The Dirichlet Integral - University of Alberta

MATH 300 Fall 2004
Advanced Boundary Value Problems I
Dirichlet’s Integral
Department of Mathematical and Statistical Sciences
University of Alberta
In this note, we show that
as follows. For each n ≥ 1, define
un =
π
2
0
∞
0
sin t
t
dt = π
2
sin 2nx cot x dx and vn =
0
π
2
0
sin 2nx
x
(a) First we show that un+1 = un = π
for all n ≥ 1.
2
(b) Next we show that v = lim
n→∞ vn
∞
sin x
=
0 x dx.
(c) Finally, we integrate by parts and use L’Hospital’s rule to show that lim
n→∞ (un − vn) = 0.
(d) From all of the above, we have
∞
sin x π
dx =
x 2 .
Solution:
(a)
un+1 − un =
π
2
0
π
2
= 2
0
=
1
2(n + 1)
therefore un+1 − un = 0 for all n ≥ 1.
Now,
and therefore un+1 = un = π
2
π
2
[sin 2(n + 1)x − sin 2nx] cot x dx = 2
0
cos x cos(2n + 1)x dx =
sin 2(n + 1)x
u1 =
=
π
2
0
π
2
0
= π
2 ,
for all n ≥ 1.
π
2
0
+ 1
2n
π
2
0
dx
sin x cos(2n + 1)x cot x dx
[cos(2n + 2)x + cos 2nx] dx
sin 2nx
π
2
π
2
sin 2x cot x dx = 2
0
(1 + cos 2x) dx = π
2
0
= 0,
cos 2 x dx
+ sin 2x
2
π
2
0

π
2
(b) Make the substitution t = 2nx in the integral
0
and therefore
(c) Integrating by parts, we obtain
and therefore
un − vn =
un − vn = (−1)n
nπ
π
2
0
cos 2nx
= −
2n
Using L’Hospital’s rule, we have
and
vn =
π
2
0
sin 2nx
x
sin 2nx
x
nπ
dx =
v = lim
n→∞ vn
∞
=
0
cot x − 1
sin 2nx dx
x
cot x − 1
x
π
2
0
0
− 1
π
2
2n 0
dx, then
sin t
t dt.
+ lim
p→0 +
cos 2np
cot p −
2n
1
−
p
1
π
2
2n 0
lim
p→0 +
cot p − 1
= 0,
p
lim
x→0 +
csc 2 x − 1
x2
= 1
3 .
sin t
t dt,
csc 2 x − 1
x2
cos 2nx dx,
csc 2 x − 1
x2
cos 2nx dx. (∗)
Therefore, we may redefine the integrand on the right-hand side of (∗) to be continuous on the interval
], and hence bounded there. Thus, there exists a constant M > 0 such that
[0, π
2
π
2
0
csc 2 x − 1
x2
Letting n → ∞ in (∗), we obtain
and therefore,
cos 2nx dx
≤
≤
π
2
0
π
2
0
≤ πM
2 .
lim
n→∞ (un − vn) = 0,
csc 2 x − 1
x2
cos 2nx
dx
csc2 x − 1
dx
lim
n→∞ un = lim
n→∞ (un − vn + vn) = lim
n→∞ (un − vn) + lim
n→∞ vn,
so that ∞
0
sin t
t
dt = π
2 .
x 2

Next we show that
as follows.
Let n be a positive integer,
(a) First we show that
(b) Next we show that
(c) Finally, we show that
(d) From all of the above, we have
Solution:
(a)
(n+1)π
0
sin x
x
0
∞
sin t
t dt = +∞
0
(n+1)π
sin x
x dx =
0
n
π
k=0
(n+1)π
sin x
2
x dx ≥
π
∞
k=0
dx =
=
=
1
= lim
k + 1 n→∞
0
n
k=0
0
n
k=0
sin x
x + kπ dx.
1
k + 1 .
1
= +∞.
k + 1
∞
sin x
x dx = +∞.
n
(k+1)π
k=0
kπ
n
π
k=0
0
n
π
k=0
0
sin x
x
(−1) k sin t
t + kπ
sin t
t + kπ dt,
dx =
dt =
where the last equality follows since sin t ≥ 0 for 0 ≤ t ≤ π.
1
(b) For 0 < t < π, we have
t + kπ >
1
, so that (a) implies that
π(k + 1)
(c) For t ≥ 1,
(n+1)π
0
sin x
x
dx >
n
k=0
x
log x =
for x ≥ 1, and replacing x by k+2
k+1 , we have
1
log
n
π
k=0
0
n
π
k=0
π
1
sin t dt =
π(k + 1) 0
2
π
x
1
dt ≤ 1 dt = x − 1
t 1
k + 2
≤
k + 1
1
k + 1
0
sin(t + kπ)
t + kπ
| sin t|
t + kπ dt
n
k=0
1
k + 1 .
dt

(d)
for all k ≥ 0. Therefore,
and so
n
k=0
1
k + 1 ≥
lim
n→∞
n
(log(k + 2) − log(k + 1)) = log(n + 2),
k=0
n
k=0
∞
sin x
(n+1)π
0 x dx = lim
n→∞
0
≥ lim
n→∞
Therefore, ∞
1
≥ lim log(n + 2) = +∞.
k + 1 n→∞
0
2
π
sin x
2
x dx ≥ lim
n→∞ π
log(n + 2) = +∞,
sin x
x
dx = +∞.
n
k=0
1
k + 1