Real Analysis 575Remark.This is a particular case of integrals of the form
∫∞
0f(ax)−f(bx)
x dx, known as
Froullani integrals. In general, iffis continuous and has finite limit at infinity, the value
of the integral is(f ( 0 )−limx→∞f(x))lnba.
524.We do the proof in the case 0<x<1, since for− 1 <x<0 the proof is completely
analogous, while forx=0 the property is obvious. The functionf:N×[ 0 ,x]→R,
f (n, t)=tn−^1 satisfies the hypothesis of Fubini’s theorem. So integration commutes
with summation:
∑∞n= 0∫x0tn−^1 dt=∫x0dt
1 −t.
This implies
∑∞n= 1xn
n=−ln( 1 −x).Dividing byx, we obtain
∑∞n= 1xn−^1
n=−
1
xln( 1 −x).The right-hand side extends continuously at 0, since limx→ (^01) tln( 1 −t)=−1. Again we
can apply Fubini’s theorem tof (n, t)=t
n− 1
n onN×[^0 ,x]to obtain
∑∞
n= 1
xn
n^2
=
∑∞
n= 1∫x0tn−^1
ndt=∫x0∑∞
n= 1tn−^1
ndt=−∫x01
tln( 1 −t)dt,as desired.
525.We can apply Tonelli’s theorem to the functionf(x,n)=x (^2) +^1 n 4. Integrating term
by term, we obtain
∫x
0
F(t)dt=
∫x
0
∑∞
n= 1f (t, n)dt=∑∞
n= 1∫x0dt
t^2 +n^4=
∑∞
n= 11
n^2arctanx
n^2.
This series is bounded from above by
∑∞
n= 11
n^2 =π^2
6. Hence the summation commutes
with the limit asxtends to infinity. We have
∫∞0F(t)dt=xlim→∞∫x0F(t)dt=xlim→∞∑∞
n= 11
n^2
arctanx
n^2=
∑∞
n= 11
n^2·
π
2