The second integral is intended as an improper Riemann integral.

5.1 Bessel functions 51

Proof The ¬rst result is known as Cooke™s inequality. The easiest way to prove it is to use

another representation of the Bessel function, namely

∞

2 sin(ut)

J0 (t) = du;

π (u 2 ’ 1)1/2

1

see Watson [187], p. 170. This shows that

∞

1 ’ cos(r u)

r

2

J0 (t)dt = du > 0

π u(u 2 ’ 1)1/2

0 1

for all r > 0.

∞

Finally, let us discuss the second result. We know by Lemma 5.8 that 0 J0 (t)e’r t dt =

(1 + r 2 )’1/2 for all r > 0. Hence we want to let r ’ 0. Unfortunately, since J0 is not in

L 1 (R) we cannot use classical convergence arguments and so we have to be more precise.

The idea is to use the triangle inequality twice to get the bound

R R

2 ’1/2

J0 (t)(e’r t ’ 1)dt

1’ J0 (t)dt ¤ 1 ’ (1 + r ) +

0 0

∞

J0 (t)e’r t dt

+

R

for an arbitrary r ∈ (0, 1]. Next suppose that for every > 0 we can ¬nd an R0 > 0 such

that the last integral becomes uniformly less than /3 for all r ∈ (0, 1] provided R > R0 .

If this is true then we can ¬x such an R > R0 and then choose 0 < r ¤ 1 such that the

¬rst two terms also become small (using that J0 is bounded), which establishes the second

result. Hence it remains to prove this uniform bound on the last integral. To this end we use

the asymptotic expansion of J0 from Proposition 5.6, i.e.

2 cos(t ’ π/4)

J0 (t) = + S(t),

√

π t

with a remainder S(t) satisfying |S(t)| ¤ Ct ’3/2 for t ≥ 1. The remainder part of the integral

∞ ∞

in question can easily be bounded since | R S(t)e’r t dt| ¤ C R t ’3/2 dt = C R ’1/2 with a

generic constant C > 0 that is independent of r > 0. The main part is bounded by integration

by parts:

1 ’r t (1 + r ) cos t + (r ’ 1) sin t ∞

∞

e’r t π

√ cos t ’ dt = ’ √ e

1 + r2

4

t 2t

R R

∞

e’r t (1 + r ) cos t + (r ’ 1) sin t

’ √ dt.

1 + r2

8 t 3/2

R

This expansion shows that the integral on the left-hand side can also be bounded uniformly

by a constant times R ’1/2 for all r ∈ (0, 1].

After discussing Bessel functions of the ¬rst kind we come to another family of Bessel

functions, called Bessel functions of imaginary argument or modi¬ed Bessel functions of

the third kind, sometimes also Mcdonald™s functions.

52 Auxiliary tools from analysis and measure theory

De¬nition 5.10 The modi¬ed Bessel function of the third kind of order ν ∈ C is de¬ned

by

∞

e’z cosh t cosh(νt) dt

K ν (z) :=

0

for z ∈ C with |arg(z)| < π/2; cosh t = (et + e’t )/2.

It follows immediately from the de¬nition that K ν (x) > 0 for x > 0 and ν ∈ R. The

modi¬ed Bessel functions satisfy recurrence relations similar to the Bessel functions of the

¬rst kind (see Watson [187], p. 79). The only one that matters here is

dν

z K ν (z) = ’z ν K ν’1 (z). (5.2)

dz

Moreover, there exist several other representation formulas. The following one is of partic-

ular interest for us. Again, its proof goes beyond the scope of this book, so for this we refer

the reader again to Watson [187], p. 206.

Proposition 5.11 For ν ≥ 0 and x > 0 the modi¬ed Bessel function of the third kind has

the representation

∞

e’x

π ν’1/2

u

1/2

e’u u ν’1/2 1 +

K ν (x) = du.

(ν + 1/2)

2x 2x

0

This representation gives some insight into lower bounds on the decay of the modi¬ed

Bessel functions.

Corollary 5.12 For every ν ∈ R the function x ’ x ν K ’ν (x) is nonincreasing on (0, ∞).

Moreover, it has the lower bound

π e’x

K ν (x) ≥ √, x > 0,

2x

if |ν| ≥ 1/2. In the case |ν| < 1/2 the lower bound is given by

√ |ν|’1/2

e’x

π3

K ν (x) ≥ |ν|+1 √, x ≥ 1.

(|ν| + 1/2) x

2

Proof The recurrence relation (5.2) together with K ’ν = K ν and the fact that K ν’1 is

positive on (0, ∞) gives the monotonic property of x ν K ’ν (x). To prove the lower bound

we can restrict ourselves to ν ≥ 0 because of K ν = K ’ν . If ν ≥ 1/2 then we get from

Proposition 5.11

∞

e’x

π π

1/2 1/2

e’u u ν’1/2 du = e’x .

K ν (x) ≥

(ν + 1/2)

2x 2x

0

However, if 0 ¤ ν < 1/2 then we have to be more careful. We use that for u ∈ [0, 1] and

5.1 Bessel functions 53

x ≥ 1 it is obviously true that u ¤ 1 and 1 + u/(2x) ¤ 1 + 1/(2x) ¤ 3/2, so that

∞ 1

ν’1/2 ν’1/2

u u

e’u u ν’1/2 1 + e’u u ν’1/2 1 +

du ≥ du

2x 2x