Originally Posted by
Random Variable According to Maple, the series expansion of $\displaystyle \frac{1}{\ln t} $ about $\displaystyle t = 1 $ is $\displaystyle \frac{1}{t-1} + \frac{1}{2} - \frac{1}{12} (t-1) + \frac{1}{24} (t-1)^{2} - \frac{19}{720} (t-1)^{3} + ... $ (EDIT: You can derive it by starting with the Taylor series for $\displaystyle \ln t $ centered at $\displaystyle t=1 $ and using long division.)
Since we're integrating over values close to 1, $\displaystyle \frac{1}{\ln t}$ is approximately $\displaystyle \frac{1}{t-1}$.
And what happened to $\displaystyle +\frac{1}{2}$ above...?
Tonio
so $\displaystyle \lim_{x \to 1^{+}} \int^{x^{4}}_{x^{2}} \frac{1}{\ln t} \ dt = \lim_{x \to 1^{+}} \int^{x^{4}}_{x^{2}} \frac{1}{t-1} \ dt $
$\displaystyle = \lim_{x \to 1^{+}} \ln (t-1) \Big|^{x^{4}}_{x^{2}} $
$\displaystyle =\lim_{x \to 1^{+}} \Big( \ln (x^{4}-1) - \ln (x^{2}-1) \Big) $
$\displaystyle = \lim_{x \to 1^{+}}\ln \Big(\frac{x^{4}-1}{x^{2}-1}\Big) $
$\displaystyle = \lim_{x \to 1^{+}} \ln (x^{2}+1) = \ln 2$