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OpenStudy (anonymous):
what is ln(ln2) - ln(ln 1)
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OpenStudy (dumbcow):
ln 1 = 0
and
ln 0 is undefined
OpenStudy (anonymous):
so its infinty
OpenStudy (dumbcow):
in terms of limits the answer is infinity
\[\ln(\ln 2) - \lim_{x \rightarrow 1} \ln(\ln x) = \ln(\ln 2) - (-\infty) = \infty\]
OpenStudy (anonymous):
how can i find the value of p
The integral from 1 to 2 of 1/(x(ln(x)^p)) dx
i already Substitute u=ln(x) and you get 1/u^p
OpenStudy (anonymous):
\[\int\limits_{1}^{2}\frac{ dx }{ x (\ln x )^{p} }\]
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OpenStudy (anonymous):
in to \[\int\limits_{1}^{2} \frac{ 1 }{ u ^{p} }\]
OpenStudy (anonymous):
how can i find tha value of p to make it converges
OpenStudy (dumbcow):
\[\int\limits_1^2 u^{-p} du = \frac{u^{1-p}}{1-p} = \frac{(\ln x) ^{1-p}}{1-p}\]
plug in limits
OpenStudy (anonymous):
??? no idea
OpenStudy (anonymous):
why we divided by 1-p
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OpenStudy (dumbcow):
power rule for integrating a polynomial
\[\int\limits x^n = \frac{x^{n+1}}{n+1}\]
OpenStudy (anonymous):
how it acuatlly became u to thepower1-p
OpenStudy (anonymous):
aha
OpenStudy (anonymous):
that was stupid
OpenStudy (dumbcow):
as long as p does not equal 1, it will converge because ln(1) = 0
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OpenStudy (anonymous):
so the value of p should be bigger than 1
OpenStudy (dumbcow):
p can be any number other than 1
OpenStudy (anonymous):
thank you very much you made my day <3
OpenStudy (dumbcow):
yw
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