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Refer to my previous topic: A generalized integral need help I think we get this : $$\frac{\sin \theta}{1-2\cos \theta x+x^2}=\sum_{k=1}^{\infty}\sin (k\theta )x^{k-1}$$ Then $$\int_0^1\frac{\ln ^2x}{1-2\cos \theta x+x^2}\text{d}x=\frac{2}{\sin \theta}\sum_{k=1}^{\infty}\frac{\sin (k\theta)}{k^3}=\frac{2}{\sin \theta}\left(\frac{\pi ^2\theta}{6}-\frac{\pi \theta ^2}{4}+\frac{\theta ^3}{12}\right)$$ Moreover $$\begin{align} & I\left( \lambda ,\theta \right)=\int_{0}^{\infty }{\frac{{{\ln }^{2}}x}{{{x}^{2}}-2\lambda x\cos \theta +{{\lambda }^{2}}}\text{d}x}=\int_{0}^{\infty }{\frac{{{\ln }^{2}}x}{{{\lambda }^{2}}{{x}^{2}}-2\lambda x\cos \theta +1}\text{d}x} \\ & \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{\lambda }\int_{0}^{\infty }{\frac{{{\left( \ln x-\ln \lambda \right)}^{2}}}{{{x}^{2}}-2\cos \theta x+1}\text{d}x}=\frac{{{\ln }^{2}}\lambda }{\lambda }\int_{0}^{\infty }{\frac{1}{{{x}^{2}}-2\cos \theta x+1}\text{d}x}+\frac{1}{\lambda }\int_{0}^{\infty }{\frac{{{\ln }^{2}}x}{{{x}^{2}}-2\cos \theta x+1}\text{d}x} \\ & \ \ \ \ \ \ \ \ \ \ \ =\frac{{{\ln }^{2}}\lambda }{\lambda }\cdot \frac{\pi -\theta }{\sin \theta }+\frac{4}{\lambda \sin \theta }\left( \frac{{{\pi }^{2}}\theta }{6}-\frac{\pi {{\theta }^{2}}}{4}+\frac{{{\theta }^{3}}}{12} \right) \\ \end{align}$$

Can anybody verify my result? Or perhaps show more method? :)

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I have gone over the computation and I can reproduce your final expression $I(\lambda,\theta)$. I have also verified your formula for $\sum_{k=1}^{\infty}\frac{\sin (k\theta)}{k^s}$ for $s = 1,3$ through Fourier series expansion. Everything is fine except that formula only works for $\theta \ge 0$. – achille hui Feb 5 '13 at 17:22
To evaluate the integral $\int_0^1\frac{\ln ^2x}{1-2\cos \theta x+x^2}\text{d}x$, use the general techniue which I introduced to solve your previous problem. – Mhenni Benghorbal Feb 5 '13 at 23:42
up vote 2 down vote accepted

Some food for thought that hopefully one of us will pursue in more detail soon.

The function $(1-2 x \cos{\theta} + x^2)^{-1}$ is a generating function for the Chebyshev polynomials of the second kind $U_n(\cos{\theta})$:

$$(1-2 x \cos{\theta} + x^2)^{-1} = \sum_{n=0}^{\infty} U_n(\cos{\theta}) x^n$$

So your integral is

$$\int_0^1 dx \: \frac{\ln ^2x}{1-2\cos \theta x+x^2} = \sum_{n=0}^{\infty} U_n(\cos{\theta}) \int_0^1 dx \: x^n \, \log^2{x} $$

Now, I found this very interesting: It turns out that

$$\int_0^1 dx \: x^n \, \log^2{x} = \frac{2}{(n+1)^3} $$

The integral becomes

$$\int_0^1 dx \: \frac{\ln ^2x}{1-2\cos \theta x+x^2} = 2\sum_{n=0}^{\infty} \frac{U_n(\cos{\theta})}{(n+1)^3}$$


I will derive the result for the integral above. Integrate by parts, as I indicated:

$$\begin{align}\int_0^1 dx \: x^n \, \log^2{x} &= \underbrace{[x^{n+1} \log{x} (\log{x} - 1)]_0^1}_{0} - \int_0^1 dx \: x (\log{x} - 1) \frac{d}{dx} [x^n \log{x}]\\ &= -\int_0^1 dx \: x^n (\log{x} - 1) (1+n \log{x})\\ &= -n \int_0^1 dx \: x^n \, \log^2{x} + (n-1) \int_0^1 dx \: x^n \, \log{x} + \int_0^1 dx \: x^n \\\end{align}$$

so that

$$\begin{align} (n+1)\int_0^1 dx \: x^n \, \log^2{x} &= \frac{1}{n+1} - \frac{n-1}{(n+1)^2} \\ &= \frac{2}{(n+1)^2} \\ \end{align}$$

The result follows from this.


Well, in my hubris, I did not catch the defining relation of $U_n$:

$$U_n(\cos{\theta}) = \frac{\sin{[(n+1) \theta]}}{\sin{\theta}}$$

This reproduces the sum you evaluated above.

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Hmm... $\int_0^1 dx x^n \log^2{x}$ should be $\frac{2}{(n+1)^3}$ – achille hui Feb 5 '13 at 14:16
You are correct; I forgot a factor of $n$. I will fix. Thank you. – Ron Gordon Feb 5 '13 at 14:27

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