263

Integral proofs that 355/113 > πStephen K. Lucas

1Introduction

One of the more beautiful results related to approximating π is the integral
Z 1 422x (1 − x)4dx =− π.
21+x
70(1)

Since the integrand is nonnegative on the interval [0, 1], this shows that π is strictly lessthan 22/7, the well known approximation to π. The first published statement of this result
was in 1971 by Dalzell [3], although anecdotal evidence [2] suggests it was known by Kurt
Mahler in the mid-1960s. The result (1) is not hard to prove, if perhaps somewhat tedious.
A partial fraction decomposition leads to a polynomial plus a term involving 1/(1 + x2 ),which integrates immediately to the required result. An alternative is to use the substitution
x = tan θ, leading to a polynomial in powers of tan θ. We then apply the recurrence relation
for taking the integrals of powers of tan θ. Of course, the simplest approach today is tosimply verify (1) using a symbolic manipulation package such as Maple or Mathematica.
An obvious question at this point might be whether similar elegant integral results canbe found for other rational approximations for π. A particularly good approximation is355/113, which is accurate to seven digits. Our aim here is to find a variety of such integralresults.

2

More general integrals

Backhouse [1] recently generalized (1) by varying powers in the numerator of the integrand,leading toZ 1 m
x (1 − x)nIm,n =dx = a + bπ + c log 2,(2)1 + x20
where a, b, c are rationals that depend on the positive integers m and n, and a and b have
opposite sign. Backhouse [1] showed that if 2m − n ≡ 0 (mod 4), then c = 0 and a varietyof approximation to π are obtained. An integral equal to a + bπ leads to a rational approximation of π as |a/b|, and the maximum value of the integrand gives an upper bound on theerror. As m and n increase, the integrand becomes increasingly flat (Backhouse calls them“pancake functions”) and the approximation to π improved. Unfortunately, no form of (2)leads to the approximation 355/113.
Another question raised in [1, 2] is whether there are other families of integrals withpositive integrands that can be used to give increasingly accurate estimates of π. One suchclass of integrals not previously considered isZ 1 mx (1 − x)n√Jm,n =dx = a + bπ,(3)1 − x20
where a and b are rationals of opposite sign depending on the positive integers m and n.
Note that an approximation to π can be found from (3) for every choice of positive m andn. For comparison purposes, Table 1 lists Im,n and Jm,n for the first few m and n for which

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Stephen K. Lucasm nIm,n1 2π/2 − 3/2
3 219/12 − π/2π − 47/15
2 41 663/10 − 2π
π/2 − 47/30
5 24 422/7 − π
3 6 2π − 1759/280
7 2 1321/840 − π/2
2 8 3959/315 − 4ππ − 1979/630
6 41 10 8π − 15829/630
377/60 − 2π
5 69 2 π/2 − 989/630
4 8 4π − 4838/385
8 4 10886/3465 − π

I error
Jm,nJ error
1.4e − 15/3 − π/21.9e − 1
2.5e − 26/5 − 3π/85.8e − 28.3e − 349π/32 − 24/5
6.9e − 3
8.4e − 3681/35 − 99π/16
3.0e − 3
8.3e − 3104/105 − 5π/162.8e − 2
1.3e − 3323π/256 − 416/105
1.5e − 3
5.2e − 45018/315 − 649π/128
2.6e − 4
3.6e − 3272/315 − 35π/128
1.6e − 24.7e − 4 10439π/512 − 20176/315
1.0e − 4
3.2e − 4563π/512 − 1088/315
5.1e − 4
9.2e − 4 178555/693 − 20995π/256 9.4e − 57.4e − 5 48008/3465 − 1129π/256
4.7e − 5
1.9e − 3128/165 − 63π/256
1.1e − 2
3.4e − 5 36231π/2048 − 64192/1155 9.5e − 6
1.1e − 42023π/2048 − 512/165
2.2e − 4

Table 1. Comparing Im,n and Jm,nIm,n does not involve log 2, as well as the error in the approximation to π. While the errors
are similar, in no case is the approximation 355/113 forthcoming.

3

Some integral results

In what follows, we shall look at a variety of ways of forming integrals with positive integrands that evaluate to 355/113 − π.
3.1Idea 1 – set b = 1

The first idea is to multiply the integrands in (2) and (3) by a positive rational α so thatthe coefficient multiplying π becomes one, and the integral equals a0 − π. Note that thislimits our choices of m, n to when b is positive. We then require m, n large enough thata0 < 355/113. If we add β = 355/113 − a0 to both sides, we get
Z 1 m355
αx (1 − x)n+ β dx =− π,(4)21+x1130andZ 1αxm (1 − x)n355√+ β dx =− π,(5)
2
113
1−x0
where α and β (different for the two different integrals, of course) are positive rationals.
Since the integrands in (4,5) are nonnegative, this gives the required integral proof that355/113 > π. Numerical testing with the criteria of choosing β with the smallest number ofdigits leads to the best results
Z 1  10x (1 − x)85355+dx =− π,(6)4(1 + x2 )138450312
1130andZ 108192 x9 (1 − x)815409
355√dx =− π.
+114291219772564011
113
1 − x2(7)Integral proofs that 355/113 > π265

Now while these integrals do indeed have positive integrands, in a sense they are not reallygiving us an approximation involving 355/113. We have simply chosen m and n large
enough so that the approximations to π given by Im,n and Jm,n are better than 355/113,
then degraded the results to get the approximation 355/113 by adding a “fudge factor” tothe integrand. While strictly correct, they lack a certain elegance.
3.2Idea 2 – set a = 355/113

Another approach is to multiply the integrands of (2) or (3) by a positive rational α so thatthe integral equals 355/113 − b0 π. If we choose m and n large enough that b0 is a positive
rational greater than one, then we can choose β = 4(b0 − 1) and β 0 = 2(b0 − 1) so that
Z 1Z 1355
αxm (1 − x)n + βαxm (1 − x)n + β 0355√dx=−πanddx =− π.
(8)221+x113
113
1−x00R1R1 √These results use the integrals 0 1/(1+x2 ) dx = π/4 and 0 1/ 1 − x2 dx = π/2, and since
the integrands are nonnegative, once again lead to 355/113 > π. Numerical testing withvarious values of m and n lead to the simplest results in terms of the number of digits in αand β as
Z 135521747726x10 (1 − x)8 + 4dx =− π,(9)2)
86990903(1+x1130andZ 10 .p355355 m355
15409
nx (1 − x) +− π.
1 − x2 dx =10535216
345217957888
113(10)

Unfortunately, the previous argument can also be used against the elegance of these results,
its just the position of the addition to the integral to degrade the approximation that haschanged.
3.3

Idea 3 – multiply by a polynomial

Finally, we can take the approach of multiplying the integrand by a low order polynomial,
and adjusting the coefficients to return the correct result. We then choose the simplest casewhere the polynomial is nonnegative on [0, 1] – simplest in the sense of the smallest numberof digits in the coefficients. Experimenting with Im,n leads to the results
Z 1 7x (1 − x)7 (192 − 791x + 983x2 )355dx =− π,(11)23164(1 + x )113
0andZ 10x8 (1 − x)8 (25 + 816x2 )355dx =− π.
3164(1 + x2 )
113

(12)

While the powers are lower in (11), the number of characters in the expression is larger
than in (12), and it is not immediately obvious that the integrand is nonnegative. The
integral (12) is nearly as simple as (1), and it is immediately obvious that the integrand isnonnegative. The best case for Jm,n isZ 1 7x (1 − x)8 (617273 − 478592x)355√dx =− π,(13)2113
16288272 1 − x0
which only uses a linear factor. The results using quadratic factors are more complicated.

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4Conclusion

After testing a variety of families of integrals (some of which are not mentioned here), theresult (12) is the simplest formula available that represents the gap between 355/113 and π
as the integral of a positive integrand. Since 1 < 1 + x2 < 2 for the integral range, we canuse it to form the bound911335
911355−<π<−,(14)113 2630555928
113 5261111856
or 355/113 − 3.46 × 10−7 < π < 335/113 − 1.73 × 10−7 . The true result is π = 355/113 −2.67 × 10−7 , which is very close to the middle of the bound.

References[1] N. Backhouse, Note 79.36, Pancake functions and approximations to π, Math. Gazette 79 (1995),371–374.
[2] J.M. Borwein, The life of Pi, history and computation, seminar presentation 2003, available fromhttp://www.cecm.sfu.ca/~jborwein/pi-slides.pdf, March 2005.
[3] D.P. Dalzell, On 22/7 and 355/113, Eureka: the Archimedian’s Journal 34 (1971), 10–13.
School of Mathematics and Statistics, University of South Australia, Mawson Lakes SA 5095
E-mail: Stephen.Lucas@UniSA.edu.auReceived 10 February 2005, accepted 28 March 2005.