A.3. TRANSCENDENTAL NUMBERS 367

is symmetric in the β 1, · · · ,β m. Thanks to the factor vpv(m+1)p and the factor p! comingfrom j ≥ p, it is a symmetric polynomial in the vβ i with integer coefficients, each multi-plied by p with the β i roots of Q(x) = vxm + · · ·+u. By Theorem A.2.7 this is an integer.As noted earlier, it equals 0 unless j ≥ p when it contains a factor of p. Thus the sum ofthese integers is also an integer times p. It follows that

m

∑i=1

n

∑j=0

f ( j) (β i) = m2 (p) p, m2 (p) an integer.

Note that no use was made of p being a large prime number. This will come next.

Lemma A.3.4 If K and c are nonzero integers, and β 1, · · · ,β m are the roots of a singlepolynomial with integer coefficients,

Q(x) = vxm + · · ·+u

where v,u ̸= 0, then,K + c

(eβ 1 + · · ·+ eβ m

)̸= 0.

Letting

f (x)≡ v(m+1)pQp (x)xp−1

(p−1)!

and I (s) be defined in terms of f (x) as above,

I (s)≡∫ s

0es−x f (x)dx = es

deg( f )

∑j=0

f ( j) (0)−deg( f )

∑j=0

f ( j) (s) ,

it follows,

limp→∞

m

∑i=1

I (β i) = 0 (1.10)

and for n the degree of f (x) ,n = pm+ p− 1, where mi (p) is some integer for p a largeprime number.

Proof: The first step is to verify 1.10 for f (x) as given in 1.6 for p large prime numbers.Let p be a large prime number. Then 1.10 follows right away from the definition of I

(β j

)and the definition of f (x) .

∣∣∣I(β j

)∣∣∣≤ ∫ 1

0

∣∣∣β j f(

tβ j

)eβ j−tβ j

∣∣∣dt ≤∫ 1

0

∣∣∣∣∣∣∣|v|(m−1)p

∣∣∣Q(tβ j

)∣∣∣p t p−1∣∣∣β j

∣∣∣p−1

(p−1)!dt

∣∣∣∣∣∣∣which clearly converges to 0 using considerations involving convergent series which showthe integrand converges uniformly to 0. The degree of f (x) is n≡ pm+ p−1 where p willbe a sufficiently large prime number from now on.

From 1.4,

cm

∑i=1

I (β i) = cm

∑i=1

(eβ i

n

∑j=0

f ( j) (0)−n

∑j=0

f ( j) (β i)

)