# How to Prove the Riemann Hypothesis ∫

```Proceedings of the World Congress on Engineering and Computer Science 2010 Vol I
WCECS 2010, October 20-22, 2010, San Francisco, USA
How to Prove the Riemann Hypothesis
Abstract- To prove the Riemann Hypothesis is to show
that the nontrivial zeros of the Riemann zeta function,
which are complex, have real part equal to 0.5 . The proof
given herein is divided into two parts. Integral calculus is
used in the first part, while variational calculus is employed
in the second part. Given that (a) is the real part of any
nontrivial zero of the Riemann zeta function, it is assumed
that (a) is a fixed exponent in the equations 50-59 and hence
it is verified that a=0.5. In the remaining equations
beginning in equation (60) (a) is treated as a parameter
(a<0.5) and a contradiction is obtained.
At the end of the proof (from equation (73) onward), it is
verified again that a = 0.5
Index Terms-Functional Equation, L'Hospital's Rule,
Variational Caculus.
Subj-class: Functional analysis, complex variables,
general mathematics.
I. INTRODUCTION
The Riemann zeta function is the function of the complex
variable s = a + bi (i = − 1 ), defined in the half plane a >1 by
the absolute convergent series
∞
1
ζ (s) = ∑
(1)
s
1 n
and in the whole complex plane by analytic continuation.
The function ζ (s ) has zeros at the negative even integers -2, 4, … and one refers to them as the trivial zeros. The Riemann
Hypothesis states that the nontrivial zeros of ζ (s ) have real
part equal to 0.5.
II, PROOF OF THE RIEMANN HYPOTHESIS
We begin with the equation
ζ (s) =0
(2)
where
s= a+bi
(3)
i.e.
ζ ( a + bi ) = 0
(4)
It is known that the nontrivial zeros of ζ (s ) are all complex.
Their real parts lie between zero and one.
If
0 < a < 1 then
ζ (s)
∞
=s
[ x] − x
∫0 x s + 1 dx
(0 < a < 1)
(5)
[x] is the integer function
∞
∫x
Separating the real and imaginary parts we get
∞
∫x
− 1 − a ([ x] − x) cos(b log x)dx = 0
(10)
− 1 − a ([ x] − x) sin(b log x)dx = 0
(11)
0
∞
∫x
0
According to the functional equation, if
Hence we get besides (11)
∞
− 1 − a − bi dx = 0
0
∞
∫y
− 1 − a ([ y ] − y ) sin(b log y )dy = 0
∫
([ x] − x) x − 1 − a x − bi dx = 0
(13)
0
We form the product of the integrals (12)and (13).This is justified by
the fact that both integrals (12) and (13) are absolutely convergent .As
to integral (12) we notice that
∞
∫
x − 2 + a ([ x] − x) sin(b log x)dx ≤
∞
x− 2 + a
∫
0
0
([x]-x) sin (blog x)
∞
∫x
≤
dx
− 2 + a (( x))dx
0
( where ((z)) is the fractional part of z , 0 ≤ ((z))<1)
1− t
= lim(t →
0)
∫
x −1 + a dx + lim (t → 0 )
0
∞
x − 2 + a ((x))dx
∫
1+t
(t is avery small positive number) (since ((x)) =x whenever 0 ≤ x<1)
∞
=
1
− 2 + a ((x))dx
+ lim (t → 0 ) ∫ x
a
1+t
<
1
− 2 + a dx = 1 + 1
+ lim (t → 0 ) ∫ x
a
a a −1
1+t
∞
(6)
∞
As to integral (13)
(7)
≤
(8)
0
Fayez Fok Al Adeh is the President of the Syrian Cosmological Society
∫y
− 1 − a ([ y ] − y ) sin(b log y )dy
0
∞
0
∞
(12)
In (11) replace the dummy variable x by the dummy variable y
Therefore
∫ ([ x] − x) x
ζ (s) =0 then ζ (1 − s) =0.
− 2 + a ([ x] − x) sin(b log x)dx = 0
∫x
∞
∞
(9)
0
Hence
[ x] − x
∫0 x s + 1 dx = 0
− 1 − a ([ x ] − x )(cos( b log x ) − i sin( b log x )) dx = 0
y − 1 − a ([ y ] − y ) sin(b log y )
∫
dy
0
∞
≤
∫y
− 1 − a (( y ))dy
0
(Phone 00963 – 11 – 2713005; email: [email protected];
airmail address: P.O.Box 13187 Damascus Syria).
ISBN: 978-988-17012-0-6
ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online)
WCECS 2010
Proceedings of the World Congress on Engineering and Computer Science 2010 Vol I
WCECS 2010, October 20-22, 2010, San Francisco, USA
1−t
= lim (t →
0)
∫
Rewrite this integral in the equivalent form
y − a dy + lim(t → 0 )
∞∞
∫∫x
0
∞
0 0
∞∞
0 0
(t is avery small positive number) (since ((y)) =y whenever
0 ≤ y<1)
1
− 1 − a (( y ))dy
+ lim(t → 0 ) ∫ y
1− a
1+t
∞∞
∫∫x
∞
1
− 1 − a dy = 1 + 1
+ ∫ y
1 − a 1+t
1− a a
∫∫x
1
{( x 2 − 2a [ ] −
x
− 2 + a y − 1 − a ([ x] − x)([ y ] − y )
(14)
0 0
sin(b log y ) sin(b log x)dxdy = 0
Thus
∞ ∞
∫∫x
− 2 + a y − 1 − a ([ x ] − x )([ y ] − y )
(15)
0 0
(cos( b log y + b log x ) − cos( b log y − b log x )) dxdy = 0
∞ ∞
∫∫x
− 2 + a y − 1 − a ([ x ] − x )([ y ] − y )
0 0
(cos( b log xy ) − cos( b log
(16)
y
)) dxdy = 0
x
limit is given by
1
x − a y − 1 − a [ - (( y)) ] cos (blog xy ) [ - (( )) +
x
2a − 2 ]
((x)) x
(32)
Lim
0 0
The above limit vanishes ,since all the functions [ - ((y)) ] ,
∞∞
− 2 + a y − 1 − a ([ x] − x)([ y ] − y ) cos(b log y )dxdy
(17)
∫0 ∫0 x
x
cos (blog xy ), - ((
∞∞
(18)
1
− dz
z dx = z 2
∫∫
F(x,y) dx dy is bounded as follows
∫∫
F(x,y) dx dy =
I1
The integral becomes
∞ 0
(19)
I1
1
x
∞ 0
(20)
I1
This is equivalent to
If we replace the dummy variable z by the dummy variable x,
the integral takes the form
∞∞
∫∫x
0 0
− a y − 1 − a ([ 1 ] − 1 )([ y ] − y ) cos(b log xy )dxdy
x
x
ISBN: 978-988-17012-0-6
ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online)
)) + ((x))
(21)
(22)
∞
=
x − a y − 1 − a [ - (( y)) ] cos (blog
x 2a − 2
∞
∫ (∫
p
)) + ((x)) x
2a − 2
] dx dy
(34)
1
x − a y − 1 − a [ - (( y)) ] cos (blog xy ) [ - ((
x
∫∫
≤
∞∞
∫∫
(33)
I1
xy ) [ - ((
That is
− a − 1 − a ([ 1 ] − 1 )([ y ] − y ) cos(b log zy )dzdy
∫0 ∫0 z y
z
z
1
)), and ((x)) remain bounded as x → ∞
x
and y → ∞
Note that the function F (x,y) is defined and bounded in the
region I 1. We can prove that the integral
Consider the integral on the right-hand side of (17)
1 1
− ∫ ∫ z − a y − 1 − a ([ ] − )([ y ] − y ) cos(b log zy )dzdy
z
z
0 ∞
)-([x]-x)}dx dy=0
x − 2 + a y − 1 − a ([ y ] − y ) cos( b log xy )
1 x 2 − 2a
2
2
a
−
{( x
[ ]−
)-([x]-x)}
(31)
x
x
Let us find the limit of F (x,y) as x → ∞ and y → ∞ . This
− 2 + a y − 1 − a ([ x] − x)([ y ] − y ) cos(b log xy )dxdy =
∫∫x
2 − a − 1 − a ([ 1 ] − 1 )([ y ] − y ) cos(b log zy ) − dz dy
∫0 ∞∫ z y
z z
z2
x
(25)
Denote the integrand in the left hand side of (25) by F (x,y) =
∞∞
In this integral make the substitution x =
x 2 − 2a
Let p >0 be an arbitrary small positive number.we consider the
following regions in the x –y plane.
(26)
The region of integration I = [0, ∞ ) ×[0, ∞ )
The large region I1 =[ p, ∞ ) × [ p, ∞ )
(27)
The narrow strip I 2 =[ p, ∞ ) × [0,p ]
(28)
The narrow strip I 3 = [0,p]× [ 0, ∞ )
(29)
Note that
(30)
I = I1 U I 2 U I 3
That is
− 2 + a y − 1 − a ([ x] − x)([ y ] − y ) cos(b log y )dxdy
∫0 ∫0 x
x
− 2 + a y − 1 − a ([ y ] − y ) cos(b log xy)
0 0
Since the limits of integration do not involve x or y , the
product can be expressed as the double integral
∞∞
(24)
2 − 2a
− 2 + a y − 1 − a ( x 2 − 2a [ 1 ] − x
)([ y ] − y ) cos(b log xy )dxdy
∫0 ∫0 x
x
x
Write (24) in the form
∞∞
∞
<
− 2 + a y − 1 − a ([ x] − x)([ y ] − y ) cos(b log xy )dxdy =
∫∫x
1+t
=
(23)
Thus (17) becomes
y − 1 − a (( y ))dy
∫
2 − 2a
− 2 + a y − 1 − a ( x 2 − 2a [ 1 ] − x
)([ y ] − y ) cos(b log xy )dxdy
x
x
x− a
] dx dy
cos (blog xy ) [ - ((
p
x 2a − 2
] dx)
1
x
)) + ((x))
y − 1 − a [ - (( y)) ]dy
WCECS 2010
Proceedings of the World Congress on Engineering and Computer Science 2010 Vol I
WCECS 2010, October 20-22, 2010, San Francisco, USA
∞
≤
∞
∫
(
p
x− a
∫
p
x 2a − 2
∞
∞
x− a
(
p
[ - ((
x 2a − 2
∞
∫
x− a
]
dx)
[ ((
1
x
Ρ
∞
1
x
x 2a − 2
=
∞
[ ((
Ρ
1
x
∫∫
F(x,y) dx dy is bounded
∫∫
F(x,y) dx dy =
∫∫
F(x,y) dx dy
∫
x− a
[ ((
1+ t
1
x
lim(t →
0)
∫
)) + ((x))
x− a
[ ((
Ρ
∫
x− a
[ ((
1+ t
1
x
Since x >1 ,then ((
∞
lim(t →
0)
1
x
)) + ((x))
= lim(t →
0)
∫
1
x
)) =
1
x
[ ((
x− a
1+ t
∞
= lim(t →
0)
∫
x 2a − 2
≤
] dx }
] dx
∫
[
)) + ((x))
+ ((x))
≤
x 2a − 2
x 2a − 2
] dx
∞
] dx
0
x − a { ((
∫
1
ya
))- ((x))
− a { (( 1
x
1
ya
p
(40)
dx)
1
x
x 2a − 2 }
))- ((x))
x 2a − 2 }
dy
1
x
))- ((x))
1
ya
x 2a − 2 }
dy
))- ((x))
x 2a − 2
}
dx ×
dy
(This is because in this region ((y)) = y). It is evident that the
∞
integral
] dx
(39)
dy
x − a { ((
∫ ∫
(
Ρ
Ρ
∫
x − a { ((
∫
Ρ
1
x
))- ((x))
x 2a − 2
}
dx is
bounded,this was proved in the course of proving that the
[x
− a −1+ xa − 2
integral
] dx
Ρ
1
=
a(1 − a)
Hence the boundedness of the integral
∫
0
∫∫
F(x,y) dx dy is
F(x,y) dx dy is bounded .Also it is evident that
1
ya
dy
is bounded. Thus we deduce that the integral (40)
dx dy is bounded
I1
ISBN: 978-988-17012-0-6
ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online)
∫∫
the integral I1
1+ t
proved.
Consider the region
I4=I2 U I3
We know that
p
∞
and we have
1
x
(x
cos(b log xy )
] dx is bounded.
− a − 1 + ((x)) x a − 2
[x
∞
0)
∫
∫
p
≤
F(x,y) dx dy
F(x,y) dx dy
1
x
1
ya
cos(b log xy ) dx)
x 2a − 2
1
x
∫∫
x − a { ((
∞
0
x 2a − 2
∫∫
I3
≤
cos(b log xy ) dx)
1+ t
< lim(t →
F(x,y) dx dy +
p
p
)) + ((x))
)) + ((x))
1+ t
∞
∫ (∫
0
x− a
∫
(38)
I2
∞
p
] dx
is bounded, it remains to
show that lim(t → 0 )
∞
∫∫
I2
where t is a very small arbitrary positive.
number.Since the integral
1− t
(37)
I4
Remember that
0
1
x
[ ((
F(x,y) dx dy (36)
I4
dy
=
1− t
∫∫
I1
From which we deduce that the integral
] dx
x 2a − 2
)) + ((x))
1
−a
{ lim(t → 0 ) ∫ x
=
ap a
Ρ
x 2a − 2 ] dx + lim(t → 0 )
∞
I1
I4
I2
Consider the integral
x− a
∫
F(x,y) dx dy +
)) +
Ρ
1
ap a
∫∫
F(x,y) dx dy is bounded
y−1− a
∫
F(x,y) dx dy =
∫∫
y − 1 − a [ - (( y)) ]
)) + ((x))
∫∫
I
and that
dy
p
((x))
<
cos (blog xy )
0=
)) + ((x))
y − 1 − a [ - (( y)) ]
] dx)
∫ ∫
≤
1
x
cos (blog xy ) [ - ((
Hence ,according to (39),the integral
(35)
bounded.
∫∫
F(x,y)
I2
∫∫
F(x,y) dx dy is
I3
WCECS 2010
Proceedings of the World Congress on Engineering and Computer Science 2010 Vol I
WCECS 2010, October 20-22, 2010, San Francisco, USA
Now we consider the integral
∫∫
∞
F(x,y) dx dy
(41)
∫∫
p ∞
∫ (∫
y
−1− a
∞
0
0
y
∞
∫
(
0
p ∞
∞
y
−1− a
∫
−1− a
((y)) cos (b logxy) dy )
dx
1
{(( )) − x 2 a −1 }
x
xa
((y)) dy )
((y)) dy
dx
(43)
1− t
= (lim t →
0)
∫
y
−1− a
× y dy+ (lim t →
0)
0
y
−1− a
≤K
(44)
y
−1− a
((y)) cos (b logxy)dy
≤
K
(45)
p ∞
F(x,y) dx dy =
0
Since
0
∫
p
∫∫
( where t is a very small arbitrary positive number) .( Note
that ((y))=y whenever 0 ≤ y<1).
∞
Thus we have (lim t →
0)
∫
y
−1− a
((y)) dy <
y
−1− a
dy=
1+ t
1− t
and (lim t →
0)
∫
y
−1− a
1
a
((y)) cos (b logxy) dy)
0
(46)
F(x,y) dx dy is bounded, then
1
{(( )) − x 2 a −1 }
x
dx is bounded
xa
Hence the integral (43)
∫
y
−1− a
1
1− a
∫
0
1
xa
δ
{ ((
1
x
))-
G [F] =
(48)
G [F] be the variation of the integral G due to the
∫
δ
F.
F dx (the integral (49) is indefinite )
(49)
( here we do not consider a as a parameter, rather we treat it as a
given exponent)
We deduce that
δG[ F ]
=1
δF ( x)
that is
δ G [F] = δ F (x)
But we have
ISBN: 978-988-17012-0-6
ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online)
x 2a − 1 }
variation of the integrand
Since
((y)) dy is bounded.
0
(47)
We denote the integrand of (47) by
Let
× y dy=
0
∞
G=
F=
1+ t
∫
−1− a
1
{(( )) − x 2 a −1 }
x
dx is also bounded. Therefore
xa
p
0)
y
the integral
((y)) dy
∞
∫ (∫
1
{(( )) − x 2 a −1 }
x
dx
xa
1
{(( )) − x 2 a −1 }
p
x
≥ ∫ (− K )
dx
xa
0
1
{(( )) − x 2 a −1 }
0
x
=K ∫
dx
xa
p
1+ t
(lim t →
= H (x)
I3
−1− a
y
0
∫
∫
≤
∫∫
Now we consider the integral with respect to y
∞
((y)) cos (b logxy)dy
According to (42) we have
0
y
−1− a
y
0
–K
((y)) cos (b logxy) dy)
1
{(( )) − x 2 a −1 }
x
xa
∫
is a bounded function of x
( K is a positive number )
Now (44) gives us
0
∞
((y)) cos (b logxy)dy
0
p
0
∫
(42)
0
≤ ∫ (∫
−1− a
0
−1− a
y
∞
1
{(( )) − x 2 a −1 }
x
dx
xa
≤
y
. Let this function be H(x) . Thus we have
( This is because in this region ((x)) = x )
∫ (∫
∫
0
1
{(( )) − x 2 a −1 }
x
dx
xa
≤
∫
((y)) cos (b logxy) dy)
I3
−1− a
≤
((y)) cos (b logxy)dy
((y)) dy , we conclude that the integral
0
p ∞
y
∞
0
I3
and write it in the form
F(x,y) dx dy =
∫
Since
−1− a
(50)
WCECS 2010
Proceedings of the World Congress on Engineering and Computer Science 2010 Vol I
WCECS 2010, October 20-22, 2010, San Francisco, USA
δ
G[F] =
∫
dx
δG[ F ]
δF (x)
δF ( x)
(51)
p
( the integral (51) is indefinite)
Using (50) we deduce that
δ
G[F] =
∫
dx
G = lim (t
δF (x)
(52)
δG = ∫
Ρ
dF
δx
dx
dx
δF (x) = ∫
dx
0
0
= [ Fδ
x
] (at
x = p)- [ F δ x ] (at x = 0)
(54)
Since the value of [ F δ x ] (at x = p)is bounded, we
deduce from (54) that
lim (x → 0 ) F δ x must remain bounded.
(55)
Thus we must have that
(lim x
1
xa
→ 0) [ δ x
1
x
{ ((
x 2a − 1 }]
))-
(56)
x
a
×
δx
0)
d (δx)
=0
dx
x
a
= (lim x →
1
0 ) × x1− a
a
1
x
))-
1
x
) { ((
))-
x 2a − 1 }
will remain bounded in the limit as (x
dx
δG a[x]
δx
=
δG a[ x]
∫ dx
Thus (64) reduces to
p
∫
→ 0)
G – Fx (at p ) = - lim (t
x dF
(66)
t
Note that the left – hand side of (66) is bounded. Equation (63)
gives us
δ
p
G a =lim (t →
0)
∫
F
δx
x
dx
(67)
We can easily prove that the two integrals
dx
F
δx
x
∫
x dF and
are absolutely convergent .Since the limits of
integration do not involve any variable , we form the product of
(66) and (67)
→ 0)
(59)
p
∫ ∫
t
= lim(t
→ 0)∫
xdF× dx
t
→ 0 ) .Therefore , in
F ( x, a )
x
x
(60)
δG a [ x ]
δx
δx
F ( x, a )
∫ dx x δ x
ISBN: 978-988-17012-0-6
ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online)
p
FdF ×
t
∫
F
δx
x
δ xdx
(68)
t
( K is a bounded quantity )
That is
K = lim(t
→ 0) [
F2
2
( at p ) -
F2
2
(at t) ] × [
(at t) ]
We conclude from (69) that
{[
F2
2
( at p) - lim(t
is bounded .
( since lim(x
→ 0)
→ 0 )δ
δ
x (at p) -
δ
x
(69)
F2
2
(at t) ] ×[ δ
x
(at p) ] }(70)
x = 0 , which is the same thing as
(61)
→ 0 ) δ x = 0)
F2
F2
( at p) is bounded , we deduce at once that
must
Since
2
2
remain bounded in the limit as (t → 0 ), which is the same thing
as saying that F must remain bounded in the limit as (x → 0) .
(62)
Therefore .
lim(t
( the integral (62) is indefinite )
Substituting from (61) we get
=
(65)
p
F ( x, a )
x
But we have that
δG a[ x]
1
t
t 1− a (( )) – t a = 0
K = lim(t
(the integral (60) is indefinite )
Thus
=
lim (t → 0) Fx (at t ) = lim (t → 0)
p
this case (a=0.5) (56) will approach zero as (x → 0 ) and
hence remains bounded .
Now suppose that a< 0.5 .In this case we consider a as a
parameter.Hence we have
∫
Let us compute
t
x 2a − 1 }
(60) G a [x]=
x dF
t
∫
must remain bounded.
Assume that a =0.5 .( remember that we are treating a as a
given exponent )This value a =0.5 will guarantee that the
quantity
{ ((
∫
→ 0) Fx (at t) } - lim (t → 0)
t
(58)
× (lim x → 0
= { F x(at p ) – lim (t
p
We conclude from (56) that the product
0
p
p
(57)
Applying L 'Hospital ' rule we get
(lim x →
( t is a very small positive number 0<t<p)
(t is the same small positive number 0<t<p)
δx
0)
(64)
t
is bounded .
First we compute
(lim x →
→ 0) ∫ Fdx
t
( the integral (52) is indefinite)
Since G[F] is bounded across the elementary interval [0,p],
δ G[F] is bounded across this interval
(53)
From (52) we conclude that
Ρ
( the integral (63) is indefinite )
lim (x
(63)
1
(( )) − x 2 a −1
→ 0) x a
x
(71)
must remain bounded
WCECS 2010
Proceedings of the World Congress on Engineering and Computer Science 2010 Vol I
WCECS 2010, October 20-22, 2010, San Francisco, USA
But
1
(( )) − x 2 a −1
x
lim(x → 0 )
= lim(x → 0 )
xa
1
(( )) − x 2 a −1
1− 2 a
x
× x a
1− 2 a
x
x
1
x1− 2 a (( )) − 1
x
= lim(x → 0 )
x1− a
−1
= lim(x → 0 ) 1−a
x
(72)
1
( − 2)
a
lim(a → 0.5) (x → 0)
x 1− a
(0.5 + x) −1 − 2
= lim (x → 0)
x 0.5− x
(0.5 + x) −1 − 2
= lim (x → 0)
x 0.5 × x − x
(0.5 + x) −1 − 2
= lim (x → 0)
x 0.5
−x
( Since lim (x → 0) x
=1)
(80)
It is evident that this last limit is unbounded. This contradicts
our conclusion (71) that
lim (x
)
1
(( )) − x 2 a −1
→ 0) x a
x
must remain bounded (for a< 0.5
Therefore the case a<0.5 is rejected .We verify here that ,for a =
0.5 (71) remains bounded as (x → 0 ) .
We have that
1
(( )) − x 2 a −1
x
< 1- x
2 a −1
(73)
We must apply L 'Hospital ' rule
lim (x
→ 0)
(0.5 + x) −1 − 2
= lim (x
→ 0)
− (0.5 + x) −2
0.5 x −0.5
= lim (x
→ 0)
− 2 × x 0.5
(0.5 + x ) 2
Therefore
lim(a
(x
→ 0.5) (x
→ 0)
1
(( )) − x 2 a −1
→ 0) x a
x
< lim(a
→ 0.5)
1 − x 2 a −1
xa
(74)
We consider the limit
lim(a
→ 0.5) (x → 0)
We write
a = (lim x →
Hence we get
lim(a
1 − x 2 a −1
xa
(75)
0 ) ( 0.5 + x )
→ 0.5) (x → 0) x 2 a −1
(76)
= lim (x
→ 0)
x
→ 0) x 2 x = 1
x
(Since lim(x → 0) x = 1)
=lim (x
2 ( 0.5 + x ) −1
(81)
x 0.5
=0
Thus we have verified here that ,for a = 0.5 (71) approaches
zero as (x → 0 ) and hence remains bounded.
We consider the case a >0.5 . This case is also rejected, since
according to the functional equation, if ( ζ (s ) =0) (s = a+ bi)
has a root with a>0.5 ,then it must have another root with another
value of a < 0.5 . But we have already rejected the case with
a<0.5
Thus we are left with the only possible value of a which is a =
0.5
Therefore a = 0.5
This proves the Riemann Hypothesis .
(77)
References
[1] E. C. Titchmarch, The Theory of the Riemann zeta –
function. London: Oxford University Press, 1999, pp.13-14.
Therefore we must apply L 'Hospital ' rule with respect to x in the
[2] T. M. Apostol, Mathematical Analysis. Reading,
limiting process (75)
Massachusetts: Addison – Wesley Publishing Company,
1 − x 2 a −1
1974, pp. 76 – 77, pp. 391 –396.
= lim(a → 0.5)
lim(a → 0.5) (x → 0)
[3] H. M. Edwards, Riemann's zeta function. New York:
xa
Academic Press Inc., 1974, pp. 37-38.
− (2a − 1) x 2 a − 2
[4] T. M. Apostol, Introduction to Analytic Number Theory.
(78)
(x → 0)
ax a −1
New York: Springer – Verlag, 1976, pp. 249 – 273.
( − 2)
Functions. New York: Springer – Verlag, 1984, pp. 109 –
a
= lim(a → 0.5) (x → 0)
128.
x 1− a
[6] W. Geiner, and J. Reinhardt, Field Quantization. Berlin:
Now we write again
Springer – Verlage, 1996, pp. 37-39.
a = (lim x → 0 ) ( 0.5 + x )
(79)
Thus the limit (78) becomes
ISBN: 978-988-17012-0-6
ISSN: 2078-0958 (Print); ISSN: 2078-0966 (Online)
WCECS 2010
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