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Consider $$ y = \int_1^{\infty} \frac{\operatorname{li}(x)^2 (x - 1)}{x^4} dx, $$ where $\operatorname{li}(x)$ is the logarithmic integral. Is there a closed form for y ?

It appears that a good approximation is $ 10 \cdot \operatorname{Ci}\bigl( \frac{56}{19}\bigr)$, where $\operatorname{Ci}(x)$ is the Cosine integral.

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If there is No closed form would it help to allow the function $$ t(x) = \int_1^x \operatorname{li}(t)^2 dt \hspace{10mm} ??$$

Leucippus
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mick
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  • Logarithmic integral li and cosine integral Ci , in case you wonder. – mick Jun 04 '18 at 23:49
  • Maybe this helps. It looks somewhat similar. https://math.stackexchange.com/questions/513865/a-closed-form-for-int-01-frac-ln-ln-x-operatornameli2xxdx – mick Jun 05 '18 at 10:49

2 Answers2

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Using numerical integration, I found a value of $$1.3707783890401886970603459722050209910157915843390$$ Looking for this number in $ISC$, I found that it is $$10\sum_{n=1}^\infty \frac{1}{4 n^2 \binom{2 n}{n}}$$ which is $\frac{5 }{36}\pi ^2$ !!

Now, $???$

Claude Leibovici
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    What is ??? Suppose to mean ? – mick Jun 05 '18 at 06:25
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    @mick. $??? = $ How to prove it ? – Claude Leibovici Jun 05 '18 at 06:27
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    What is ??? Suppose to mean ? So can we express the indefinite integral in standard functions and Then plug in 1 and Oo and compute the limit ? Or do we use hypergeo functions and theorems ? Does ??? Mean 1) I give up bc Its hard. 2) I give up bc I found it and do not care to prove 3) I am very amazed ... or combinations ? ... Also Maybe similar integrals have been posted here on MSE , comparing might help. Kudo’s to the ISC. Besides this answer implies I lost 2 bets to my mentor. – mick Jun 05 '18 at 06:37
  • What do you think about my comments ? Leibovici ? – mick Jun 05 '18 at 06:39
  • I must say this Number looks very familiar ( the exact form , not the digits ) ... probably from a zeta kind of function or Sum harmonic / n^2 or So ... i Will come back to this. Ofc this suggests that this may be a way of attacking the hypothesis. – mick Jun 05 '18 at 06:43
  • @mick. You may be right, for sure ! The problem is that I have no idea about a way to attack the problem. I shall wait for a real answer to this interesting problem. Just out of curiosity : in which context did you find this problem ? – Claude Leibovici Jun 05 '18 at 06:58
  • Ah Yes , the Number is equal to 10 sum_1^oo harmonic(n)/n^3. – mick Jun 05 '18 at 07:04
  • The context was calculus :p . No seriously now it came from my mentor. We were talking about tetration and dynamics and calculus related to that. We considered “ new special functions “ like integral of li^2. ( I believe that has No closed form right ? Maybe hypergeo ?? ... anyway he came up with “ better ones “ and with better he mainly meant that integrals could be done in standard numbers. He gave the integral from the Op and we had a few bets. I lost the bet that it had no short closed form ... it Clearly does. Also he assumed it would contain pi. He has not tried to prove it. – mick Jun 05 '18 at 07:16
  • Also we were driven to these type of integrals by searching for ways too classify How fast functions grow ... I wonder If integral li^2 occurs in Number theory ? – mick Jun 05 '18 at 07:22
  • If anyone knows a link to a similar integral on MSE or MO ( with li^2 or li^3 ) , let me know. – mick Jun 05 '18 at 07:24
  • Differentiating under the integral sign 2 times might Also work ... – mick Jun 05 '18 at 07:30
  • An analogue of this answer might work ? https://math.stackexchange.com/questions/191008/a-curious-limit-for-frac-pi2 – mick Jun 05 '18 at 07:37
  • Maybe this helps , it looks similar : https://math.stackexchange.com/questions/513865/a-closed-form-for-int-01-frac-ln-ln-x-operatornameli2xxdx – mick Jun 05 '18 at 10:50
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For what it's worth, for $n=0$ and $n=1$ the left-hand side (LHS) and RHS are equal in the below equation, to high numeric precision. I thought I saw a pattern developing for a generalization, but have been unable to ascertain the correct one yet. For $n=0$ and using the dilogarithm at argument 1/2, the already proposed identity results. There is absolutely no rigor in my nascent identity, so I won't present the development.

$$ \int_1^\infty \frac{x-1}{x^{n+4}}\, li(x)^2 \,dx \, \, \, ( \text{for } n=0, 1) \\ =(-1)^{n+1}\Big(\, \frac{\zeta(2)}{2} + \sum_{m=0}^n \frac{(-1)^m}{m+3}\big(\log^{\,2}(m+2)+2\text{Li}_2(1/(m+2))-5\zeta(2)\big)\, \Big)$$

As far as I know, $\text{Li}_2(1/3)$ does not have a 'closed form,' so generalizations in this direction appear unlikely.

user321120
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