TITLE: Find $\lim_{x\to 1}\int_{x}^{x^2}\frac{1}{\ln {t}}\mathrm dt$.
QUESTION [11 upvotes]: Find $\lim_{x\to 1}f(x)$, where
  $$f(x) = \int_{x}^{x^2}\frac{1}{\ln {t}}\mathrm dt$$

I tried splitting it into two integrals, one from 1 to $x^2$ and the other one from $x$ to $1$. Doesn't matter how I split it, I got zero. Wolfram Alpha has some kind of a different, more sophisticated answer with functions we didn't cover at the lecture.

REPLY [5 votes]: This seems like a good problem for generalization, since the proof for the general result might actually be easier. Suppose $f$ is continuous near $1,$ $f(1)=0,$ and $f'(1)\ne 0.$ Then
$$\tag 1 \lim_{x\to 1} \int_x^{x^2} \frac{1}{f(t)}\,dt = \frac{\ln 2}{f'(1)}.$$
With $f(t)= \ln t,$ we get the answer of $\ln 2$ for the given problem.
Rough sketch of proof of $(1)$:
$$\frac{1}{f(t)} = \frac{1}{t-1}\cdot \frac{t-1}{f(t)}.$$
Now the second fraction has limit $1/f'(1)$ as $t\to 1.$ So given $\epsilon>0,$ this fraction lies between the constants $(1-\epsilon)/f'(1)$ and $(1+\epsilon)/f'(1)$ for $t$ close enough to  $1.$ Move those constants outside the integral, evaluate the easy integral remaining, and the result follows.

REPLY [2 votes]: Setting $t = e^s$ we get
$$
f(x) 
= \int_{\ln x}^{2 \ln x} \frac{e^s}{s}\ ds
= \int_{\ln x}^{2 \ln x} \frac{1+s+O(s^2)}{s}\ ds
= \int_{\ln x}^{2 \ln x} \left(\frac{1}{s}+1+O(s)\right) \ ds \\
= (\ln(2 \ln x) - \ln(\ln x)) + (2\ln x - \ln x) + O((\ln x)^2) \\
= \ln 2 + \ln x + O((\ln x)^2) 
\to \ln 2
$$
as $x \to 1$.