Download THE METHOD OF STATIONARY PHASE 1. A crash course on

LECTURE 5: THE METHOD OF STATIONARY PHASE
1. A crash course on Fourier transform
¶ Some notions.
• For j = 1, · · · , n,
– ∂j = ∂x∂ j .
– Dj = 1i ∂j .
• For any multi-index α = (α1 , · · · , αn ) ∈ Nn .
– |α| = α1 + · · · + αn .
– α! = α1 ! · · · αn !.
– xα = xα1 1 · · · xαnn .
– ∂ α = ∂xα11 · · · ∂xαnn .
– Dα = D1α1 · · · Dnαn .
• For any ϕ ∈ C ∞ (Rn ),
– Gradient vector ∇ϕ = (∂1 ϕ, · · · , ∂n ϕ).
– Hessian matrix d2 ϕ = [∂i ∂j ϕ]n×n .
¶ Schwartz functions.
Definition 1.1. A function ϕ ∈ C ∞ (Rn ) is called a Schwartz function (or a rapidly
decreasing function) if
sup |xα ∂ β ϕ| < ∞
(1)
Rn
for all multi-indices α and β.
We will denote the set of all Schwartz functions by S (Rn ), or by S for simplicity.
Obviously S is an infinitely dimensional vector space. It is easy to see
• C0∞ (Rn ) ⊂ S (Rn ) ⊂ Lp (Rn ) for any 1 ≤ p ≤ ∞.
2
• For any λ ∈ C with Re(λ) > 0, ϕ(x) = e−λ|x| ∈ S (Rn ).
• If ϕ is Schwartz, so are xα Dβ ϕ, Dα xβ ϕ, where α, β are any multi-indices.
Recall that a semi-norm on a vector space V is function | · | : V → R so that
for all λ ∈ C and all u, v ∈ V ,
(1) (Absolute homogeneity) |λv| = |λ||v|.
(2) (Triangle inequality) |u + v| ≤ |u| + |v|.
On S one can define, for each pair of multi-indices α, β a semi-norm
(2)
|ϕ|α,β = sup |xα ∂ β ϕ|.
Rn
1
2
LECTURE 5: THE METHOD OF STATIONARY PHASE
We say that ϕj → ϕ in S if
|ϕj − ϕ|α,β → 0
as j → ∞ for all α, β. The topology induced by these semi-norms turns S into a
Fréchet space.
Equivalently, one can define a metric d on S via
∞
X
kϕ − ψkk
(3)
d(ϕ, ψ) =
,
2−k
1 + kϕ − ψkk
k=0
where for each k ≥ 0,
kϕkk = max sup |xα Dβ ϕ|
|α|+|β|≤k x∈Rn
is a norm on S . Then we have a metric topology on S which coincides with the
one described above.
Exercise 1. Under the topology described above, C0∞ (Rn ) is dense in S .
¶ Fourier transform on S .
Let ϕ ∈ S be a Schwartz function. By definition its Fourier transform Fϕ is
Z
e−ix·ξ ϕ(x)dx.
(4)
Fϕ(ξ) = ϕ̂(ξ) =
Rn
For example, one can prove
F(e−
(5)
|x|2
2
n
) = (2π) 2 e−
|ξ|2
2
.
By defintion we have the obvious inequality
kFϕkL∞ ≤ kϕkL1 .
It is easy to check that for any ϕ ∈ S and any multi-index α,
F(xα ϕ) = (−1)|α| Dξα (F(ϕ))
(6)
and
F(Dxα ϕ) = ξ α F(ϕ).
(7)
As a consequence, we see
ϕ ∈ S =⇒ Fϕ ∈ S .
From definition it is also easy to check
Z
Z
(8)
Fϕ(ξ)ψ(ξ)dξ =
Rn
ϕ(x)Fψ(x)dx.
Rn
Lemma 1.2. Suppose ϕ ∈ S .
(1) For any λ ∈ R+ , if we let ϕλ = ϕ(λx), then ϕλ ∈ S and
ξ
1
(9)
Fϕλ (ξ) = n Fϕ( ).
λ
λ
LECTURE 5: THE METHOD OF STATIONARY PHASE
3
(2) For any a ∈ Rn , if we let Ta ϕ(x) = ϕ(x + a), then Ta ϕ ∈ S and
(FTa ϕ)(ξ) = eia·ξ Fϕ(ξ).
(10)
Now we can prove
Theorem 1.3. F : S → S is an isomorphism with inverse
Z
1
−1
(11)
F ψ(x) =
eix·ξ ψ(ξ)dξ.
n
(2π) Rn
Proof. We have
Z
Z
Z
x
1
Fϕ(ξ)ψ(λξ) = n ϕ(x)Fψ( )(x)dx = ϕ(λx)Fψ(x)dx.
λ
λ
Letting λ → 0, we get
Z
Z
Fϕ(ξ)dξ = ϕ(0)
ψ(0)
In particular, if we take ψ(ξ) = e−
(12)
Fψ(x)dx.
|ξ|2
2
, then we get
Z
1
ϕ(0) =
Fϕ(ξ)dξ.
(2π)n
Finally we replace ϕ by Ta ϕ in the above formula, then
Z
1
ϕ(a) = Ta ϕ(0) =
eia·ξ Fϕ(ξ)dξ.
(2π)n
As a consequence, we get
kϕk2L2 =
1
kϕ̂k2L2 .
(2π)n
1 T
Qx
¶ The Fourier transform of the Gaussian e− 2 x
.
Theorem 1.4. Let Q be any real, symmetric, positive definite n × n matrix, then
(13)
1 T
Qx
F(e− 2 x
)(ξ) =
(2π)n/2 − 1 ξT Q−1 ξ
e 2
.
(det Q)1/2
Proof. Let’s first assume n = 1, so that Q = q is a positive number. By definition,
Z
1
2
− 12 xT qx
F(e
)(ξ) =
e− 2 qx −ixξ dx.
R
4
LECTURE 5: THE METHOD OF STATIONARY PHASE
Taking ξ derivative, we get
Z
Z
d
i
d − 1 qx2 −ixξ
− 12 qx2 −ixξ
− 12 xT qx
F(e
)(ξ) = −ix e
dx =
(e 2 )e
dx
dξ
q R dx
R
Z
1 T
1
ξ
−i
2 d
(e−ixξ )dx = − F(e− 2 x qx )(ξ).
=
e− 2 qx
q R
dx
q
It follows
ξ2
1 T
F(e− 2 x qx )(ξ) = Ce− 2q ,
where
r
Z
1 T
1
2π
2
C = F(e− 2 x qx )(0) =
e− 2 qx dx =
.
q
R
Next suppose Q is a diagonal matrix. Then the left hand side of (13) is a product
of n one-dimensional integrals, and the result follows from the one-dimensional case.
For the general case, one only need to choose an orthogonal matrix O so that
OT QO = D = diag(λ1 , · · · , λn ),
where λi ’s are the eigenvalues of Q, then if we set y = O−1 x,
Z
Z
1 T
T
− 21 xT Qx
− 12 xT Qx−ix·ξ
F(e
)(ξ) =
e
dx =
e− 2 y Dy−iy·O ξ dy
Rn
Rn
n/2
(2π)
(2π)n/2 − 1 ξT Q−1 ξ
− 12 ξ T OD−1 OT ξ
=
e
e 2
=
.
(det D)1/2
(det Q)1/2
Now suppose Q is an n × n complex matrix such that Q = QT and Re(Q) is
positive definite. We observe that the integral on the left hand side of (13) is an
analytic function of the entries Qij = Qji of Q in the region ReQ > 0. The same is
true for the right hand side of (13), as long as we choose det1/2 Q to be the branch
such that det1/2 Q > 0 for real positive definite Q. So we end up with
Theorem 1.5. Let Q be any n × n complex matrix such that Q = QT and Re(Q)
is positive definite, then
1 T
Qx
F(e− 2 x
(14)
)(ξ) =
(2π)n/2 − 1 ξT Q−1 ξ
e 2
.
(det Q)1/2
Another way to describe this choice of det1/2 : Since Q = QT , for any complex
vector w ∈ Cn we have
Re(w̄T Qw) = w̄T (ReQ)w.
It follows that all eigenvalues λ1 , · · · , λn of Q has positive real part. Then
1/2
(det Q)1/2 = λ1 · · · λ1/2
n ,
1/2
where λj
is the square root of λj that has positive real part.
LECTURE 5: THE METHOD OF STATIONARY PHASE
5
¶ The Fourier transform on S 0 .
Definition 1.6. Any linear continuous map u : S → C is called a tempered distribution. The space of tempered distributions is denoted by S 0 .
So S 0 is the dual of S . The topology of the space S 0 is defined to be th weak-*
topology, so that uj → u in S 0 if
∀ϕ ∈ S .
uj (ϕ) → u(ϕ),
For example,
• For any a ∈ Rn , the Dirac distribution δa : S → C defined by
δa (ϕ) = ϕ(a)
is a tempered distribution.
• More generally, for any a ∈ Cn and any multi-index α, the map u : S → C
defined by
u(ϕ) = Dα ϕ(a)
is a tempered distribution.
• For any bounded continuous function ψ on Rn defines a tempered distribution
by
Z
ϕ(x)ψ(x)dx.
u(ϕ) =
Rn
Note that if ψ ∈ S , then the distribution defined by the above formula is
non-zero unless ψ = 0. It follows that S ⊂ S 0 .
Definition 1.7. Let u ∈ S 0 . We define
(1) xα u ∈ S 0 via (xα u)(ϕ) = u(xα ϕ).
(2) Dα u ∈ S 0 via (Dα u)(ϕ) = (−1)|α| u(Dα ϕ).
(3) Fu ∈ S 0 via (Fu)(ϕ) = u(Fϕ).
Now let Q be any real, symmetric, non-singular n × n matrix. Then the function
i T
e 2 x Qx is not a Schwartz function, but we can think of it as a distribution. We would
like to calculate its Fourier transform.
Recall that the signature of Q is
sgn(Q) = N + (Q) − N − (Q),
(15)
where N ± (Q) = the number of positive/negative eigenvalues of Q.
Theorem 1.8. For any real, symmetric, non-singular n × n matrix Q,
π
(16)
i
F(e 2 x
T Qx
)=
(2π)n/2 ei 4 sgn(Q)
| det Q|
1
2
i T −1
Q ξ
e− 2 ξ
.
6
LECTURE 5: THE METHOD OF STATIONARY PHASE
Proof. Let’s first assume n = 1, so that Q = q 6= 0 is a non-zero real number. For
any ε > 0 we let qε = q + iε. Applying theorem 1.5 we get
i
2
1
2
F(e 2 qε x ) = F(e− 2 (ε−iq)x ) =
(2π)1/2 − 1 (ε−iq)ξ2
e 2
,
(ε − iq)1/2
the square root is described after theorem 1.5. Note that when ε → 0+, we have
√ − iπ
p
iπ
qe 4 q > 0
1/2
= |q|e− 4 sgn(q) ,
(ε − iq) −→ √
iπ
−qe 4 q < 0
so the conclusion follows.
For general n, one just proceed via the diagonalization method as before.
2. The method of stationary phase
¶ Oscillatory integrals.
The method of stationary phase is a method for estimating the small ~ behavior
of the oscillatory integrals
Z
ϕ(x)
(17)
I~ =
ei ~ a(x)dx,
Rn
where ϕ ∈ C ∞ (Rn ) is called the amplitude, and a ∈ Cc∞ (Rn ) is called the phase. For
simplicity we will assume ϕ is real-valued.
To illustrate, let’s start with two extremal cases:
• Suppose ϕ(x) = c is a constant. Then
I~ = eic/~ A,
R
where A = Rn a(x)dx is a constant independent of ~.
• Suppose n = 1 and ϕ(x) = x. Then by definition I~ = F(a)(− ~1 ). Since a is
compactly supported, F(a) is Schwartz. It follows
I~ = O(~N ),
∀N.
Intuitively, in the second case the exponential exp(i x~ ) oscillates fast in any interval
of x for small ~, so that many cancellations take place and thus we get a function
rapidly decreasing in ~. This is in fact the case at any point x which is not a critical
point of ϕ. On the other hand, near a critical point of x the phase function ϕ doesn’t
change much, i.e. “looks like” a constant, so that we are in case 1. According to
these intuitive observation, we expect that the main contributions to I~ should arise
from the critical points of the phase function ϕ.
We will use the following notions:
Definition 2.1. Let f = f (~).
LECTURE 5: THE METHOD OF STATIONARY PHASE
(1) We say f ∼
P∞
k=0
ak ~k if for each nonnegative integer N ,
f−
(18)
7
N
X
ak ~k = O(~N +1 ),
~ → 0.
k=0
(2) We say f = O(~∞ ) if f ∼ 0, i.e.
f = O(~N ),
(19)
~→0
for all N .
So in the second extremal case we discussed above, we can write I~ = O(~∞ ).
P
k
Remark. We don’t require the series ∞
k=0 ak ~ to be convergent!
¶ Non-stationary phase.
Proposition 2.2. Suppose the phase function ϕ has no critical point in a neighborhood of the support of a. Then
I~ = O(~∞ ).
(20)
Proof. Let χ be a smooth cut-off function which is identically one on the support of
a so that ϕ has no critical point on the support of χ. Then a(x) = χ(x)a(x), and
χ(x)
is a smooth function. Define an operator L by
|∇ϕ(x)|2
Lf (x) =
χ(x) X
∂j ϕ(x)∂j f (x).
|∇ϕ(x)|2 j
Then it is easy to check
L(ei
ϕ(x)
~
)(x) =
χ X
i∂j ϕ i ϕ(x)
i i ϕ(x)
~
∂
ϕ
e
=
χe ~ .
j
|∇ϕ|2 j
~
~
It follows
~
I~ =
i
Z
L(e
Rn
i
ϕ(x)
~
~
)a(x)dx =
i
Z
ei
ϕ(x)
~
(L∗ a)(x)dx.
Rn
Repeating this N times, we get
N Z
ϕ(x)
~
I~ =
ei ~ (L∗ )N a(x)dx = O(~N ).
i
Rn
8
LECTURE 5: THE METHOD OF STATIONARY PHASE
¶ Morse lemma.
Recall that a point p is called a critical point of a smooth function ϕ if
∇ϕ(p) = 0.
(21)
A critical point p of ϕ is called non-degenerate if the matrix d2 ϕ(p) is non-degenerate:
2
∂ ϕ
2
(22)
det d ϕ(p) = det
(p) 6= 0.
∂xi ∂xj
A smooth function all of whose critical points are non-degenerate is called a Morse
function.
Theorem 2.3 (Morse lemma, form 1). Let ϕ ∈ C ∞ (Rn ). Suppose p is a nondegenerate critical point of ϕ. Then there exists a neighborhood U of 0, a neighborhood V of p and a diffeomorphism f : U → V so that f (0) = p and
1
(23)
f ∗ ϕ(x) = ϕ(p) + (x21 + · · · + x2r − x2r+1 − · · · − x2n ),
2
where r is the number of positive eigenvalues of the Hessian matrix d2 ϕ(p).
We shall prove the following form of the Morse lemma which is obviously equivalent to the above statement:
Theorem 2.4 (Morse lemma, form 2). Let ϕ0 and ϕ1 be smooth functions on Rn
such that
• ϕi (0) = 0,
• ∇ϕ0 (0) = ∇ϕ1 (0) = 0.
• d2 ϕ0 (0) = d2 ϕ1 (0) is non-degenerate.
Then there exist neighborhoods U0 and U1 of 0 and a diffeomorphism f : U0 → U1
such that f (0) = 0 and f ∗ ϕ1 = ϕ0 .
Proof. Let ϕt = (1 − t)ϕ0 + tϕ1 . Applying Moser’s trick, it is enough to find a
time-dependent vector field Xt such that Xt (0) = 0 and
LXt ϕt = ϕ0 − ϕ1 .
(24)
Let Xt =
P
Aj (x, t) ∂x∂ j , where Aj (0, t) = 0. According to the second and the third
2
ϕt
conditions, [ ∂x∂ i ∂x
(0)] is non-degenerate. It follows that the system of functions
j
{
∂ϕt
| i = 1, 2, · · · , n}
∂xi
t
form a system of coordinates near 0 with ∂ϕ
(0) = 0. According to the next lemma,
∂xi
one can find functions Aij (x, t) with Aij (0, t) = 0 so that
X
∂ϕt
Aj (x, t) =
Aij (x, t)
.
∂xi
LECTURE 5: THE METHOD OF STATIONARY PHASE
9
Inserting this into the equation (24), we get
X
∂ϕt ∂ϕt
(25)
Aij (x, t)
= ϕ0 − ϕ1 .
∂xi ∂xj
On the other hand, the three conditions implies that ϕ0 − ϕ1 vanishes to 2nd order
at 0, so according the next lemma, one can find functions Bijk (x, t) so that
X
∂ϕt ∂ϕt ∂ϕt
.
ϕ0 − ϕ1 =
Bijk (x, t)
∂xi ∂xj ∂xk
In other words, the equation (25) is solvable with Aij (0, t) = 0. This proves the
existence of the diffeomorphism f .
Lemma 2.5. Let ϕ be a smooth function with ϕ(0) = 0 and ∂ α ϕ(0) = 0 for all
|α| < k. Then there exists smooth functions ϕα so that
X
(26)
ϕ(x) =
xα ϕα (x).
|α|=k
Proof. For k = 1, one just take
Z
1
∂ϕ
(tx)dt.
0 ∂xi
For larger k, apply the above formula and induction.
ϕi (x) =
By exactly the same method, one can prove the following Morse lemma with
parameters:
Theorem 2.6 (Morse lemma with parameter). Let ϕsi ∈ C ∞ (Rn ) be two family of
smooth functions, depending smoothly on the parameter s ∈ Rk . Suppose
• ϕsi (0) = 0,
• ∇ϕs0 (0) = ∇ϕs1 (0) = 0,
• d2 ϕs0 (0) = d2 ϕs1 (0) are non-degenerate.
Then there exist an ε > 0, a neighborhood U of 0 and for all |s| < ε an open
embedding fs : U → Rn , depending smoothly on s, so that fs (0) = 0 and fs∗ ϕs1 = ϕs0 .
¶ The stationary phase formula for quadratic phase.
Associated to any polynomial
X
p(ξ) =
pα ξ α
we have a constant coefficient differential operator
X
p(D) =
pα D α .
According to the Fourier transform formula that for any a ∈ S ,
F(Dα a) = ξ α Fa
we immediately get
10
LECTURE 5: THE METHOD OF STATIONARY PHASE
Lemma 2.7. For any a ∈ S ,
F(p(D)a)(ξ) = p(ξ)Fa(ξ).
(27)
In particular, associated to any n × n symmetric matrix A = (aij ) one has a
quadratic function
X
pA (ξ) = ξ T Aξ =
aij ξi ξj
and thus a second order constant coefficient differential operator
X
(28)
pA (D) :=
akl Dk Dl .
It follows that
F(pA (D)a)(ξ) = (ξ T Aξ)Fa(ξ).
More generally, for any nonnegative integer k,
F(pA (D)k a)(ξ) = (ξ T Aξ)k Fa(ξ).
(29)
Suppose the phase function is a quadratic function, i.e. ϕ(x) = 12 xT Qx for some
symmetric real n × n matrix Q. Then the only critical point of ϕ is the origin x = 0.
More over, the non-degeneracy condition implies that Q is non-singular.
Theorem 2.8. Let Q be a real, symmetric, non-singular n × n matrix. For any
a ∈ Cc∞ (Rn ), one has
k
Z
π
i T
ei 4 sgn(Q) X k 1
i
x
Qx
n/2
(30)
e 2~
a(x)dx ∼ (2π~)
~
− pQ−1 (D) a(0).
| det Q|1/2 k
k!
2
Rn
Proof. We start with the formula
Z
Z
Fϕ(ξ)ψ(ξ)dξ = ϕ(x)Fψ(x)dx.
i
Recall from theorem 1.5 that if ϕ(x) = e 2 x
Fϕ(ξ) =
T Qx
, then
i π4 sgn(Q)
(2π)n/2 e
| det(Q)|1/2
i T −1
Q ξ
e− 2 ξ
.
It follows
Z
π
~n/2 ei 4 sgn(Q)
−~ 2i ξ T Q−1 ξ
e
a(x)dx =
e
Fa(ξ)dξ.
(2π)n/2 | det Q|1/2 Rn
Rn
Using the Taylor’s expansion formula for the exponential function, we see that for
any non-negative integer N , the difference
Z
Z
N
X
1
i~ k
−~ 2i ξ T Q−1 ξ
e
Fa(ξ)dξ −
(− )
(ξ T Q−1 ξ)k Fa(ξ)dξ
k!
2
Rn
k=0
Z
i T
x Qx
2~
is bounded by
N +1
~
1
N
+1
2
(N + 1)!
Z
Rn
|(ξ T Q−1 ξ)k Fa(ξ)|dξ.
LECTURE 5: THE METHOD OF STATIONARY PHASE
11
Note that
(ξ T Q−1 ξ)k Fa(ξ) = F(pQ−1 (D)k a)(ξ),
so by the Fourier inversion formula,
1
pQ−1 (D) a(0) =
(2π)n
k
The conclusion follows.
Z
(ξ T Q−1 ξ)k Fa(ξ)dξ.
¶ The stationary phase formula for general phase.
As we have seen, only the critical points of the phase function ϕ give an essential
contribution to the oscillatory integral
Z
ϕ(x)
I~ =
ei ~ a(x)dx.
Rn
In what follows we will assume that ϕ ∈ C ∞ (Rn ) admits only non-degenerate critical
points in a neighborhood of the support of a. Since non-degenerate critical points
must be discrete, ϕ has only finitely many critical points in the support of a. Thus
one can find a partition of unity {Ui | 1 ≤ i ≤ N + 1} of the support of a so that
each Ui , 1 ≤ i ≤ N , contains exactly one critical point pi of ϕ, and UN +1 contains
no critical points of ϕ. A partition of unity arguments converts the asymptotic
behavior of I~ to the sum of the asymptotic behavior of
Z
ϕ(x)
pi
I~ =
ei ~ χi (x)a(x)dx,
Ui
where χi is a function compactly supported in Ui and equals 1 identically near pi .
According to the Morse lemma, for each Ui (one can always shrink Ui in the above
integral if necessary) one can find a diffeomorphism f : Vi → Ui so that f (0) = pi
and
1
f ∗ ϕ(x) = ϕ(pi ) + (x21 + · · · + x2ri − x2ri +1 − · · · − x2n ) =: ψpi (x),
2
where ri is the number of positive eigenvalues of the matrix d2 ϕ(pi ). It follows from
the change of variable formula that
Z
ψpi (x)
pi
I~ =
ei ~ a(f (x))χi (f (x))| det df (x)|dx.
Vi
Moreover, since ψpi (x) has a unique non-degenerate critical point at 0, modulo
O(~∞ ) one has
Z
ψpi (x)
pi
(31)
I~ =
ei ~ a(f (x))| det df (x)|dx + O(~∞ ).
Rn
The asymptotic of this integral is basically given by theorem (2.8).
In particular, to get the leading term in the asymptotic expansion above, one
only need to evaluate | det df (0)|, which follows from
12
LECTURE 5: THE METHOD OF STATIONARY PHASE
Lemma 2.9. Let ψ, ϕ be smooth functions defined on V and U respectively, and
suppose 0 is a non-degenerate critical point of both ψ and p a non-degenerate critical
point of ϕ. If f : V → U a diffeomorphism so that f (0) = p and f ∗ ϕ = ψ. Then
df T (0)d2 ϕ(p)df (0) = d2 ψ(0).
(32)
Proof. We start with the equation ϕ ◦ f = ψ. Taking derivatives of both sides one
gets
∂ϕ
∂fk
∂ψ
(f (x))
=
.
∂yk
∂xj
∂xj
Taking the second order derivatives at 0 of both sides, and using the conditions
f (0) = p and ∇ϕ(p) = 0, one gets
∂fl
∂fk
∂ 2ψ
∂ 2ϕ
(p)
(0)
(0) =
(0),
∂yk ∂yl
∂xi ∂xj
∂xi ∂xj
in other words
df T (0)d2 ϕ(p)df (0) = d2 ψ(0).
It follows that in our case,
| det df (0)| = | det d2 ϕ(pi )|−1/2 .
(33)
In conclusion, we have
Theorem 2.10. As ~ → 0, the oscillatory integral has the asymptotic expansion
X p
(34)
I~ ∼
I~ i ,
dϕ(pi )=0
where
(35)
I~pi ∼ (2π~)n/2 ei
ϕ(pi )
~
iπ
2 ϕ(p ))
i
e 4 sgn(d
X
a(pi )
~j Lj (a)(pi ),
2
1/2
| det d ϕ(pi )| j≥0
where Lj = Lj (x, D) is a differential operator of order 2j (which depends on the
phase function ϕ) with L0 = 1. In particular, the leading term is
X
ϕ(pi ) iπ
n
a(pi )
2
+1
2
+
O(~
(36)
I~ = (2π~)n/2
ei ~ e 4 sgn(d ϕ(pi ))
).
| det d2 ϕ(pi )|1/2
dϕ(pi )=0
Remark. There are more complicated version of the stationary phase method where
the phase function is allowed to have degenerate critical points.