I know how deduce for $x>1$ that $$-\log\left(1-\frac{1}{x^2}\right)=\sum_{n=0}^\infty\frac{1}{n+1}\frac{1}{x^{2n+2}},$$ and from a online tool that $$\int -\log\left(1-\frac{1}{x^2}\right)dx =-x\log\left(1-\frac{1}{x^2}\right)+\log(1-x)-\log(1+x)+\text{constant}.$$

Question 1. Can you prove this fact, it is a/previous closed form for $$\int_2^x-\log\left(1-\frac{1}{t^2}\right)dt?$$

I don't know how evaluate the limit


When I type the integral or previous closed form for the integral, or the limit for the symbolic integral I obtain from the online resource one time $\infty$ and other time $i\pi$. By assumption of previous closed form I compute $\lim_{x\to\infty}-x\log\left(1-\frac{1}{x^2}\right)=0$, when I assume convergence in previous series expansion, and I deduce that $\lim_{x\to\infty}\log\left({\frac{1-x}{1+x}}\right)$ doesn't exists. Perhaps I need show a kind of convergence to combine with some integral theorem, is it?

I need refresh this computations.

Question 2. Can you help me to know where were the mistakes? I say, can you compute previous $$\lim_{x\to\infty}\int_2^x-\log\left(1-\frac{1}{t^2}\right)dt?$$

Thanks in advance. I need this facts, when I did this afternoon some computations using the Hermite-Hadamard inequality for this function, for wich I showed in my computations that is convex $x>1$.


First question hint

$$\log\left(1- \frac{1}{t^2}\right) = \log\left(\frac{t^2-1}{t^2}\right) = \log(t^2 - 1) - \log(t^2) = \log(t^2-1) - 2\log(t)$$

Split the integral in the two parts, use log properties and integrate by part as you usually do when you have simple logarithms.

Second question Hint

By the result of the integration, you have:

$$\left[-x\log\left(1 - \frac{1}{x^2}\right) + \log(1-x) - \log(1+x)\right]\bigg|_2^{\infty} = \left[-x\log\left(1 - \frac{1}{x^2}\right) + \log\left(\frac{1-x}{1+x}\right)\right]\bigg|_2^{+\infty}$$

You observe that since $x\to \infty$ you have

$$\lim_{x\to \infty} -x\log\left(1 - \frac{1}{x^2}\right) = 0$$ $$\lim_{x\to \infty} \log(1-x) - \log(1+x) = \lim_{x\to \infty}\log\left(\frac{1-x}{1+x}\right) = i\pi$$

When $x\to 2$ you simply have $i\pi - \log\left(\frac{27}{16}\right)$

Then do the math.

Explanation of the limit

$$e^{\pi i} + 1 = 0 ~~~~~ \to ~~~~~ -1 = e^{i\pi}$$


$$\ln(-1) = \ln(e^{i\pi}) = i\pi$$

  • $\begingroup$ Very thanks much, now I am taking notes to study your answers. $\endgroup$ – user243301 Feb 16 '16 at 18:55
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    $\begingroup$ I will try your hints, very thanks much for the trick about $\lim_{x\to2}$ that removes $i\pi$ and by the explanation of the limit. $\endgroup$ – user243301 Feb 16 '16 at 19:23
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    $\begingroup$ @user243301 When you try to compute the limit for $x\to 2$, just use the properties of logarithm to write: $$-x\log\left(1 - \frac{1}{x^2}\right) + \log\left(\frac{1-x}{1+x}\right) = -\log\left(\left(1 - \frac{1}{x^2}\right)^x\right) + \log\left(\frac{1-x}{1+x}\right)$$ In which you used simply the fact that $a\log(b) = \log(b^a)$. Now if you use the property $\log(X) - \log(Y) = \log(X/Y)$ you get $$\log\left(\frac{\frac{1-x}{1+x}}{\left(1 - \frac{1}{x^2}\right)^x}\right)$$ $\endgroup$ – Von Neumann Feb 16 '16 at 20:41
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    $\begingroup$ Now if you substitute $x = 2$ you will get $$\log\left(-\frac{16}{27}\right)$$ Which can be seen as $$\log\left(-1\cdot \frac{16}{27}\right)$$ Using the property $\log(a\cdot b) = \log(a) + \log(b)$ you get: $$\log(-1) +\log(16/27)$$ Now $\log(-1) = i\pi$ and you can see (just because) $\log(16/27) = -\log(27/16)$ and you will have the result $$i\pi - \log(27/16)$$ $\endgroup$ – Von Neumann Feb 16 '16 at 20:44
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    $\begingroup$ Now I believe that understand your answer, now I do your computations as you say, very thanks much. $\endgroup$ – user243301 Feb 16 '16 at 20:47

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