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 August 31st, 2019, 11:38 PM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 Evaluate integral Evaluate $\displaystyle I_{x}= \int_{1}^{x} \ln \ln(t) dt$ .
 September 1st, 2019, 05:25 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 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). Thanks from idontknow
 September 1st, 2019, 06:09 AM #3 Senior Member   Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 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! Thanks from idontknow
 September 1st, 2019, 10:28 AM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 642 Thanks: 91 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 : $\displaystyle \lim_{x\rightarrow 1} [li(x)-xlnln(x)]$.
September 1st, 2019, 11:02 PM   #5
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Quote:
 Originally Posted by idontknow 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?

 September 1st, 2019, 11:55 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 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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