398 APPENDIX B. CLASSIFICATION OF REAL NUMBERS

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 B.3.2 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

You see that this is unchanged from switching two variables. For example, switch x1 andx2. By assumption, nothing changes in the terms after the first two. The first term thenbecomes

h(x2,x2,x1...,xn)xr2 = h(x2,x1,x2, ...,xn)xr

2

and the second term becomes

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

1

which are the same two terms, just added in a different order. The situation works the sameway with any other pair of variables. ■

Recall that every algebraic number is a root of a polynomial having integer coefficients.