$\lim\limits_{x \to 1^{+}} \int_x^{x^2} \frac{\,dt}{\log t}$
$\lim\limits_{x \to 1^{+}} \int_x^{x^2} \frac{\,dt}{\log t}$
Calculate the limit (if it exists): $$\lim\limits_{x \to 1^{+}} \int_x^{x^2} \frac{\,dt}{\log t}$$
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Re: $\lim_{x \to 1^+} \int_x^{x^2} \frac{dt}{\log t}$
Hello RD,r9m wrote:Calculate the limit (if it exists): $$\lim\limits_{x \to 1^{+}} \int_x^{x^2} \frac{\,dt}{\log t}$$
1st solution: We are using the double (and pretty well known inequality)
$$1- \frac{1}{x}< \ln x < x- 1, \;\; \forall x >1 $$
Hence:
$$\frac{1}{t-1}< \frac{1}{\ln t}< \frac{t}{t-1}\Rightarrow \int_{x}^{x^2}\frac{{\rm d}t}{t-1}< \int_{x}^{x^2}\frac{{\rm d}t}{\ln t}< \int_{x}^{x^2}\frac{t}{t-1}\, {\rm d}t$$
Then as we can easily compute the integrals at the edges we get that our result is $\ln 2$.
2nd solution: We just bound the function using the limits of the integral. Indeed we can see that as $x\leq t \leq x^2 \Rightarrow \frac{x}{ t \ln t} \leq \frac{1}{\ln t} \leq \frac{x^2}{t \ln t}$ thus integrating we have that:
$$\int_{x}^{x^2}\frac{x}{t \ln t}\, {\rm d}t \leq \int_{x}^{x^2} \frac{{\rm d}t}{\ln t} \leq \int_{x}^{x^2}\frac{x^2}{t \ln t} \, {\rm d}t$$
Now, we are just evaluating the edge integrals (simply by recalling that $\int \frac{{\rm d}t}{t \ln t} =\ln (\ln t)+c $) we get again that our proposed limit is $\ln 2$.
3rd solution: We apply the change of variables $u= \ln t$ thus our initial limit is equal to
$$\lim_{x\rightarrow 1+} \int_{x}^{x^2}\frac{{\rm d}t}{\ln t} = \lim_{y \rightarrow 0+}\int_{y}^{2y}\frac{e^u}{u}\, {\rm d}u $$
Following the same procedure as in solution 2 by just bounding the integral we get that the last limit is also equal to $\ln 2$ , hence evaluating the initial one.
P.S: Sorry for not keeping the initial notation on the logarithm but I am much more familiar with $\ln $ rather than $\log$, although the second is used more often.
Imagination is much more important than knowledge.
Re: $\lim_{x \to 1^+} \int_x^{x^2} \frac{dt}{\log t}$
Nice solutions!!
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