The problem of determining the value of $\zeta (2)$ also known as t...
Tom Mike Apostol was an American analytic number theorist and profe...
If you want to learn more about Apéry's proof click [here](https://...
Here the author uses the formula for the sum of a geometric series ...
Note that $\int_{0}^{1} x^n dx= \left[ \frac{x^{n+1}}{n+1}\right]_{...
With this change of variables we change the region and limits of in...
Alternatively we could note that $$ \frac{d}{du} \arctan \frac{...
A Proof that Euler Missed"
Evaluating the Easy Way
Tom M. Apostol
R. Ap6ry [1] was the first to prove the irrationality of
symmetry of this square about the u-axis we find
| 1 l/v~ u
~(3)= ~. 1=4s ( s
dv )du
,=1 2 - u 2 + v 2
Motivated by Ap6ry's proof, F. Beukers [2] has given
a shorter proof which uses multiple integrals to
establish the irrationality of both ~(2) and ~(3). In this
i.* [ r*-u dv
note we show that one of the double integrals con- + 4 |
sidered by Beukers, -,l/V~ 2 -
u 2 + v 2 du.
J0
J
1 fol 1
I = 1 x------y dxdy,
can be used to establish directly that ~(2) = 7r2/6. This
evaluation has been presented by the author for a
number of years in elementary calculus courses, but
does not seem to be recorded in the literature.
The relation between the foregoing integral and ~(2)
is obtained by expanding the integrand in a geometric
series and integrating term by term. Thus, we have
I 1 oo
fo fo xay
n=O
Since
X
fo dt 1 x
a 2 +t 2 - a arctan-~-
we have
u dv 1 u
2-u 2+v 2 ~ X/2-u 2
arc
tan
1 oo n oo
dy = (n +
1) 2 -~(2)"
Now we evaluate the integral another way and show
that I = ~ra/6. We simply rotate the coordinate axes
clockwise through an angle of 7r/4 radians by intro-
ducing the change of variables
U--V Uq-V
x- v~ ,y=--~--
so that 1-xy = (2-u 2 +v2)/2. The new region of
integration in the
uv-plane
is a
square with two
oppo-
site
vertices at (0,0) and (X/-2,0). Making use of the
Tom M. Apostol
THE MATHEMATICAL INTELLIGENCER VOL. 5, NO. 3 9 1983 Springer-Verlag New York 59
and
,/2-u
fo
hence
dv 1
2-u 2+v 2 = ~ arc tan
V2- u
X/2 - u s
1 2
The substitution u = y + -~-x(y - 1) in the integral
with respect to y converts this to
1 1
f f0
P +Q = du l+2ux +x 2 "
1
~tv~- u du
P
I
4
J0 arc tan V~-u s X/2-u s
v~ V~- u du
+ 4 | arc tan
t.b
/,a X/~- u s X/2- u s
J1
-- 11 + I2,
say. Put u = V2 sin 0 in
IlSO
that du = X/2 cos 0 dO
= X/2 - -u ~ dO, and tan 0 = u/X/2
- u 2.
This gives us
(6) 2
11 = 4 OdO =2
Now put u = cos ~ so that (sin ~)/(1 + 2ux + x 2) =
d ( x+cos~),hence
dx arc tan sin
lr
P + Q - 2 ~ d~P = -4 "
which, together with P = 2Q, implies P = 7ra/6.
References
In 12 we put u = X/2 cos 20 so that
du = -2X/2 sin 20 dO = -2X/2 X/1 -cosZ20 dO
=
-2X/2 X/1 - u2/2 dO -- -2V'2 - u 2 dO,
and
X/2 - u V~(1 - cos 20)
V~-_/./2
- X/2-2cos 220
~
1-cos20 _ ;__
1 ~ cos 20
2 sin 2 0
- tan 0,
2 cos 2 0
hence
Iz = 8f0 0d0 =4
(T) 2
Therefore I = I1 + 12 = 6 = 6 "
Note. Another evaluation of ~(2) using double in-
tegrals in a less straightforward manner was given by
F. Goldscheider [3] in response to a problem proposed
by P. St~ckel. He considers the two double integrals
p =
1 1
dx dy
1 1
dx dy
fo fo --andQ= fo fo
1 - xy 1 + xy
1p
and shows first thatP-Q = 2 sop =2Q. Onthe
other hand,
1 1
f fo
P +Q = 1 dy l+xy"
1. R. Ap6ry (1979) Irrationalit6 de ~(2) et ~(3). Astdr-
isque 61:11-13. Paris: Societ6 Math~matique de France.
2. F. Beukers (1979) A note on the irrationality of ~(2) and
~(3), Bull. Lon. Math. Soc. 11:268-272.
3. F. Goldscheider (1913) Arch. Math. Phys. 20:323-324.
California Institute of Technology
Pasadena, California 91125
Ergebnisse der Mathematik,
3. Folge
A Series of Modern Surveys in
Mathematics
Edited by:
S. Feferman, N.H. Kuiper, P. Lax,
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THE MATHEMATICAL INTELLIGENCER VOL. 5, NO. 3, 1983

Discussion

Tom Mike Apostol was an American analytic number theorist and professor at the California Institute of Technology. ![](http://dlmf.nist.gov/about/bio/photo_apostol.jpg) Note that $\int_{0}^{1} x^n dx= \left[ \frac{x^{n+1}}{n+1}\right]_{0}^1=\frac{1}{n+1} $ With this change of variables we change the region and limits of integration ![](http://i.imgur.com/FSkUgUA.jpg) and so $$I= \int \int \frac{1}{1-xy} dxdy = \int \int \frac{2}{2-u^2+v^2} dudv $$ If we look at the integral in uv-space it doesn't change if $u \rightarrow -u $ or $v \rightarrow -v $, and so the integral in the region above the u-axis will be equal to the integral below the u-axis. ![](http://i.imgur.com/bdyxlfK.jpg) $$ I = 2I_{A} = 2 \left( \int_0^{1/\sqrt{2}} \int_{0}^{u} \frac{2}{2-u^2+v^2}dvdu + \\ \int_{1/\sqrt{2}}^{\sqrt{2}} \int_{0}^{\sqrt{2}-u} \frac{2}{2-u^2+v^2}dvdu \right) $$ If you want to learn more about Apéry's proof click [here](https://someclassicalmaths.wordpress.com/2011/10/14/roger-aperys-proof-that-zeta3-is-irrational/) and for Beuker's original proof click [here](http://www.math.wvu.edu/~mays/545/Bull.%20London%20Math.%20Soc.-1979-Beukers-268-72.pdf). Alternatively we could note that $$ \frac{d}{du} \arctan \frac{u}{\sqrt{2-u^2}} = \frac{1}{\sqrt{2-u^2}} \\ \frac{d}{du} \arctan \frac{\sqrt{2}- u}{\sqrt{2-u^2}} =-\frac{1}{2}\frac{1}{\sqrt{2-u^2}} $$ and so the integral becomes $$ I= 4 \int_{0}^{1/\sqrt{2}}f(u)f'(u) du + 4 \int^{\sqrt{2}}_{1/\sqrt{2}}g(u)(-2)g'(u) du \\ = 4\left(\left[f(u)^2/2\right]_{0}^{1/\sqrt{2}}-2\left[g(u)^2/2\right]^{\sqrt{2}}_{1/\sqrt{2}}\right ) $$ Since $f(0)=g(\sqrt2)=0$ and $f(1/\sqrt{2})=g(1/\sqrt{2})=\pi/6$ $$ I = 2\pi^2/6^2+4 \pi^2/6^2= \pi^2/6 $$ The problem of determining the value of $\zeta (2)$ also known as the Basel problem was first posed by [Pietro Mengoli](https://en.wikipedia.org/wiki/Pietro_Mengoli) in 1644. It consisted of determining the value of the following series $$ \zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} $$ A lot of famous mathematicians tried to attack this problem but was Euler at 28 in 1735 who proved for the first time that $\zeta(2) = \frac{\pi^2}{6}$ (the problem is named after Basel, hometown of Euler as well as of the Bernoulli family who unsuccessfully attacked the problem). To prove it Euler used the Taylor expansion of the sine function and also expressed it as a product of the roots of an infinite polynomial (which at the time was still not known to be valid!). In this paper Tom Apostol gives an alternative proof based on solving a double integral which avoids Euler's expansions of the sine function. The Basel problem and Euler's ideas were later used by Riemann in his seminal 1859 paper "On the Number of Primes Less Than a Given Magnitude", in which he defined his zeta function $$ \zeta (s) = \sum_{n=1}^{\infty} \frac{1}{n^s} $$ one of the most important functions in mathematics. Here the author uses the formula for the sum of a geometric series $$ s = \sum_{n=0}^{\infty} x^n = 1 + x +x^2 + ... $$ if we multiply both sides by $x$ we have $$ sx = x+ x^2 + x^3+ ... \\ sx = s -1 \\ s = \frac{1}{1-x}, \text{ for $|x|<1$} $$ And so $\sum_{n=0}^{\infty} (xy)^n= \frac{1}{1-xy}$