

#2

Post by galactus »

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:


$$\boxed{-2\left(\zeta(2)-\frac{\pi}{2}a+\frac{a^{2}}{4}\right)}$$
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