Mfrhtyj

Consider the function $f: (0, infty) to R $

$$ f(x)= begin{cases}
int_{x}^{x^2} frac{dt}{log(t)} & xneq 1 \
l & x=1 \
end{cases}
$$

Determine $l$ such that the function $f$ should be continuous at 1. For the determined value of $l$, is the function differentiable?

I started out by trying to calculate the following limit, however, I had trouble continuing.

$$lim_{xto1}int_{x}^{x^2}frac{mathrm dt}{log(t)}$$

Thank you!

To calculate the limit itself (and not address the differentiability question)...

Below, consider $x>1$ as fixed, and make (as Tommaso does) the substitution $t = e^s$. Then

$$f(x) = int_x^{x^2} {dt over ln t} = int_{ln x}^{2ln x} {e^sover s}, ds.$$
Make the substitution $s = u ln x$:
$$ f(x) = int^2_1 e^{uln x} {duover u}.$$

As $xdownarrow 1$, we see that
$$ f(x) to int^2_1{duover u}= ln 2.$$
(Bringing the limit inside the integral is valid - by dominated convergence, for instance.)

Addendum/Edit:

1) The same argument works starting with $0<x<1$, and taking $xuparrow 1$.

2) If one sets $f(1) = ln 2$, one has that

$$ f(x) = int^2_1 e^{u ln x} {duover u}$$

identically, for $xin (0,infty)$. So, on condition that you can show that to calculate limits, one can perform the same corresponding calculations on the integrand, you can show that $f$ is differentiable.

First change variables $t = e^s$ and get
$$ int_{ln(x)}^{ln(x^2)} e^s s^{-1}ds. $$
Now expand the exponential around zero $e^s = sum_{ngeq 0} s^n/n!$ and pass the series outside the integral as $s^{-1} s^n/n! |_{(ln(x),ln(x^2)}$ converges uniformly to $0$. Pay attention to the first term, as $s^0/0!equiv 1$
begin{align*} int_{ln(x)}^{ln(x^2)} (1 + sum_{ngeq 1}s^n/n!) s^{-1}ds &= left. ln(s) + sum_{ngeq 1} s^n/(n!n) right|_{ln(x)}^{ln(x^2)} \
& = ln(ln(x^2)/ln(x)) + sum_{ngeq 1} (2^n-1)ln(x)^n/(n!n).
end{align*}
The term on the right goes to zero as $xto 0$ because the series converges absolutely in a right neighborhood of $1$. Therefore
$$ lim_{xto 1^+} f(x) = lim_{xto 1^+} ln(ln(x^2)/ln(x)) = ln(lim_{xto 1^+} ln(x^2)/ln(x)) = ln(lim_{xto 1^+} x^{-2}2x/(x^{-1}) = ln(2). $$