I recently stumbled upon an 'almost rational' integral, namely:

$$\int_0^{\pi/2} x \frac{\sqrt{\sin x}-\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} dx=0.231231222\dots \approx 0.231231231\dots= \frac{77}{333}$$

The error here is only:

$$\frac{77}{333}-I \approx 9 \cdot 10^{-9}$$

I don't think this integral has any known closed form (although I could be mistaken).

Do you know any integrals without a known closed form which are 'almost rational', meaning their value can be approximated by a rational number with the conditions:



I hope there are some non-trivial integrals like the example I provided.

I tried to be as clear as possible, because the MathWorld page about almost integers is filled with 'questionable' entries.

  • $\begingroup$ @orion, there are no integrals on these pages (there is one, but not what I asked) $\endgroup$ – Yuriy S May 10 '16 at 12:12
  • $\begingroup$ There's a section on Evaluation of Integrals in this paper on experimental mathematics: ams.org/notices/200505/fea-borwein.pdf . I don't see one there just like yours, but yours may be more than a curiousity. $\endgroup$ – Ethan Bolker May 10 '16 at 12:18
  • $\begingroup$ @EthanBolker, thank you for the link. I especially like the section 'Coincidence and Fraud' $\endgroup$ – Yuriy S May 10 '16 at 12:27
  • $\begingroup$ Im running Do[{y = Rationalize[ Integrate[((Pi^.5)/2)*Exp[h + c*x^(2)], {x, i, j, h, c}], 0], If[Denominator[y] < 200, If[Numerator[y] < 200, If[y != 0, Print[y]]]]}, {i, 1, 100}, {j, 1, 100}, {h, 1, 100}, {c, 1, 100}] in mathmatica which will search a well known non closed form Integral $\endgroup$ – shai horowitz May 18 '16 at 2:32

One of the most remarkable integrals for me is the Borwein sequence. They do have closed form, but somehow I feel they should be mentioned here.

More interesting integrals.


At least your integral turns out to have a closed form:

$$ \int_{0}^{\pi/2} x \, \frac{\sqrt{\sin x} - \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, \mathrm{d}x = G + \pi \left( \frac{1+2\sqrt{2}}{4} \log 2 - \log (1+\sqrt{2}) \right), $$

where $G$ is the Catalan's constant. So it seems to me that the 'almost rationality' of this integral is just a coincidence.

  • $\begingroup$ Any reference for the solution? $\endgroup$ – Yuriy S May 18 '16 at 8:16
  • $\begingroup$ @YuriyS, At this point I haven't made any serious attempt to prove it as I literally have no time to try it. I performed integration by parts and made some substitutions to get another integral representation. Then I punched it into Mathematica and got this answer. I guess it is related to dilogarithms, and will update my answer as soon as I can afford to do it. $\endgroup$ – Sangchul Lee May 18 '16 at 10:33

Integral of $e^{-2x^2}$ from $-2$ to $62$ is about $1/8$ which has a $.003$ error. Also maybe cheating but its possible to get this integral arbitrarily small giving really small errors to something without closed form. This has implication that you could integrate any probability distributions most of which have no close form to 1 plus an arbitrary small error.

Integral of $e^{-2x^2}$ from $2(e^{\pi}-\pi)/100$ to $2(e^{\pi}-\pi)$ = .5 with .006 error

  • $\begingroup$ Regrettably, none of these cases fit the requirements I presented in the OP $\endgroup$ – Yuriy S May 19 '16 at 6:34
  • $\begingroup$ I'm sorry for not giving you a bounty, but Sangchul put a lot of work in his answer (even though you are the only one who tried to honestly answer my question). I give you +1 of course, and I'll accept the answer if you find some integral fitting with my requirements $\endgroup$ – Yuriy S May 19 '16 at 18:41
  • $\begingroup$ note this is as simple as slightly altering the bounds of the derivative to make the aproximation more exact. however I wouldn't define that as interesting. I like the trancedental bounds. $\endgroup$ – shai horowitz May 19 '16 at 18:47
  • 1
    $\begingroup$ I'm fully against arbitrarily altering the bounds. I want some meaningful and relatively simple bounds (like $0,\pi/2$ for trig functions or $0,1$ for other functions). It's not a test question after all, so I can happily wait years until I find something truly interesting $\endgroup$ – Yuriy S May 19 '16 at 18:55
  • $\begingroup$ do you like my bounds for the second? $\endgroup$ – shai horowitz May 19 '16 at 18:58

Consider any function $f$ such that $\int_{a}^{b} f(x)dx$ does not have a closed form, now choose $b-a$ small enough that the integral is very close to zero.

Eg: $\int_{0}^{1} e^{-x^{100}} dx$ perhaps does not have closed form and is surely very close to zero.

Any such function can be modified to make the integral very close to any given rational integer.

  • $\begingroup$ I do think I made it clear that I want interesting cases, not something specifically created to fulfill the criteria. Especially not something like 0. However, turns out my own example has closed form, so who am I to talk $\endgroup$ – Yuriy S May 18 '16 at 8:22

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