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In the book, "Pi and the AGM" by Borwein and Borwein, it is mentioned that Gauss computed the following integral to the eleventh decimal palce.

$\int_0^1 \frac{1}{\sqrt{1-x^4}}dx$

How did he do it? Personnally, I looked at a Taylor expansion of

$\frac{1}{\sqrt{1-x}}$

Where I substituted $t^4$ for $x$, and integrated term by term, but this gives a series that converges really slowly. Is there an obvious transform to make this computation faster?

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    $\begingroup$ en.wikipedia.org/wiki/Gauss%27s_constant $\endgroup$
    – Lucia
    Aug 18, 2013 at 17:37
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    $\begingroup$ Gauss had a numerical value for the integral before he knew that it could be computed by the AGM (that is how he discovered the connection). According to Borwein and Bailey, Mathematics by Experiment, p. 13, "One day in 1799, while examining tables of integrals provided originally by James Stirling, he noticed that the reciprocal of the integral [...] agreed numerically with the limit of the rapidly convergent arithmetic-geometric mean iteration [...]". If I read it correctly, it was actually Stirling who computed the integral numerically. No idea how, though. $\endgroup$ Aug 18, 2013 at 18:54
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    $\begingroup$ This is an elliptic integral. Gauss observed that arithmetic-geometric means can be described by certain elliptic integrals. You can turn this around and conclude that certain elliptic integrals can be computed via arithmetic-geomtric means en.m.wikipedia.org/wiki/Arithmetic%E2%80%93geometric_mean $\endgroup$ Aug 18, 2013 at 20:24
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    $\begingroup$ I don't know the history, either. But what we now call Simpson's rule would have been known to Stirling, and I suspect that you don't need to break [0,1] into very many intervals to get 11 digits of accuracy. (Yes, I'm being lazy by not actually doing it!) $\endgroup$ Aug 18, 2013 at 21:03
  • $\begingroup$ Thanks, Fredrik Johansson, it is Stirling's computation that I am interested in. To Joe Silverman: if I understand well, Simpson's rule can be used for definite integrals, but this one is improper, so I will use a few variable changes, most of which are in Borwein&Borwein, to establish links between this one and proper integrals, before applying Simpson's rule. $\endgroup$
    – Yves
    Aug 21, 2013 at 10:39

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A good place to look is pages 405 and 413 of the Nachlass section of Gauss's Werke III, which can be found online through Google Books. On page 405, he gives the following formula for "$\text{arc sin lemn }x$":

$$\text{arc sin lemn }x= x+{1\over2}\cdot{1\over5}x^5 + {1\cdot3\over2\cdot4}{1\over9}x^9+{1\cdot3\cdot5\over2\cdot4\cdot6}{1\over13}x^{13}+{1\cdot3\cdot5\cdot7\over2\cdot4\cdot6\cdot8}{1\over17}x^{17}+\cdots$$

One page 413, he computes the value of $\int_0^1{dx\over\sqrt{1-x^4}}$, presumably using the expansion above, but explicitly citing the formula

$$\text{ arc sin lemn }{7\over23}+2\text{arc sin lemn }{1\over2}$$

obtaining

$$1.3110287771\quad460599052\quad320.7$$

He also notes there that Stirling had obtained the value $1.3110287771\ 4605987$. This is in reference to the calculations on pages 57-58 of Stirling's Methodus differentialis from 1730, which can also be found through Google Books. It might be worth noting that even Gauss was slightly off in the last couple of decimal places. A more accurate value, which I took from here, is

$$1.3110287771\quad460599052\quad324197949$$

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Here is one solution to compute $\int_0^1 \frac{1}{\sqrt{1-x^4}}dx$, with a big accuracy, without using arithmetic-geometric means.

The variable change x = sin u gives

$\int_0^1\frac{1}{\sqrt{1-x^4}}dx = \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1+sin^2 u}}du$

The second integral can be computed quickly using Simpson's rule, 8 intervals are enough for eleven decimal places. Thanks to Joe Silverman for pointing me in the right direction.

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How Gauss (supposedly) calculated this integral in terms of AGM (namely $M(1,\sqrt{2})$) is outlined in http://home.sandiego.edu/~langton/gaussagm.pdf (Gauss, recurrence relations, and the AGM, by Stacy G. Langton). The final formula is $$\int\limits_0^1\frac{dx}{\sqrt{1-x^4}}=\frac{\pi}{2}\frac{1}{M(1,\sqrt{2})}.$$

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I can't speak for Stirling, but if you break up the integral into an integral from 0 to 1/2 and an integral from 1/2 to 1, both of those converge very fast.

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