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August 31st, 2019, 11:38 PM   #1
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Evaluate integral

Evaluate $\displaystyle I_{x}= \int_{1}^{x} \ln \ln(t) dt$ .
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September 1st, 2019, 05:25 AM   #2
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This can't be done without resort to special functions (which wouldn't help, as you'd need a special function that is defined as an integral).
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September 1st, 2019, 06:09 AM   #3
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The Idontknow function, named for one of its proponents and denoted $\displaystyle I(x)$ (alternately, $\displaystyle I_x$), is defined as
$\displaystyle I(x)= \int_{1}^{x} \ln \ln(t) dt = \int_{1}^{e^x} e^s ln(s) ds$.

It was made famous by the countless variations of the inevitable humorous introduction in calculus classrooms:
Professor: Can you tell me how we evaluate this integral?
Student: I don't know.
Professor: That's right!
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September 1st, 2019, 10:28 AM   #4
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I agree .

$\displaystyle I_x =t\ln\left(\ln\left(t\right)\right)-\operatorname{li}\left(t\right)+\lim_{x\rightarrow 1} [li(x)-xlnln(x) ]$.

Now only the limit is left : $\displaystyle \lim_{x\rightarrow 1} [li(x)-xlnln(x)]$.
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September 1st, 2019, 11:02 PM   #5
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Quote:
Originally Posted by idontknow View Post
Evaluate $\displaystyle I_{x}= \int_{1}^{x} \ln \ln(t) dt$ .
I just watched Sal Khan of Khan Academy find the antiderivative of $\ln x$ using integration by parts. It's $x \ln x - x$. Might be worth playing around with except that Skipjack already said it can't be done by elementary means. How do we know that, if I may ask?
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September 1st, 2019, 11:55 PM   #6
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W|A gave an answer using a special function that implies the integral can't be done, but a formal proof might be tedious, as it would rely on a formal proof that some particular special function can't be defined in closed form.
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