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In the question: Integral $\int_1^\infty\frac{\operatorname{arccot}\left(1+\frac{2\pi}{\operatorname{arcoth}x-\operatorname{arccsc}x}\right)}{\sqrt{x^2-1}}\mathrm dx$, the value of that integral was conjectured to be $\frac{\pi\,\ln\pi}4-\frac{3\,\pi\,\ln2}8$.

Apart from the fact that this integral is interesting, I want to avoid this question to be a copy of that one, and my attention went to the statement:

the value is correct up to at least 900 decimal digits

My (imho) intuitive question is this: How many decimal digits are necessary to be considered a valid formal proof?

I know this is not the usual way to prove an integral, but since we have computers that can check many digits I think it is time to ask ourselves such questions !

I am aware of "mathematical coincidence," however I assume there must be limits to this coincidence.

I also understand that the amount of digits needed to be considered a proof depends on the "length" of the integrand and the conjectured value of the integral. By "length" I mean some measure of data size. For example the amount of functions, additions, multiplications, constants, etc. used.

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mick
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    All of them.... – abiessu Jan 08 '14 at 21:33
  • @abiessu How do you know that ? – mick Jan 08 '14 at 21:35
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    Because a formal proof means that it is fully proven that the result of the integral is $\frac{\pi,\ln\pi}4-\frac{3,\pi,\ln2}8$, which means that every digit must be accounted for. It might be "accepted that the result is probably $\frac{\pi,\ln\pi}4-\frac{3,\pi,\ln2}8$", but never fully proven unless every digit is verified, or the integral is proven to have that result directly. – abiessu Jan 08 '14 at 21:38
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    The idea is that after say 10^500 digits the result must be true ; not because it has been " explained " but because it cant be anything else ... – mick Jan 08 '14 at 21:57
  • To give more weight to the previous comment : the idea that it cant be anything else assumes that there are a finite amount of potential candidate answers. Why ? From the idea the the output data size is bounded by a function of the imput data size. For example the resulting value must have data size = exp(data size integrand) or less. See what I mean ? – mick Jan 08 '14 at 22:00
  • @mick: This is just ludicrous. Your reasoning suggests that $0 = 10^{-n}$ for sufficiently large but finite $n$. Don't confuse "close" with "equal". Sorry to be harsh, but well...grrrr... – MPW Jan 08 '14 at 22:07
  • @MPW No, It's more like this , suppose the integrand is relatively short in notation and contains no integers larger than 8 or any nonstandard functions. Now if the integral appears to be $0.000000...$ for the first $10^{10000}$ digits , we are tempted to conclude the result is exactly $0$. – mick Jan 08 '14 at 22:11
  • @mick: What about \begin{equation}\int_1^{\infty} \chi_{[1,2]}(x)\cdot 8^{-8^{8^8}}dx\end{equation} – MPW Jan 08 '14 at 22:15
  • What about it ? – mick Jan 08 '14 at 22:16
  • You still would need to formally prove that $n$ digits are enough, which probably isn't any easier than proving the correctness of the integral in another way. – JiK Jan 08 '14 at 22:16
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    Yes Jik , But thats exactly what the OP asks , a theory to compute $n$ or check $n$ or even prove the existance of $n$ or bound $n$ from above and below. – mick Jan 08 '14 at 22:21
  • Perhaps related : http://mathworld.wolfram.com/ConstantProblem.html – mick Jan 08 '14 at 23:09

2 Answers2

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The existence of such a criterion is clear: For a given "complexity" $L$ of mathematical expressions (given by the number of symbols etc.) there are only finitely many possible values. These form a discrete subset of $\mathbb R$ (or $\mathbb C$ or whatever - and ignoring any undefined expressions). Hence there exists an $\epsilon=\epsilon(L)>0$ such that two expressions that evaluate to values at most $\epsilon$ apart are in fact equal. However, it is more or less hopeless to explicitly determine such an $\epsilon$ for any interesting values of $L$ (e.g. big enough to allow both $\int_1^\infty\frac{\operatorname{arccot}\left(1+\frac{2\pi}{\operatorname{arcoth}x-\operatorname{arccsc}x}\right)}{\sqrt{x^2-1}}\,\mathrm dx$ and $\frac{\pi\ln\pi}4-\frac{3\pi\ln2}8$)

Hagen von Eitzen
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Consider your question under the assumption that the value of the integral winds up being

$$\frac{\pi\,\ln\pi}4 \; -\; \frac{3\,\pi\,\ln2}8 \; + \; 10^{-901}$$

or

$$\frac{\pi\,\ln\pi}4 \; -\; \frac{3\,\pi\,\ln2}8 \; + \; 10^{-10000}$$

or

$$\frac{\pi\,\ln\pi}4 \; -\; \frac{3\,\pi\,\ln2}8 \; + \; 10^{-10^{100}}$$

Dave L. Renfro
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  • (+1) - even though I can't really conceive of an integral having a true value of $\frac{\pi,\ln\pi}4 ; -; \frac{3,\pi,\ln2}8 ; + ; 10^{-10^{100}}$! – Old John Jan 08 '14 at 22:00
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    Nevertheless , that $10^{-1000}$ must come from somewhere. As I just wrote in the comment , Assume the output data size is a function of the imput data size.

    For instance it is unlikely that an integrand that contains only one exponential gives a final value that contains $10^{-10^{10^{10^{10}}}}$.

     – mick Jan 08 '14 at 22:04
  • See my comment to MPW. – mick Jan 08 '14 at 22:12
  • As an example : suppose I said the integral from 2 to + infinity of $\frac{dx}{\Gamma(x)}$ equals $3 + 100^{-100^{100^{100}}}$ then nobody would believe that ! And statistically that is almost impossible. – mick Jan 08 '14 at 22:26
  • @mick: Keep in mind that I could just as well have replaced the large negative powers of $10$ with comparably small numbers that are not rational, or even with numbers that are not expressible using standard mathematical constants and operations. – Dave L. Renfro Jan 08 '14 at 22:38
  • I know Dave , I know. But still. – mick Jan 08 '14 at 22:55