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In this question, Evaluating the integral $\int_{-\infty}^\infty \frac {dx}{\cos x + \cosh x}$ , robjohn evaluates the integral to a nice summation with an approximate value. When plugged into W|A, it gives a possible closed form as $\dfrac{1512835691 \pi}{1983703776}$, correct to at least 20 decimal digits. When subtracting the two in W|A, it gives a nice result of $0$. (1) Can we prove that it equals the conjectured closed form?

Community
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Teoc
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    Note that you have approximated a $20$ digit number with the ratio of two $10$ digit numbers. Standard approximation theorems say that this is not unusual at all. – robjohn May 29 '15 at 23:58
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    For example, with 40 digit numerator and denominator, we can get $$\frac{2613603931557187475711267138492580362408} {3427084658857667283712884858756158507063} ,\pi$$ which is correct to $80$ places. – robjohn May 30 '15 at 00:15

2 Answers2

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This is not the correct value. The value of the integral with $70$ digits precision is $$\underline{2.395878633914562092}453189586490058501300873812447750729519628041973530$$

And the conjectured expression is $$\underline{2.395878633914562092}589756143323601049570678313084832875204827253396682$$

The two values are not equal.

user153012
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  • It's still an amazing coincidence, isn't it? Just so I know how impressed I should be, what are the odds that a random real number cn be approximated to 20 digit accuracy by the ratio of two 10-digit numbers? – bof May 29 '15 at 23:54
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    @bof these are not random real numbers. WolframAlpha has algorithms and databases to get possible closed-form expressions. Ten or twenty digits of consistency is nothing. The Inverse Symbolic Calculator is much better for conjecturing closed-form expressions. – user153012 May 30 '15 at 00:02
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    @bof Interesting question! I'm not an expert on this, but I suppose Dirichlet's approximation theorem tells us that it's common to get $2N$ digits from an $N$-digit denominator. – Chris Culter May 30 '15 at 00:14
  • (+1) However, your number appears to be $70$ digits instead of $80$ – robjohn May 30 '15 at 00:19
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Another expression for it, BTW, is $$ 4 \pi \sum_{j=0}^\infty \sigma_0(2j+1) (-1)^j \exp(-(j+1/2)\pi) $$ where $\sigma_0$ is the number of divisors function. If your number is $\pi x$, then the continued fraction of $x$ starts

$$[0; 1, 3, 4, 1, 2, 3, 4, 3, 1, 4, 12, 2, 4, 3, 104, 1, 2, 5, 1, 1, 21, 4, 1, 2, 1, 3, 1, 17, 1, 6, 5, 2, 2, 59, 1, 8, 3, 42, 15, 5, 1, 1, 2, 3, 1, 2, 12, 2, 3, 1, 2, 8, 1, 4, 2, 2, 3, 1, 2, 5, 1, 1, 18, 1, 1, 4, 2, 2, 1, 1, 2, 2, 2, 1, 20, 5, 10]$$

(i.e. $0 + 1/(1 + 1/(3+1/(4+1/\ldots)))$), which shows no sign of terminating; there's no reason to think it is rational, nor is it unusually well approximated by rationals.

Robert Israel
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