200 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

The proof makes use of the following identity. For f (x) a polynomial,

I (s)≡∫ s

0es−x f (x)dx = es

deg( f )

∑j=0

f ( j) (0)−deg( f )

∑j=0

f ( j) (s) . (9.4)

where f ( j) denotes the jth derivative. It is like the convolution integral discussed earlierwith Laplace transforms. In this formula, s ∈ C and the integral is defined in the naturalway as ∫ 1

0s f (ts)es−tsdt (9.5)

The identity follows from integration by parts.∫ 1

0s f (ts)es−tsdt = ses

∫ 1

0f (ts)e−tsdt

= ses[−e−ts

sf (ts) |10 +

∫ 1

0

e−ts

ss f ′ (st)dt

]= ses

[−e−s

sf (s)+

1s

f (0)+∫ 1

0e−ts f ′ (st)dt

]= es f (0)− f (s)+

∫ 1

0ses−ts f ′ (st)dt

≡ es f (0)− f (s)+∫ s

0es−x f ′ (x)dx

Continuing this way establishes the identity since the right end looks just like what westarted with except with a derivative on the f .

Lemma 9.2.2 Let (x1, ...,xn)→ g(x,x1, ...,xn) be symmetric and let

x→ g(x,x1, ...,xn)

be a polynomial. Thendm

dxm g(x,x1, ...,xn)

is symmetric in the variables {x1, ...,xn}. If (x1, ...,xn)→ h(x,x1, ...,xn) is symmetric, thenfor r some nonnegative integer,

n

∑k=1

h(xk,x1, ...,xn)xrk

is symmetric. In particular,n

∑k=1

dl

dxl g(·,x1, ...,xn)(xk)xrk

is symmetric in {x1, ...,xn}.

Proof: The coefficients of the polynomial x→ g(x,x1, ...,xn) are symmetric functionsof {x1, ...,xn} . Differentiating with respect to x multiple times just gives another polyno-mial in x having coefficients which are symmetric functions. Thus the first part is proved.For the second part, the sum is of the form

h(x1,x1, ...,xn)xr1 +h(x2,x1, ...,xn)xr

2 + · · ·+h(xn,x1, ...,xn)xrn