Module MA1132 (Frolov), Advanced Calculus
Tutorial Sheet 11
To be solved during the tutorial session Thursday/Friday, 7/8 April 2015
Name:
1. Compute the integral
Z
In (A, B) =
∞
−∞
Z
∞
exp −
···
−∞
n
X
xi Aij xj + 2
i,j=1
n
X
!
Bi xi
dx1 · · · dxn ,
i=1
where A = (Aij ) is a positive-definite symmetric n × n matrix.
Solution: Let us introduce the column x with entries xi and the column B with entries
Bi . Then we can write in matrix form
n
n
X
X
−
xi Aij xj + 2
Bi xi = −xt · A · x + 2B t · x ,
(1)
i,j=1
i=1
where · denotes the matrix multiplication, and t denotes matrix transposition. Then using
that At = A we can write (complete the square)
−xt · A · x + 2B t · x = −(x − A−1 · B)t · A · (x − A−1 · B) + B t · A−1 · B
= −y t · A · y + B t · A−1 · B ,
where y = x − A−1 · B. Thus, the integral In (A, B) is equal to
Z ∞
Z ∞
t
t
−1
In (A, B) =
···
e−y ·A·y+B ·A ·B dy1 · · · dyn
−∞
Z−∞∞
Z ∞
t
B t ·A−1 ·B
=e
···
e−y ·A·y dy1 · · · dyn .
−∞
(2)
(3)
−∞
Since A is symmetric it can be diagonalised by an orthogonal matrix
A = O · Λ · Ot ,
Λ = diag(λ1 , . . . , λn ) ,
λi > 0 ∀i ,
O : O · Ot = I .
(4)
Now, we make the following change of variables
y =O·z
⇔
yi =
n
X
Oij zj .
(5)
j=1
The Jacobian matrix is (∂yi /∂zj ) = Oij . Since O is orthogonal one gets
J = det(∂yi /∂zj ) = ±1 . Thus the integral takes the form
Z ∞
Z ∞
t
t
B t ·A−1 ·B
In (A, B) = e
···
e−y ·O·Λ·O ·y dy1 · · · dyn
Z−∞
Z−∞
∞
∞
t
−1
t
= eB ·A ·B
···
e−z ·Λ·z dz1 · · · dzn
Z−∞
Z−∞
∞
∞
Pn
t
−1
2
= eB ·A ·B
···
e− i=1 λi zi dz1 · · · dzn
−∞
−∞
Z ∞
n
t
−1
B t ·A−1 ·B
e
eB ·A ·B n/2
−z 2
=√
e dz
=√
π .
λ1 · · · λn
λ1 · · · λn
−∞
1
(6)
Since λ1 · · · λn = det A, we finally get
In (A, B) = eB
π n/2
√
.
det A
t ·A−1 ·B
(7)
2. Find the volume Vn,h of an n-simplex ∆n
x1 ≥ 0 , . . . , x n ≥ 0 ,
∆n :
x1 + x2 + · · · + xn ≤ h .
(8)
Solution: From the definition of ∆n we get
0 ≤ xn ≤ h − x1 − x2 − · · · − xn−1 .
(9)
Thus, the projection of ∆n onto the hyperplane xn = 0 is the n − 1-simplex ∆n−1 , and
Z h−x1 −···−xn−1
Z
Z
Z
Z
dx1 · · · dxn = · · ·
dxn dx1 · · · dxn−1 . (10)
Vn,h = · · ·
∆n−1
∆n
Repeating the procedure we get
Z
Z h Z h−x1
Vn,h =
···
0
0
0
h−x1 −···−xn−1
dxn · · · dx2 dx1 .
(11)
0
Making the change
x1 = hξ1 , x2 = hξ2 , . . . , xn = hξn ,
(12)
one gets
n
Vn,h = h αn ,
Z
1
1−ξ1
Z
αn ≡ Vn,1 =
Z
···
0
0
1−ξ1 −···−ξn−1
dξn · · · dξ2 dξ1 .
(13)
0
On the other hand comparing αn with (11), one sees that αn is equal to
Z 1−ξ1 −···−ξn−1
Z 1 Z 1−ξ1
Z 1
αn =
···
dξn · · · dξ2 dξ1 =
(Vn−1,1−ξ1 ) dξ1
0
0
0
0
Z 1
αn−1
.
= αn−1
(1 − ξ1 )n−1 dξ1 =
n
0
(14)
Taking into account that α1 = 1, one finds
αn =
1
,
n!
(15)
Vn,h =
hn
.
n!
(16)
and finally
2