Good day! Im now dealing with some dilemma regarding on how to get the standard deviation of a function in a mathematical way. Using the function $$\frac{2}{π} \ln (n)$$ and some mathematics software, I found $$0.778\times n^{0.15}$$ as the standard deviation of the function $$\frac{2}{π} \ln (n).$$
I used the formula below for finding the standard deviation of a continuous random variable and that is $$\sqrt{\int_0^1{(x - u)^2 f(x) dx}}$$ where $$u = \int_0^1 {xf(x)}$$ and $f(x)$ is the probability density function which is $$\frac{p(x)}{En},$$ where $$p(x) = \frac{4}{π}\sqrt{\frac{1}{(t^{2} - 1)^2} - \frac{(n+1)(t^{2n})}{(t^{2n+2} - 1)^2}}$$ and En is the $$\int_0^1 {p(x)}$$ which is approximately $$\frac{2}{π} \ln (n)$$.
I intend to compute the standard deviation from the following given stated above. I used the approximation of integration of p(x) from 0 to 1 which is $$\frac{4}{π} \left(\int_0^{1-1/n}{\frac{dx}{1 - x^2}} + \int_{1-1/n} ^1{\frac{\sqrt{2n - 1}dx}{\sqrt{1 - x}}}\right)$$ because there is no direct integration available to $p(x)$. And as I go along with my calculation,with the best of my knowledge, I end up to $$\text{sd}(n) =\sqrt{\left(\frac{2\ln (2n - 1)}{\ln (n)}\right)^{2} - 1}.$$ But the problem is it does not fit my findings that it should be somehow close to $$0.778\times n^{0.15}.$$ What am i missing?
Along the process,I rejected terms with numerator less than the denominator $\ln (n)$ for I thought that as $n$ approaches infinity it will eventually become $0$,become insignificant in the totality of the equation. please, any idea will be of great help,thanks!