POW 2019 06 - KAIST

POW 2019 ? 06

2017xxxx Keonwoo Kim

There are several available proofs. (In proofs shown below, many integrals are defined improperly or as the Lebesgue integral without explicit notations, and and ~ are assumed to be a (small) positive real. Also, Lemmas are gathered at the end of this document.)

Proof 1 (Everyone might have tried this). Let

/2

I=

log (2 cos x) dx.

0

Note that cos

2

-

x

=

sin x,

hence

substituting

with

t

=

2

-

x,

we

have

0

/2

I = log 2 cos - t ? (-1) dt =

log (2 sin t) dt.

()

/2

2

0

Therefore,

2I = I + I

/2

/2

=

log (2 cos x) dx +

log (2 sin x) dx

0

0

/2

=

log (4 sin x cos x) dx

0

/2

=

log (2 sin (2x)) dx

0

1

=

log (2 sin u) du

20

(u = 2x)

Note

that

sin u

is

symmetric

about

the

vertical

line

x

=

2

.

Hence

1

/2

log (2 sin u) du =

log (2 sin u) du = I.

20

0

Because 2I = I, we get I = 0.

1

Proof 2 (With a product of sines). We have (). By Lemma 1,

/2

n-1

j

I = lim

log (2 sin x) dx = lim

log 2 sin

.

0

n n

n

j=1

Also, there is a famous identity (Lemma 2):

n-1

j

n

sin n = 2n-1 .

j=1

Therefore,

n-1

j

I = lim

log 2 sin

n n

n

j=1

= lim log 2n-1 n-1 sin j

n n

n

j=1

= lim log n = 0.

n n

Furthermore, there is another way to represent this integral. Note that log |cos x| is

an antiderivative of tan x, and cos x is positive on

0,

2

-

.

Thus,

x

log (cos x) = - tan t dt,

0

/2-

x

I = lim

00

log 2 - tan t dt dx.

0

By exchanging two integral signs1,

/2-

/2-

log 2

/2-

I = lim

log 2 -

tan t dx dt =

- lim

- t tan t dt.

00

t

2

00

2

Also,

by

substituting

x

=

2

-

t,

we

have

/2-

I1 := lim

00

-t

2

/2

tan t dt = lim

t cot t dt.

0

So, it suffices to show I1 = ( log 2)/2.

1Tonelli's theorem works since tan t is positive; see, for instance, [1, 2.37].

2

Proof 3 (Using the Leibniz integral rule). Define an integral with a parameter as

follows:

/2

I() =

f (, x) dx

0

where

tan-1 ( tan x)

f (, x) =

.

tan x

Then, we have I(0) = 0 and I(1) = lim 1 I() = I1. Using the Leibniz integral rule on

R

=

{(x, y)

:

x

2

- ~,

0

y

x},

since

f

and

1

f (, x) = ( tan x)2 + 1

are continuous on a region containing R,

/2-~

/2-~

f (, x) dx =

f (, x) dx

/2-~

1

=

( tan x)2 + 1 dx.

Calculating the integral, (0 < < 1)

1

1

( tan x)2 + 1 dx = (u2 + 1)(2u2 + 1) du

1 = 2 - 1

2 2u2 + 1 du -

tan-1 ( tan x) - x

=

2 - 1

+ C.

1 u2 + 1 du

(u = tan x) (partial fraction)

Thus,

/2-~

tan-1 ( tan x) - x /2-~

f (, x) dx =

2 - 1

tan-1 =

tan

2

-

~

- tan-1 2 - 1

(

tan )

-

2

+

+

~ .

Note that the limit of the above converges uniformly to (/2-/2)/(2-1) = /(2(+1)) as , ~ 0. Therefore, we can exchange the limit and the differentiation2:

/2-~

/2-~

= lim

f (, x) dx = lim

f (, x) dx = I().

2( + 1) ,~0

,~0

Now, by the fundamental theorem of calculus,

I() = I() - I(0) =

d = log( + 1),

0 2( + 1)

2

I(1) = 2 log 2.

2See, e.g., [2, Theorem 7.17].

3

Proof 4 (Using the Riemann?Lebesgue lemma). (Lemma 3), we have the following:

By a simple summation of sines

Thus,

N

cos ((2N + 1)x)

2 sin (2nx) = cot x -

.

sin x

n=0

/2

N /2

/2 cos ((2N + 1)x)

x cot x dx = 2

x sin (2x) +

x

dx.

0

n=0 0

0

sin x

Note that whence

cos ((2N + 1)x) = cot x cos (2N x) - sin (2N x)

sin x

/2 cos ((2N + 1)x)

/2

/2

x

dx =

x cot x cos (2N x) dx -

x sin (2N x) dx.

0

sin x

0

0

Here, the RHS converges to 0 as N by the Riemann?Lebesgue lemma. Consequently,

/2

/2

(-1)n-1 log 2

x cot x dx = 2

x sin (2nx) dx =

=

.

0

n=0 0

2n

2

n=0

Lemmas

Lemma 1 (convergence of Riemann sums for improper integrals for a monotone function). Let f be a real-valued monotone function defined on a half-open interval. Suppose, wlog, f : (0, 1] R is nonnegative and decreasing. (O/w, consider C - f (kx) for an appropriate constant C and k.) Suppose further that there is a singularity at x = 0, but f is Riemann integrable on [c, 1] for all 0 < c 1 and the improper integral is convergent:

1

1

f (x) dx = lim f (x) dx < .

0

c 0c

Moreover, assume x f (x) 0 as x 0. Then we have:

1n

k

1

f

= f (x) dx.

n

n

k=1

0

4

Proof. Since f is decreasing,

1k

(k+1)/n

1 k+1

f

f (x) dx f

nn

k/n

n

n

so that which implies

1 n-1

k

1

1n

k

f

f (x) dx

f

,

n

n

k=1

1/n

n

n

k=2

1

1

1n

k

1

11

f (x) dx + f (1)

f

f (x) dx + f

.

1/n

n

n

n

k=1

1/n

nn

Since f (1)/n and f (1/n)/n tend to 0 by the assumptions, by the squeeze theorem,

1n

k

1

1

lim

f

= lim f (x) dx f (x) dx.

n n

n

k=1

n 1/n

0

Lemma 2.

n-1

j

n

sin n = 2n-1 .

j=1

Proof. Consider a root of unity = ei 2/n. Note that

|1 - k| =

2k

2k

1 - cos + i sin

=

2k

k

2 - 2 cos = 2 sin .

n

n

n

n

Note that

1

+

z

+

???

+

zn-1

=

zn

-

1

=

n-1

(z

-

k ).

z-1

k=1

Evaluating the limit as z 1,

n-1

n = (1 - k),

k=1

n

=

|n|

=

n-1

|1

-

k|

=

2n-1

sin

k n

.

k=1

Lemma 3.

N

cos ((2N + 1)x)

2 sin (2nx) = cot x -

.

sin x

n=0

5

Proof. When N = 0, both sides are the same with 0. Also, an identity

cos ((2N - 1)x) - cos ((2N + 1)x) 2 sin (2N x) =

sin x

completes the induction step:

N

N -1

2 sin (2nx) = 2 sin (2nx) + 2 sin (2N x)

n=0

n=0

cos ((2N - 1)x)

= cot x -

+

sin x

cos ((2N + 1)x) = cot x -

sin x

cos ((2N - 1)x) - cos ((2N + 1)x) sin x

References

[1] Folland, G. B. 1999. Real analysis: Modern techniques and their applications. New York: Wiley.

[2] Rudin, W. 1964. Principles of mathematical analysis. McGraw-Hill.

6

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