DEFINITE INTEGRALS by MANNOS Submitted in Partial Fulfillment of

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DEFINITE INTEGRALS
by
MURRAY MANNOS
Submitted in Partial Fulfillment of
the Requirements for the Degree of
Bachelor of Science in Mathematics
Massachusetts Institute of Technology
June 1937
Professor in Charge:
Professor R. D. Douglass
INTRODUCTION
The first two chapters consist of the evaluation
and extension of a number of definite integrals; whereas, in
the third chapter, an attempt is made to show a possible
method for obtaining the sum of a finite series by means of
definite integrals somewhat of the form used in Chapters I
and II.
It is perhaps advisable to outline the useful
notions and formulae which are used repeatedly in the evaluation of the definite integrals.
This will enable the
reader to follow the steps more readily without an undue
repetition of the same theorems, etc., between equations.
The notions and formulae, the proofs of which may be found
in practically any text on Functions of a Complex Variable,
will merely be stated as it is not deemed necessary to
repeat such proofs since this would merely take up space.
Methods of finding residues:
(1) Simple pole as of function
Then
Lim (z - a)
z-a
if the Limit
$
(z) =
1
f(z)
z - a
=
(z)
f (a) is the residue at
is a definite number.
a
tHospital t s rule that, if
(2) It follows from L'
S(z)
repeated factor of
and if
z - a
and
is a nonat
the residue of
V/(z),
_4 ial_
is
a
are holomorphic at
W(z)
a
- (a)
( f h (a)
(3) If we have poles of higher multiplicity, then we use
Taylorts expansion
where
f (z)
a
(z -a)
is real or complex.
$' (a) + (z -
(a) + (z - a)
(z - a)
If
n = 1
If
n = 2
If
n = 3
$
then
+
....
e...
the residue at
is
(a)
a) 2 •_a)
21
"1'(a)
n nna
_2!
gn-1l
If n = n
(n-1) .
Two useful theorems for showing that
(where
R--> oo and
lim
Z
If
lim
(z - a)
f (z)
around
r--
r
z f (z)
= K
CO
where
=k
f (z) dz
o
or
around large circle
k
is a constant,
z-- a
lim
R
are the following:
e91) K
m SR f () dz = i (92
R--oo
and
0
=
r--o)
I
=
(@2 -
1)
k
then
]
The following useful formulae are given:
z
eiz + e-iz
Cos z
=
-
e,-.---
eiz
sin z
e
+ e
e+e
2
cosh z
-
e-iz
sin i z
=
i sinh z
cos i z
=
cosh z
sin
29
S 7T
e z - e -Z
sinh z
2i
2
=
1
=
where
0
- z
z2
g
- 1
7r
2
TABLE OF CONTENTS
Chapter I
Chapter II
Evaluation of Definite Integrals . . . .
-
Extension of the Evaluation of One
Defirnite Integral to That of Many
Others
Chapter III
-
..
.....
.
20
. . . . .
23
. ..............
Summation of a Finite Series by
Means of Definite Integrals
Chapter I
EVALUATION OF DEFINITE INTEGRALS
cos mx
0 x4 + a4
z.z za,
a[Cos
+ i sin jvr---4+ 2k7Q]
=- cos (7r+-2k7TI
+ 1~~sin
(T--r
where
k = 0, 1, 2, 3
Consider
imz
z 4 + a4
dz
77
zi
1 = ae
z2
=
z3
=
e
=
a e
a
4
4
i 577
z4
4
I
777
%
4
Integral around
r
0
9R-
R4 -
<R 4 --a 4 S0 2 e
for
im(R cos 9 + 1 sin 9)
------------------
- mR
-iR
SRe - sin 9 Q
0
R = 0
a4
2mR
r
d
2 R
a4
R4 _
- 2R
< ------- 4=
ReigdQ
-T
o
2mR (R - a )
-mR
sin 9
a0 as R-->
im a
Residue at 0(
e
Residue at ae 4
ima(cos T+i sT)
sin
4
4
e
2t
4a
S7r
Residue at ae
e
4
3
ma
4a3 (- -+i -
e
ima(cos 37T
3- + i sin 37T)
11)
4
4
4a 3 e
- ma
-F2 " e
e
4 a0 (As
4
ma
,2-
+ i1e)
Adding residues, we have
g
ma
e 42
4 as
ma
ma
i
e
1
e-/2e
1
1
1
-2
41
S-1-1)
i-J2
ma
7
-1
4 a3 (+ 1 +
2
ma
3
-()e
ma
a
i i
-
2
ema
J2-2 e---F (e -7
4 a
iea-3
2
sin
(f4
ma
+r2)
4
(-+i
+ 1 i)
2
,-iJ
(-J) e4
mx
ei
0
- 00 x 4
2
e
-
3
-----------a
=
+ a4 dx
ma
ý sin( (a +7C)
ma
2
Oo cos mx dx
-0 x 4 + a 4
sin (E + -- )
e
..............
=
--- 4
3
a
ma
cos mx dx =
4
co
o
x4 + a
e
qT sin (Sý + -
2a
0o sin mx 4dx
-So x+
0
0
=
x4 + a4
-co
S00
)
. . . . . s.
m
x sin mx dx
x4 + a4
and
a
positive
Consider
Seimz
z4 +-----dz
z4 + a 4
R = 0
Integral around
ei
Re
mR
for
(cos 9 + i sin g) iRei 9 dg
R4 e 419 + a4
0
r
T2 -mR sin 9
2R 2
e-mRsiln 9
-- s2
e
------------- 4d9 e4
4
4
4
0
R -a
R -a
0
2
d9
7T
T7
S------S2
R4
-
a4
2mR9
e
0
=
- a4 )2
(R427T
229mR
e
dQ---2-R
2mR (R4 _ a4)
-7TR- 4 [1 - e - R ]
4 _ a
R
-a
as
0
R--7
7
0
e
S
Residue at OC
i ma (- 1 + i
iv
2
4a
47
e
ma
4 a2 1
+
ma
1)
+
ma
4 a
i ma
. 7E
2
4ma
(e
4-
Residue at ae
1 )
e------
4
Residue at ae
im Q
-4
ma
a2 1
e
Adding residues
ma
ma
e
2
=
e---
sin ma
2 a2
si
a2
ma
'3
7 ie
Sx ei mx
x + adx
a
sin
2
JE
ma
oO x sin mxdx
o
oooxx4-------cos
4 dx
+ amx
x
S x
+a
o0
-
2
17fe
sin-
a
+ a
=
xx- o+ a dx =
X 4cos mx
4 dx
0
ma
o
0
x- sin mx dx
x4
+
a4
=
2a
2
sin ma
--
2E
ma
JE
aŽ
o00x 3 sin mx
4
x 44 + a4
0
S --------- dx
z 4 eim
dz
z4 + a--------dz
z + a4
0,
R
S -----------------------------R e i + a4
R
4
-mR sin 9
< 7 R--e 4
0
R
0
-R
7r R3 e3iG eimR (cos 9 + i sin 9)
0
m
4
2 --
S
R4 - a 4
0
d9 Z
- a4
4
---- 2
2 -R
R4 7r
---[
4)
R
m
2
a
(R4 .
i e
d
2mRg
r7
R - e - mR]
A
dg
e
0
R - -, o
as
Sia eim(
Residue at o<
4 o(
71
Residue at ae
e ma (+
e 3ma
e i ma
3
i)
e--------r-----4-2--
4
4r
Residue at ae
I ma (- -L+ 1
e------ 4---------A)
-
ma
ee4
41
4
Adding residues
ma
e
42
2
mama
F
e __2 + e
2
Sx3 ei MX
_- x4 - 4 dx
x + a
ma
2e
cos
2
=
Tie
ma
J
cos
ma
JE
ma
--
ma
e
,
10
3
co X
x s cos mx
---------cX S x4 + a44
°on
+i
ma
x s3 sin mx
--
-dx+i
--
=
[email protected]
x4 ++aa4
o x 3 cos
mx dx
dx
+ a0
-x
eJ
1dx e7e4
=
x +a
ma
sina4mxdx
0o0 x3x-4----dx =+ 2
cos
dIT
X2 dx
o00
x 4 + 5x2 + 4
0
Consider
z2 dz
z4 + 5a2
z
z
+
2
=
- 1
= + 1
- 4
or
or
G:4z 3 +2 10
residue at pole
4
----------5+•125- 16
2
zA
IoCI
+ 21
-------------40<, 3 + 106K
4 c\2 + 10
=--
residue at
S
ma
--
"
+ 1
= ----2
2i
4 (-4) + 10
= 6
=-
i
3
Since the exponent of the numerator is 2
less than that of the biggest exponent
in denominator
2 the ;• = 0as
(i.e.,
(i.e.,
z0 as R-'>co)
R->
i
\
o ma=i7
COS
x 2 dx
--- 03ox 4 + 5x2+ 4
-
2
=
i
i
i
since we have an even function
- x 2 dx
x 4 + 5x 2 + 4
2 So
0
00
x4
0
If
M > 0
i
27Fi
7/-
-i
6
x2 dx
67
5x 2
6
+
+ 4
evaluate
soo cos mx dx
0 1 + x2 + x4
consider
eimz dz
2
1 +
+ z
1 + z2 + z
6
=
z
z
4
2
1 + z + z
wherever
z = cos (0 + 2k-T)
6
+ i
*"*
z->9
e
eim z
z + 4z 3a
0
, 4, 5)
i 2(
u
,
e
3
above
i 2TT
i
e
+
00
==
(k = 0, 1, 2,
6
has poles at
the x axis ) the residue at
Lim
z >e-3
=
sin (0+-2k6f)
eimZ
+ - z4
61 - 1
.. .
- 12
- 1
the residue at
+
eimz
Lim
2i 7E
2 3
2z + 4z
e
3
12
im ( +
i 2
e -- 22
---- - - t)-----2
3 (1 +
e
2e
)
)
------e 7_E 2----- 2------41
+
2i
2 e
+ i --
im (-
e
(1 + 2 e
Li -M45
2
-i
3)
-m
2
+ ------- e--- 2[ e ----2(1 + 7 1)[1+2(+ 1 +47
i))]
2(1
+ 2 1)2[1+g(12
2
2
2
2
2
e------------
e 2
-----------
-m 27
-m 2
1
-----4 ]
1))4]
2
ei 2
------- -- -- + -- - -------------
(1 + JE i) zJ1 i
e -
i4
-m42/
e
e
2
2e
T (1 - i JE)
+ e
e
(i +
1
-i L (1 + 1
2
(1 - i JZ)
(1 + 1 J4)
S_-2 - 1 --3
i4X
(- 1 + JE 1)(- JZ)
-m
+ 2e
Se
4t (i
--- 2 cos (
JE, i
21
on the circle
S
coo
=
R ei
dz
=
R i eig dG
imx
Se-m R sin
2+---- 4 dx +
0
2
9
z
e im R cos 9
e
+dQ
---------- Ri e
2
2
1 + R e g + R e4 1
i2 e
i JE
2
cos (7L - )
3
2
13
0--Owhen R -- co
the second
-m
-co 1
Cos
C
+ x2
x
dx + i s
x4
+
sin mx
-00oo
Scos
mx
0+x-+"0 i + x +
S
and
e
sin mx
dx
1 + x 2+ xi
0
= 0
0 z-a 4 1
a-i
x-- dx
i - x
PS
cos (I -- )
3
2
2
-m cos- ( - )
2
3
2
jj
dx
x4
dx = 27Te
+x2+x4
a-1
3 ALz -
1
dz
z
=
r ei
dz
=
r i ei
Lim
r --> 0
C0
9
ra-
dQ
1
ei(a-1 ) 9 r i ei
r
2IT
ei 9
Lim
Lim
R->oo
R0- oo
S 27
0
2
dQ
- 1
ra
-
9
ir r+
ra
1
-
r + i
0
S
d9 -- >'
0
dG --
0
Ra-i ei(a-!)9
R
ei
i ei9 [email protected]
----------9
R
Re
27r
3'
ES
- 0
-1
Ra
R+I
d-
Ra
R+I1
27r
14
i-£
a-1
S
x - 1
1
i-
_ e 2 ari
dx
S-----
1
x-
0
a-1
dx+
a-1
&
x-
dx+
1
xa-1
-dx
1 + f., xe1
5'
0
,r
- 1) P S 0
Xa1-X
-1 dx
0 1-x
(e 2 aITi
pS0 0
0
=
L
e27(a-1) 1 +
7T i (e 2
1-:x-
e
ra
eo
(a-1)
i +
(eaTri + e-aTri)
l1 ----------------I--
xa-1
e2a7ri
=
1
-1
-a7r i
e a 771i - e
2i
=
s nos a7
sin a7
0
00
0
sinh ax
-sinh
sinli
7w
Consider
since
d
71ctn a7
x
eaz
sinh
dz
7T Z
7Tz
e-- V
7Tz
sinh z = e 7-r
2
we have poles when
e
e2
when
z = i
7w Z
z
Z
-7p
=e-
1
i.e.,
when z = 0 and z = ki
need only concern us.using above contour.
15
z
=
= i dy
dz
1
e
iay
- e-R e-iy -
0 eFR eiy
R+iy
1 /2e
aR
0
-
eR
1
2
e-R/ =- - a--- --e
S dy
tends to 0 as R--> 0o
2
z
=
-R+iy
dz
=
i dy
tends to 0 as R --> 00
--e
0
e ax
a
dx
o0osinh 7 x
sinh Tr (x +i)
2z eaz
Lim
0
zr
Z
Lim
z->
dx
dx
ax
-- dx
-rx
sinh
dx + S-oo
2az eaz +
e e7 z - e-
ea(x+i)
snh
co sliiTy
+
0
ea(x+i)
-00
S0
z -> 0
+
2 eaz
7e z + T e--rz
1
(V ±
_2(_z_- ) eaz
ez
residue about little semicircle
-
-i
ei a
Z
e
=
-1-7T
=
i
7T
Combining
i )a
(+ eia)S
oo
0
eax
e
sih
7 x
ia)
dx + (1 + eia
0
eax
----- dx + i-ieia= 0
sinh 7T x
-7X
16
(1+ e ia)
sinh ax
00
7rx) dx = - -2 [1 - e i a ]
(sinh e-a
2eaX
so
d
e -i -
2 [i + eia ]
sinh-rx
0
ia
_ia
[1 - eia
e
a
2
2
e
2
+e
2
sin a
2
_ 1
2
dbi
9 - a cot -------2
S0
tan a
2
a
2cos
2
COS 9 - --a -bi
2Scos-=
d9
9 - a2 - bi
z
=
eiQ
dz
=
ie
dz =
iz
2T-
i ±9-a-bl
dG
dG
-j.
e
+
e
d
dG
S ------------------
=
lu
2Tr
ei
- ia + b
+ 1 dQ =
i e---------------1 dQ
ei-ia+b
-
1
e
ia-b
dz
z +z e
b-b
i
a
z
e
z-
=
-2
=
- 27ri + 4 7
=
2 7Ti
i
If b / 0
eia-b is outside
unit circle
i
ia-b
eia-b is inside
unit circle
If
b ; 0
17
Lim
z
e
bz(z - eia b) z
z--O
S2T cot
- a- bi
dG =
S- cot- 9---------rw
2ýi
2-i
when b ; 0
- 27ri
=
when b
-0
0 < a <2
Soo
0
xa-1
-------
dx
1 + x + x
Consider
a- 1
1+
a
+
1 + z + z'
dz
z2 + z + 1
Z (1 + z+ Z2
z (1 + z + z9)
dz
1-+J32
2-
Z
z
z
=
=
-
2
+
-3
2
inside figure
i) ] za
[ (_ 21 J+
2
(-12 + 2 )][z - 2
Lim
2
z [z
2
e
2Z
I'
Y4t
z za
------------
z0
-0
-
1 +x+x
Z (1 + z +
0
xa-0i
-
.
=
Ti lirm
•.
)1)
3
v-
Integrated around r
2
2
dx + S
0
a-i
-------1
=
0
dx
dx
=
z2 )
+ x + X2
27iResidue
27Ti.Residue
18
co
S
xa- 1
xa-1
ri(a-1)
-dx
+ S
1 - x + x'
0
---------- 2 dx
1 + x + x
0
2V
ST
Cos 2_Uq-a + i sin _2-ffa
cos 2
+ i sin 2-
3 +i
83
3
(cos
(cos 82L
+
sing
(
3
sin 82ELE)(cos
j
'3
(cos 2
2--
+ i sin
[cos (27Fa
oe 1x
0
iS
0
0a-I
xa-1
----------1 - x +a
sin-
a-1
S
2
- x + x2
0 1----------
1 log (x + 1)
+ x2
Consider
S around R = 0,
sin
-~
for
dz
z f (z) = 0
lim
Z
-, co
j-
27)]
3
O
3
7
a-------------
sin -r a
2
+
2l•(•
z2 + 1
277-
?Th)
1 sin 2;_)
2-7-
sin 27a 3
log 2
-
x
2
dx
=
sin
j
i sin (27 ra - 27)
3
3
a dx
x=
-
3-
cosT a dx + S-x----- 2 dx
0 1+x+x
2
0
---
a-)(cos
?27
)+ i sin (27 a
3
a- 1
X+ Xx
0
2-
3
JEa
7
2,2[)
3
19
z_•ogl
+__
_
=
+ log (z + i)
-Z
z+i
z+
------ 2 z
1 + -_z +-
-
~z+i
z
=
-- - - - - -2- - - --
Residue at
=
i
-
z--i_og_
+i)
z.-> 1 ((z - i) (z + 1)
sgx
+
- ---x2 +1
J
-0o
r?)
lim
-0
di
=
rr log
1_o_
2=
2i
2i
-74--
A-o2--.-
/
/
2-
/
--- v
-
21
20
Chapter II
EXTENSION OF THE EVALUATION OF ONE DEFINITE INTEGRAL
TO THAT OF MANY OTHERS
The following example will show from the evaluation of one definite integral how we can obtain a great
number of other integrals:
a ' 0
oo
a cos x + x sin
-o00
x
dx
+ ad
Consider
S --
iz
- dz
z - ai
77r
eiR(cos 9 + i sin 9) Riei9 do
S0
<
R ei
r
7r
0
9
- ai
-ResinG
~
Re --------s
2R
---R + a
-71-2
R+a
eR
R+a
R+
a0
9
0
- e-R]
(R+ a) 2R[1
S+ _-_22R
Limr
Residue at
z -->ia
0
as
z -ai eiz
z - ai
R--> oo
=
e
-a
ix
oo
adx
-c0 xx :-ai
Soo
Scos
---S00
S~ca
•o -
sin
s2 -2 Re
w-
=
x + iiaxsin x
i-
at
27ri e- a
dx = 27Ti e - a
(cos
x + i sin x)
x + ail dx = 2Y-i e -a
+---------------------co
(x
-
ai)(x + at)
21
o0
x cos x-a
sin x
0o
()
+ i
-------
---
=
2i
27.i
x---- a---dx
T e
=
- - - 2- - -2 - - - d
- oo
x +a
I
(2)
dx
- a
a cos x + x sin x
------
'.
a cos x+x sin
00
i-d
+
= 0 because odd function
S00
x cos x dx - a
dx
i
x +a
So
(3)
00
sin-x
-a o x 2 +a
co0 x+
sin x
.co x
=
+ a
a >
+oo+
-
- 00
-a
cos x + x sinx dx
-+--------
0
=
0
0
0
=
a------x" + a
dx
a'
eiz
Consider
0
-0o
z + ai
ix
e0
x+ia
=
dx
-00
(4)
(x + ia)(x-
(5)
x cos x + a
---------+ a
5
0
cosx + i sin x (x
0c0
-co
S oo
dz
co
sin
x dx+
----
i
-
ia)
- c0
a
- o
00
-00
x 2 + a2
+ a
=
d
x cos x2 + a 2 sin x dx
x
0
- a cos x + x sin x
-------------dx = 0
x2 + a
-_a-cos x + x sin x dx
-----------------
=
=
-a
e-a
22
Adding (1)and (4)
yo 0
(6)
x sin 2
X
-0
2
a
+
00 x-sin-
(7)
x2
0
27 e- a
x=
dx
+
dx
rs-- a
=
a2
2
Adding (3) and (6)
(8)
oo
x +2 1
x
- 0o
sin
+ a2
dx
e- a
=
Adding (6) and (8)
oo (x-dx+-1 sin x d-a
=
So
x
-o
+ a2
2
eare
Generalizing,
(9)
oo
nx
-co
X
12
2
+
dx = nTTe - a
s
a
2
where
n
is an integer
23
Chapter III
SUMMATION OF A FINITE SERIES
BY MEANS OF DEFINITE INTEGRALS
The scheme consists of evaluating
0
--x
- dx
x2n + 1
by a standard method; then, knowing the answer, we compare
this with the finite summations in the real and imaginary
parts of an equation gotten by taking a number (n) of poles
on the unit circle, where we may let these poles approach
arbitrarily close by increasing
n
1
p
0
p - 1a
00 x-A-IF
d
C
0
sufficiently.
1 +x
S P-1dz
z
1+
Lim
z f (z)
=
0
z f
=
0
1?
RL
Lim
z-Ž>O
z-->
(z)
0
oo
S1
xP-1
+x
0 (x e 2 _
dx + So
z -Lim-
since
OO
IO
0
p-1
--
1 + x
dx +
1 + xe
+ z
1--(z-+
p-1
-K-
z) p -
0 e2pVri
00
=
=
(e
2iri e ( p - 1)Ti
1
p-1
----------1 + x
dx
)
p-1
dx = 27i e
(p-1)
ri
24
(1- e 2 p
•0
0
i
- - dx
+ x
xp-i
l+x
and
n
T7r
sin py
e-Prri + ep•)
then
thn
m
- 27Ti epi
+ 2v~i
(-
If
=
1
0
S
• x
--d
1
7rsin piT
1 7--+X dx=
are positive integers, and
m 4 n
S00 x2m
2+dx
0
=y
x 2n + 1
m
2nx 2 n-1dx
1-2n
2n
1AS o00 n --- ..-
2n
-- S
2n 0
Since
0
dy
- )
2n -----1 +y
2m +
dy
2n
dy
1-2n
2n
dx
=
Y...
2n
X
=
y~n
dy
1
m
1
2n
dy
dx
1+y
(2m+1
=
= yn
x2m
(2m+i1
1
2n--
----------
dy
771
=
1+y
)
2n sin
-__T
---------------2n sin 2+
2n
-1
I
J
S
0
m+
2n
x2mn
dx
x2n + I
We now attack the problem in a different manner which will
permit us to get the sum of a finite series by means of
definite integrals.
We have the advantage now of knowing the
dx
value of the definite integral
S2m
------ dx+
25
m and n positive integers
with
m 4 n
oo
S-- x2m
---dx
0 x2n + 1
S
z2m
dz
-n
z2 n
/~I~;
-A
+ I
z2n
z = 1 (Cos
I+ 2kTT
2n
2L
2n
+ i sin
.'. we have poles at angles
=
(k = 0, 1, 2 ...
--
.----
2t --- ..
2n' 2n
2n-1)
2n
on unit circle
The poles from
.....
2n
2n
-
o unit circle are
on
n
2n
poles in upper half of the plane and as long as
n
is finite
there are no poles on the x-axis.
Lim
S
R -- o0 AB
(
f (z) dz =
If f (z)
- 91)
a
b
__
(
V(z)
where
Y (z) is a polynomial of degree n and
(z)
is a
polynomial of degree less than n; where a and b are the
coefficients of zn - 1 and zn in
(z)
and /-(z) respectively.
Since the degree of
.'. the integral of
$
(z)
is
Z-n - 2,
a = 0
and
f (z) round semicircle of radius R is 0.
R
2m
Sdx
Taking pole at
27T i 'Y Residues
=
-R x 2 n + 1
e 2n
2)m
---- ----
P21
2n (e
p
2n
=
p=1n
2n-
1
r
2n e p
)
.....
2 n e p -rr
2n
e
2n - 1.
Substitute
KLm+1112p-1i7T
p=l
2n
-- - - -
, 3,
e
K2m+1j
--- Lp.Z
P 7
2n
=
2p - 1
2+1 (2p-1)7
n
2n
2n
p
e 2n
p=1
n
(cos 2m +
2n
2n
p=1
d
(A)
S0
x2
x2 +1I
(2p - 1)M7
+ i sin 2m+t 1 (2p - 1)7F)
2n
2
=
2 771
2n
2m + 1 (2p - 1)7
2n
p=1
p=1
S2772n--
p=7
p=IA
sin
2m + 1
2n
(2p - 1)7F
since we have an even function
oo
n
2m
2m +
S dx
2n
p=1
(2p -
1) 7
27
n
x2m
0
x 2n +
2n
sin
m+ 1 (2p - 1) r
2n
p=1
2n sin 2m + 1
2n
7
from comparison with the answer obtained in the first method.
cos 2m +1
Since
(2p - 1)7
=
2n
p=l
for the imaginary must vanish in equation (A) as we know
from the answer obtained in the first way.
Checking the results of this specific finite series:
n
cos (ap + b)
ei(ap + b)
p=1
(only real part of it)
p=1
eib
eiap
p=1
Using geom. series
n
eia
eia(n+1)
iap
p=1
e
=
eia e - (2m+1) e ia -1
ii
-1
a
S2e
2
_a
i.*-12
a
e e 2_- e _- 2
ia
e
2
sin a
2
28
'7 cos
sin
ia
=
-e--e----
(b + a)
ib
(ap + b)
sin -g
2
p1j
2-----
sin a
2
2
sin
0
s------
--
0
2
n
1
(2p -
sin 2m-+l
2n
sin 2m --+ 1 7n
p=1l
for
ei(ap+b)
sin (ap + b)
p=1
(only imag.
coeff.)
p=l
n
=
=
eiap
e b
b
eib
ie i
sin a
2
p=1
cos (b + a)
2
sin a
2
cos O
-
sin (r_+
2n
.17T
1
sin
2n
7-T
We note the peculiarity of the finite series
n
cos m
(2 - )7 =
that is, no matter how many terms we have in the series given
by
n
the sum is always zero (i.e., if
n
is a finite
29
integer no matter how large
n
is (with m < n) then
wherever the series is broken off (at n = 5, 1000, 1010, or
any such integral number) the sum is zero.
Checking the result in special cases
Suppose for s~implicity
=
M
1
For n = 2
2
+ cos 977
cos
1)7F =
(2p-
cos
p=1
-
2
1-
4
= 0
2
For n = 3
3
Z
cos
p=1
6
(2p-
1)I
=
cos-
(2p-
1)7/ = cos
2
+ cos
2
+ cos5
=
2
0
For n = 4
4
cos
8
8
+ cos -•
8
+ cos 15
8
p=1
+
8
=I10
Since from the diagram and
knowledge of sign of cosine
being positive in the first and
third quadrants and negative in
the
A
second
and
fourth
quadrants.
30
cos 37r
--- + cos 217C
-- T
8
8
cos --r + cos---15=
8
8
etc., for
n > 4
-
=
0
0
(continuing the same process).
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