# Caculation a limit ?

• Jun 2nd 2010, 03:26 AM
bigli
Caculation a limit ?
How do I show $\displaystyle\lim_{x\rightarrow 1^+ }\int_{x^2}^{x^4}\!\frac{1}{\ln t}\, dt=\ln 2$ ?
• Jun 2nd 2010, 03:48 AM
tonio
Quote:

Originally Posted by bigli
How do I show $\displaystyle\lim_{x\rightarrow 1^+ }\int_{x^2}^{x^4}\!\frac{1}{\ln t}\, dt=\ln 2$ ?

If you've already done power series and stuff, we get (Bronshtein-Semendiaev, "Manuel of Mathetmatics") that

$\int\frac{dt}{\ln t} =\ln\ln t+\ln t+\sum^\infty_{n=2}\frac{(\ln t)^n}{n\cdot n!}$

Tonio
• Jun 2nd 2010, 04:00 AM
bigli
It doesn't work.Is there any easier way?
• Jun 2nd 2010, 08:55 AM
tonio
Quote:

Originally Posted by bigli
It doesn't work.Is there any easier way?

Oh, it works and very fine , thank you! Perhaps there's an easier way but I can't see it.

Tonio
• Jun 2nd 2010, 12:41 PM
Random Variable
According to Maple, the series expansion of $\frac{1}{\ln t}$ about $t = 1$ is $\frac{1}{t-1} + \frac{1}{2} - \frac{1}{12} (t-1) + \frac{1}{24} (t-1)^{2} - \frac{19}{720} (t-1)^{3} + ...$ (EDIT: You can derive it by starting with the Taylor series for $\ln t$ centered at $t=1$ and using long division.)

Since we're integrating over values close to 1, $\frac{1}{\ln t}$ is approximately $\frac{1}{t-1}$.

so $\lim_{x \to 1^{+}} \int^{x^{4}}_{x^{2}} \frac{1}{\ln t} \ dt = \lim_{x \to 1^{+}} \int^{x^{4}}_{x^{2}} \frac{1}{t-1} \ dt$

$= \lim_{x \to 1^{+}} \ln (t-1) \Big|^{x^{4}}_{x^{2}}$

$=\lim_{x \to 1^{+}} \Big( \ln (x^{4}-1) - \ln (x^{2}-1) \Big)$

$= \lim_{x \to 1^{+}}\ln \Big(\frac{x^{4}-1}{x^{2}-1}\Big)$

$= \lim_{x \to 1^{+}} \ln (x^{2}+1) = \ln 2$
• Jun 2nd 2010, 02:23 PM
tonio
Quote:

Originally Posted by Random Variable
According to Maple, the series expansion of $\frac{1}{\ln t}$ about $t = 1$ is $\frac{1}{t-1} + \frac{1}{2} - \frac{1}{12} (t-1) + \frac{1}{24} (t-1)^{2} - \frac{19}{720} (t-1)^{3} + ...$ (EDIT: You can derive it by starting with the Taylor series for $\ln t$ centered at $t=1$ and using long division.)

Since we're integrating over values close to 1, $\frac{1}{\ln t}$ is approximately $\frac{1}{t-1}$.

And what happened to $+\frac{1}{2}$ above...?

Tonio

so $\lim_{x \to 1^{+}} \int^{x^{4}}_{x^{2}} \frac{1}{\ln t} \ dt = \lim_{x \to 1^{+}} \int^{x^{4}}_{x^{2}} \frac{1}{t-1} \ dt$

$= \lim_{x \to 1^{+}} \ln (t-1) \Big|^{x^{4}}_{x^{2}}$

$=\lim_{x \to 1^{+}} \Big( \ln (x^{4}-1) - \ln (x^{2}-1) \Big)$

$= \lim_{x \to 1^{+}}\ln \Big(\frac{x^{4}-1}{x^{2}-1}\Big)$

$= \lim_{x \to 1^{+}} \ln (x^{2}+1) = \ln 2$

.
• Jun 2nd 2010, 02:32 PM
Random Variable
As $t$ approaches $1$ from the right, $\frac{1}{t-1}$ is much greater than $\frac{1}{2}$. You can include it if you want. You'll still get the same answer.
• Jun 4th 2010, 10:57 PM
Drexel28
It is fairly easy to verify that for $t>0$ we have that $\frac{t-1}{t}\leqslant\ln(t)\leqslant t-1$ and so $\frac{1}{t-1}\leqslant\frac{1}{\ln(t)}\leqslant\frac{t}{t-1}$. It follows that $\int_{x^2}^{x^4}\frac{dt}{t-1}\leqslant \int_{x^2}^{x^4}\frac{dt}{\ln(t)}\leqslant\int_{x^ 2}^{x^4}\frac{t}{t-1}dt$. Note though that the LHS is just $\ln\left(x^4-1\right)-\ln(x^2-1)=\ln(x^2+1)\overset{x\to 1^+}{\longrightarrow}\ln(2)$ and $\int_{x^2}^{x^4}\frac{t}{t-1}dt=\left(x^4-x^2\right)+\left(\ln(x^4-1)-\ln(x^2-1)\right)\overset{x\to 1}{\longrightarrow}\ln(2)$. The conclusion follows from the Squeeze Theorem.
• Jun 5th 2010, 08:07 PM
bigli
Thanks DREXEL28. Every thing is OK!. But only, your inequalities should be strictly and they are correct for $t>1$ .