What is the antiderivative of (ln n)^2/n^2? Any help would be greatly appreciated. Thanks.

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- Apr 8th 2010, 09:39 PMvlodgeAntiderivative
What is the antiderivative of (ln n)^2/n^2? Any help would be greatly appreciated. Thanks.

- Apr 8th 2010, 09:51 PMharish21
For $\displaystyle \frac{log^2(n)}{n^2}$, integrate by parts,

$\displaystyle \int f dg = f g- \int g. df$, where

$\displaystyle f = log^{2}(n) $ , $\displaystyle dg = \frac{1}{n^2} dn$,

$\displaystyle df = \frac{2 log(n)}{n} dn $, $\displaystyle g = \frac{-1}{n}$

Try doing the rest and post if you have problems. - Apr 8th 2010, 09:58 PMProve It
You need to use integration by parts.

I.e. $\displaystyle \int{u\,dv} = u\,v - \int{v\,du}$.

Let $\displaystyle u = (\ln{n})^2$ so that $\displaystyle du = \frac{2\ln{n}}{n}$.

Let $\displaystyle dv = \frac{1}{n^2}$ so $\displaystyle v = -\frac{1}{n}$.

Therefore $\displaystyle \int{\frac{(\ln{n})^2}{n^2}\,dn} = -\frac{(\ln{n})^2}{n} - \int{-\frac{2\ln{n}}{n^2}\,dn}$

$\displaystyle = -\frac{(\ln{n})^2}{n} +2 \int{\frac{\ln{n}}{n^2}\,dn}$.

Now you'll need to use integration by parts again.

Let $\displaystyle u = \ln{n}$ so that $\displaystyle du = \frac{1}{n}$.

Let $\displaystyle dv = \frac{1}{n^2}$ so that $\displaystyle v = -\frac{1}{n}$.

So $\displaystyle \int{\frac{(\ln{n})^2}{n^2}\,dn} = -\frac{(\ln{n})^2}{n} +2 \int{\frac{\ln{n}}{n^2}\,dn}$

$\displaystyle = -\frac{(\ln{n})^2}{n} + 2\left(-\frac{\ln{n}}{n} - \int{-\frac{1}{n^2}\,dn}\right)$

$\displaystyle = -\frac{(\ln{n})^2}{n} - \frac{2\ln{n}}{n} + 2\int{\frac{1}{n^2}\,dn}$

$\displaystyle = -\frac{(\ln{n})^2}{n} - \frac{2\ln{n}}{n} - \frac{2}{n} + C$