PHY6095/PHZ6166: homework assignment #2
SOLUTIONS
2
1. Problem 1
a) The first step is to re-scale the integration variable
Z
∞
dx
e−x/a
x(x
+
b)
0
Z ∞
dy
p
(y = x/a) =
e−y
y(y + c)
0
I=
(1)
p
(2)
where c ≡ b/a.
i) a b → c 1. Assume that typical y are c. Trying to put c = 0 in the integral leads to a logarithmic
divergence at y = 0
∞
Z
0
dy −y
e .
y
This is not bad as the assumption y c breaks at y ∼ c and the integral needs to cut off at y ∼ c. Since the
integral is only logarithmically divergent, ambiguity in the cutoff choice (y = c, y = c/2, etc) is not relevant
within the Main Log Approximation (MLA). Continuing along these lines
∞
Z
I=
c
dy
exp(−y).
y
The exponential factor cuts off the integral at y ∼ 1. Again, a particular choice of cutoff in not important within
the MLA, and we can drop e−y and cut the integral off at y = 1
Z
I=
c
1
dy
1
a
= ln = ln .
y
c
b
The result can also be obtained beyond the MLA. To this end, introduce a new variable z =
by parts Integrate by parts:
Z
√
y and integrate
∞
p
2
2 ∞
dz
e−z = 2 ln z + z 2 + c e−z | 0
z2 + c
0
Z ∞
Z ∞
p
p
2
2
+4
dz ln z + z 2 + c ze−z = ln c + 4
dz ln z + z 2 + c ze−z .
√
I=2
0
(3)
(4)
0
In the remaining integral one can let c → 0, because the integral is convergent. Upon that, the integral reduces
to a pure number
Z
√
∞
4
2
dz ln (2z) ze−z = −
0
πγ
≈ −0.256.
4
(5)
where γ is the Euler constant ≈ 0.577 . . . . Thus
b
I = ln −
a
√
πγ
a
= ln
4
1.291b
(6)
ii) a b → c 1. Neglecting compared y compared to c in the denominator gives
1
I=√
c
Z
0
∞
dy
2
√ e−y = √
y
c
Z
0
∞
2
dze−z =
r
πa
.
b
In the last integral, y ∼ 1 and thus neglecting y compared to c was justified.
b)
(7)
3
i) a b
Z
∞
dx
sin (x/a)
2 + b2 )
x
(x
0
Z ∞
1
dy
b
= (y = x/b) = 2
sin y
b 0 y (y 2 + 1)
a
Z
dy
1 ∞
b
π 1
≈ 2
y =
b 0 y (y 2 + 1) a
2 ab
I=
(8)
(9)
(10)
ii) a b.
Z
∞
dx
sin (x/a)
+ b2 )
0
Z ∞
dy
1
= (y = x/a) = 2
sin y
a 0 y (y 2 + b2 /a2 )
I=
x (x2
Neglect y 2 compared to b2 /a2 1 in the denominator. The remaining integral
Z ∞
dy
sin y = π/2
y
0
(11)
(12)
(13)
is dominated by y ∼ 1. Thus
I≈
1 π a2
π
= 2.
a2 2 b2
2b
(14)
Problem 2.
a)
• a 1.
Z
∞
1
x
√
2 + a2 1 +
x
x
0
= (y = x/a)
Z ∞
dyy
1
1
√
=√
√
a 0 y 2 + 1 y + 1/ a
Z ∞
1
1
dyy 1
=√
√
√
a 0 y 2 + 1 y 1 + 1/ ay
!
Z ∞
√
dy y
1
1
1
1
≈√
1− √ +
−
ay ya (ay)3/2
a 0 y2 + 1
Z ∞
1
1
1
1
dy 1
= c1 √ − c2 + c3 3/2 − 2
,
a
a 0 y2 + 1 y
a
a
I(a) =
dx
(15)
(16)
(17)
(18)
(19)
(20)
where c1−3 ∼ 1. More precisely,
Z
∞
c1 =
Z0 ∞
c2 =
0
Z
c3 =
0
∞
√
√
dy y
2π
=
,
2
y +1
2
dy
π
= ,
y2 + 1
2
√
dy
1
2π
= c1 .
√ =
2
y +1 y
2
(21)
(22)
(23)
The last term diverges at small y. When expanding the denominator, we have assumed that ay 1. This
assumption breaks down when y ∼ 1/a. Cutting of the divergence at y ∼ 1/a, we obtain the fourth term
4
within the MLA
1
a2
1
a2
∼
∞
Z
1/a
Z 1
1/a
dy 1
+1y
(24)
y2
dy
ln a−1
.
=
y
a2
(25)
Thus, for a 1,
√
2π 1
π1
√ −
+
2
2a
a
I=
√
2π 1
ln a
− 2 .
2 a3/2
a
(26)
• a 1. Within MLA,
1
Z
I(a) =
a
dx
1
= ln .
x
a
b) Function J(a) is defined as
Z
∞
J(a) =
0
dy −y
e .
y + a2
Find the first two terms of the expansion of J(a) for a 1.
Integrating by parts,
2
J(a) = ln(y + a
∞
)e−y | 0
∞
Z
2
+
dy ln(y + a )e
0
−y
1
= 2 ln +
a
Z
∞
dy ln(y + a2 )e−y
0
The integral term is convergent for a = 0, so can put a = 0 in this term which gives the next-to-leading term
Z ∞
dy ln ye−y = −γ.
0
Thus
J(a) = 2 ln
1
− γ + ...
a
Problem 3.
a)
Z
I=
∞
dx
0
x
cos ωx.
sinh x
(27)
Expand the function x/ sinh x for x → 0 :
x
1
7 4
= 1 − x2 +
x + O x5
sinh x
6
360
(28)
We see that the function itself and all its even derivative are finite at x = 0 whereas odd derivatives vanish.
Now integrate by parts
sin ωx x ∞
|
ω Z sinh x 0
d x 1 ∞
dx
sin ωx
−
ω 0
dx sinh x
cos ωx d x =
|x=0 −
ω 2 dx sinh x
Z ∞
1
d2 x dx
cos ωx
ω2 0
dx2 sinh x
= ...
I=
(29)
(30)
(31)
(32)
(33)
5
Even derivatives always come in combinations with sin ωx, which vanishes at x = 0, whereas odd derivatives
vanish themselves. Thus the integral does not have a power-law expansion in 1/ω. On the other hand, the
integrand is analytic on real axis. To get rid of the factor of x, re-write the intergal as
Z ∞
Z ∞ iωx
sin ωx
1 ∂
e
1 ∂
∂
dx
=
Im
≡
J(ω)
I=
∂ω 0
sinh x
2 ∂ω
sinh
x
2 ∂ω
−∞
The poles of the integrand are at z = iπn, n = 0, ±1, ±2 . . . To take into account only one pole, choose the
contour as shown in Fig. 1, where semi-circles circumvent poles at z = 0 and z = iπ. The vertical segments of
the countour (B and D) can be moved to ±∞ and, because the integrand does decay along the real axis, the
integrals over parts B and D go to zero. The integral over A is equal to the original integral J in the limit of
r → 0. On part D, z = x + iπ, sinh z = − sinh x, and
Z
Z −∞
eiωz
−πω
dxeiωx = e−πω J.
dz
= −e
sinh z
∞
D
By Cauchy theorem,
J(1 + e
−πω
Z
Z
)+
+
C1
where
R
C1,2
= 0,
C2
are integrals over semi-circles C1,2 . On C1 , z = reiφ , dz = ireiφ dφ, sinh z ≈ reiφ , eiωz ≈ 1, so
Z
Z
0
= ir
C1
dφeiφ
π
1
= −iπ.
reiφ
On C2 , z = iπ + reiφ , dz = ireiφ dφ, sinh z ≈ −reiφ , eiωz ≈ e−πω , so
Z
Z −π
1
= iπe−πω .
= ire−πω
dφeiφ
−reiφ
C2
0
Therefore,
J = iπ
1 − e−πω
1 + e−πω
and
ImJ = π
1 − e−πω
.
1 + e−πω
For ω 1, e−πω 1 and the fraction in the previous equation can be expanded in e−πω . Neglecting terms of
order e−2πω e−πω , we obtain
ImJ ≈ π 1 − 2e−πω
and
1 ∂
J(ω) ≈ π 2 e−πω .
2 ∂ω
I=
b)
Z
I=
∞
dx
0
x2 −x
e sin ωx.
1 + x2
(34)
The integrand is a non-analytic function on the entire real axis, so we should be able to obtain the result via
integration by parts
Z
f (0)
1 ∞
f (0) f 00 (0)
I=
+
dxf 0 (x) cos ωx =
−
+ ...
ω
ω 0
ω
ω3
6
FIG. 1. Contour of integration for Problem 2a.
where
f (x) ≡
x2 −x
1
5
e = x2 − x3 − x4 + x5 + O x6 .
1 + x2
2
6
(35)
Obviously, f (0) = 0 and f 00 (0) = 2. Thus
I≈−
2
+ O(ω −5 ).
ω3
(36)
c)
∞
Z
I=
dx
0
cos(ωx)
(x2 + 1)
2/3
.
(37)
Re-write the integral as
1
I=
2
Z
∞
dx
−∞
exp(iωx)
(x2 + 1)
2/3
.
The integrand has a branch cut at x = i, where x2 + 1 = 0. Choose the contour as shown in Fig. 2. By Cauchy
theorem,
Z
Z
Z
I+
+
+
= 0,
C
BC1
BC2
where C is a circle around
i and BC1,2 are the branch cuts. On the circle, z = i + reiφ , dz = ireiφ dφ,
R
2
2/3
2/3
1/3
(z +1)
∝r
and C ∝ r
→ 0 | r→0 . On BC1, z = is+δ with δ = 0+ , and (z 2 +1)2/3 = (−s2 +1+iδ)2/3 =
2
2/3 2iπ/3
(s − 1) e
. On BC2, z = is − δ, and (z 2 + 1)2/3 = (s2 − 1)2/3 e−2iπ/3 . Therefore,
Z
Z 1
Z ∞
idse−ωs
dse−ωs
−2iπ/3
−2iπ/3
=
−ie
=e
2
2/3
(s2 − 1)2/3
1
BC1
∞ (s − 1)
7
y
δ
C i
r
x
FIG. 2. Contour of integration for Problem 2b.
Z
=e
2iπ/3
BC2
Z
∞
1
and
I=
idse−ωs
(s2 − 1)2/3
√ Z ∞
dse−ωs
3
.
2 1 (s2 − 1)2/3
In the limit of ω → ∞, s tries to be as small as possible, therefore, we change the variable as s = 1 + t with
t 1, and replace (s2 − 1)2/3 ≈ 22/3 t2/3 . Thus
√
√
Z
3Γ( 13 ) e−ω
3 −ω ∞ dte−ωt
I = 5/3 e
=
.
2
t2/3
25/3 ω 1/3
0
Problem 4
Z
∞
Γ(z + 1) =
dxxz e−x
(38)
dx exp (−x + z ln x) .
(39)
0
Z
=
∞
0
f (x) ≡ z ln x − x
f 0 = z/x − 1 →
x0 = z, f (x0 ) = z ln z − z
f 00 (x0 ) = −1/z
f 000 (x0 ) = 2/z 2
f (IV ) (x0 ) = −6/z 3
(40)
(41)
(42)
(43)
(44)
(45)
8
Expanding the argument of the exponential, one gets
Z
∞
Γ(z + 1) ≈
dx exp (−x + z ln x)
Z ∞
1
1
3
2
dx exp(− [x − z] + 2 [x − z]
= z z e−z
2z
3z
0
1
4
− 3 [x − z] ).
4z
(46)
0
(47)
(48)
Keeping only the 2nd order term in the exponential gives the result we obtained in class
√
Γ2 (x) = z z e−z 2πz.
To evaluate the effect of the 3rd order
term, expand the exponential to leading order in this term. Rescaling
√
the integration variable as x → x/ z and changing the variable to y = x2 /2, we obtain:
Z
z z e−z ∞
1
2
3
dx exp(− [x − z] ) (x − z)
2
3z
2z
0
Z
1
2 z −z ∞
dyye−y ≈ z z+1 e−z/2 .
= z e
3
3
z/2
Γ3 (z + 1) =
(49)
Up to third-order terms, we have
√
1 z 3/2
Γ(z + 1) = Γ2 + Γ3 = z z e−z 2πz 1 + √ e−z/2 .
3 2π
(50)
Notice that the 3rd order correction is exponentially small and will be neglected in what follows.
The 4th correction is obtained by expanding the exponential 1st order in the 4th-order derivative and to second
order in the 3rd-orderderivative (these two terms give the same order in z −1 )
∞
x3
x4
1
dx exp − x2 + 2 − 3
2z
3z
4z
−∞
2 !
Z ∞
1 x3
x4
z −z
−x2 /2z
dxe
1− 3 +
≈z e
4z
2 3z 2
−∞
Z
Z
∞
h ∞
2
2
1
dxx4 e−x /2z
= z z e−z
dxe−x /2z − 3
4z
−∞
−∞
Z ∞
i
2
1
+
dxx6 e−x /2z
18z 4 −∞
5/2 Z ∞
√
2
z z e−z (2z)
= 2πzz z e−z −
dxx4 e−x
4z 3
−∞
7/2 Z ∞
z −z
2
z e (2z)
+
dxx6 e−x
18z 4
−∞
Γ(z + 1) = z z e−z
Z
Recalling that
Z
∞
2
−∞
∞
Z
3√
π,
4
15 √
=
π,
8
dxx4 e−x =
dxx6 e−x
−∞
2
we get
Γ(z + 1) =
√
z −z
2πzz e
1
1+
12z
(51)
(52)
(53)
9
or
Γ(z) = z −1 Γ(z + 1)
√
1
z−1/2 −z
e
1+
.
= 2πz
12z
(54)
(55)
Note that the combined correction due to fourth derivative and (third derivative)2 is larger than due to the
third derivative alone.
Problem 5.
a)
G(a) =
∞
X
n=0
1
(n2
2
+ a2 )
i) a 1 → the n = 0 term dominates and
G(a) ≈
1
a4
ii) a 1. Large number of terms (n ∼ a) contributes to the sum which therefore can be approximated by an
integral
∞
X
n=0
Z
1
(n2
+
∞
≈
2
a2 )
0
=
1
2 = a3
2
2
(n + a )
dn
Z
∞
0
dx
(x2
+ 1)
2
π
4|a|3
(the absolute value of a occurs because the integrand is an even function of a). Both of this approximations are
good only in giving us the leading terms. To construct regular series, use the Poisson formula. Re-arranging
the sum using the formula valid for an even function f (n)
∞
X
f (n) = −f (0) + 2
n=−∞
∞
X
f (n),
n=0
and applying the Poisson formular, we obtain
1
G(a) =
2
∞
X
1
2 + a4
2
2
n=−∞ (n + a )
∞
∞
X Z ∞
X
1
ei2πkn
=
dn
2
2
2
2
(n2 + a2 )
n=−∞ (n + a )
k=−∞ −∞
1
!
.
∞
X
π
(1 + 2π|k||a|) e−2π|k||a|
2|a|3
k=−∞
"
#
∞
X
π
=
1+2
(1 + 2πk|a|) e−2πk|a|
2|a|3
k=1
i
π h
−2π|a|
1
+
2
(1
+
2π|a|)
e
(a 1) ≈
2|a|3
π
1
π 2 −2π|a|
G(a) ≈
+
+
e
+ ...
4|a|3
2a4
2a2
=
In fact, the series over k can be summed exactly (choosing a > 0 for brevity)
∞
X
k=1
(1 + 2πka) e−2πka
10
= 1 − 2πa
=
1
e2πa
−1
d
d (2πa)
"X
∞
#
e
−2πka
−1
l=0
2πa
2πae
+
(e2πa − 1)
2
1
π
πa
G(a) = 4 + 3 coth πa +
2a
4a
sinh2 πa
For small a,the expression in the square brackets can be expanded [. . .] ≈
2 −1
πa
1
π
2
2 3 3
+
+
π
a
.
.
.
2a4
4a3 πa
45
4
π
1
= 4+
+ ...
a
90
G(a) ≈
b)
K(a) =
∞
X
n2 exp(−an6 ).
n=0
i) a 1 → K(a) ≈ exp(−a).
ii) a 1 →
∞
X
2
Z
6
n exp(−an ) ≈
∞
dnn2 exp(−an6 )
0
n=0
Z
1 ∞ 3
dn exp(−an6 )
3 0
Z
1 ∞
=
dy exp(−ay 2 )
3 0
r
1 π
=
6 a
=
Again, higher-order terms can be obtained via the Poisson formula.
+
2 3
45 π
a3 + O a5 , thus