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§2.5 Mellin Transform Methods

Contents

§2.5(i) Introduction

Let f(t) be a locally integrable function on (0,), that is, ρTf(t)t exists for all ρ and T satisfying 0<ρ<T<. The Mellin transform of f(t) is defined by

2.5.1 (f;z)=0tz-1f(t)t,

when this integral converges. The domain of analyticity of (f;z) is usually an infinite strip a<z<b parallel to the imaginary axis. The inversion formula is given by

2.5.2 f(t)=12πc-c+t-z(f;z)z,

with a<c<b.

One of the two convolution integrals associated with the Mellin transform is of the form

2.5.3 I(x)=0f(t)h(xt)t,
x>0,

and

2.5.4 (I;z)=(f;1-z)(h;z).

If (f;1-z) and (h;z) have a common strip of analyticity a<z<b, then

2.5.5 I(x)=12πc-c+x-z(f;1-z)(h;z)z,

where a<c<b. When x=1, this identity is a Parseval-type formula; compare §1.14(iv).

If (f;1-z) and (h;z) can be continued analytically to meromorphic functions in a left half-plane, and if the contour z=c can be translated to z=d with d<c, then

2.5.6 I(x)=d<z<cres[x-z(f;1-z)(h;z)]+E(x),

where

2.5.7 E(x)=12πd-d+x-z(f;1-z)(h;z)z.

The sum in (2.5.6) is taken over all poles of x-z(f;1-z)(h;z) in the strip d<z<c, and it provides the asymptotic expansion of I(x) for small values of x. Similarly, if (f;1-z) and (h;z) can be continued analytically to meromorphic functions in a right half-plane, and if the vertical line of integration can be translated to the right, then we obtain an asymptotic expansion for I(x) for large values of x.

Example

2.5.8 I(x)=0Jν2(xt)1+tt,
ν>-12,

where Jν denotes the Bessel function (§10.2(ii)), and x is a large positive parameter. Let h(t)=Jν2(t) and f(t)=1/(1+t). Then from Table 1.14.5 and Watson (1944, p. 403)

2.5.9 (f;1-z)=πsin(πz),
0<z<1,
2.5.10 (h;z)=2z-1Γ(ν+12z)Γ2(1-12z)Γ(1+ν-12z)Γ(z)πsin(πz),
-2ν<z<1.

In the half-plane z>max(0,-2ν), the product (f;1-z)(h;z) has a pole of order two at each positive integer, and

2.5.11 resz=n[x-z(f;1-z)(h;z)]=(anlnx+bn)x-n,

where

2.5.12 an =2n-1Γ(ν+12n)Γ2(1-12n)Γ(1+ν-12n)Γ(n),
2.5.13 bn =-an(ln2+12ψ(ν+12n)+ψ(1-12n)+12ψ(1+ν-12n)-ψ(n)),

and ψ is the logarithmic derivative of the gamma function (§5.2(i)).

We now apply (2.5.5) with max(0,-2ν)<c<1, and then translate the integration contour to the right. This is allowable in view of the asymptotic formula

2.5.14 |Γ(x+y)|=2π-π|y|/2|y|x-(1/2)(1+o(1)),

as y±, uniformly for bounded |x|; see (5.11.9). Then as in (2.5.6) and (2.5.7), with d=2n+1-ϵ (0<ϵ<1), we obtain

2.5.15 I(x)=-s=02n(aslnx+bs)x-s+O(x-2n-1+ϵ),
n=0,1,2,.

From (2.5.12) and (2.5.13), it is seen that as=bs=0 when s is even. Hence

2.5.16 I(x)=s=0n-1(cslnx+ds)x-2s-1+O(x-2n-1+ϵ),

where cs=-a2s+1, ds=-b2s+1.

§2.5(ii) Extensions

Let f(t) and h(t) be locally integrable on (0,) and

2.5.17 f(t)s=0astαs,
t0+,

where αs>αs for s>s, and αs+ as s. Also, let

2.5.18 h(t)exp(κtp)s=0bst-βs,
t+,

where κ is real, p>0, βs>βs for s>s, and βs+ as s. To ensure that the integral (2.5.3) converges we assume that

2.5.19 f(t)=O(t-b),
t+,

with b+β0>1, and

2.5.20 h(t)=O(tc),
t0+,

with c+α0>-1. To apply the Mellin transform method outlined in §2.5(i), we require the transforms (f;1-z) and (h;z) to have a common strip of analyticity. This, in turn, requires -b<α0, -c<β0, and either -c<α0+1 or 1-b<β0. Following Handelsman and Lew (1970, 1971) we now give an extension of this method in which none of these conditions is required.

First, we introduce the truncated functions f1(t) and f2(t) defined by

2.5.21 f1(t) ={f(t),0<t1,0,1<t<,
2.5.22 f2(t) =f(t)-f1(t).

Similarly,

2.5.23 h1(t) ={h(t),0<t1,0,1<t<,
2.5.24 h2(t) =h(t)-h1(t).

With these definitions and the conditions (2.5.17)–(2.5.20) the Mellin transforms converge absolutely and define analytic functions in the half-planes shown in Table 2.5.1.

Table 2.5.1: Domains of convergence for Mellin transforms.
Transform Domain of Convergence
(f1;z) z>-α0
(f2;z) z<b
(h1;z) z>-c
(h2;z) z<β0

Furthermore, (f1;z) can be continued analytically to a meromorphic function on the entire z-plane, whose singularities are simple poles at -αs, s=0,1,2,, with principal part

2.5.25 as/(z+αs).

By Table 2.5.1, (f2;z) is an analytic function in the half-plane z<b. Hence we can extend the definition of the Mellin transform of f by setting

2.5.26 (f;z)=(f1;z)+(f2;z)

for z<b. The extended transform (f;z) has the same properties as (f1;z) in the half-plane z<b.

Similarly, if κ=0 in (2.5.18), then (h2;z) can be continued analytically to a meromorphic function on the entire z-plane with simple poles at βs, s=0,1,2,, with principal part

2.5.27 -bs/(z-βs).

Alternatively, if κ0 in (2.5.18), then (h2;z) can be continued analytically to an entire function.

Since (h1;z) is analytic for z>-c by Table 2.5.1, the analytically-continued (h2;z) allows us to extend the Mellin transform of h via

2.5.28 (h;z)=(h1;z)+(h2;z)

in the same half-plane. From (2.5.26) and (2.5.28), it follows that both (f;1-z) and (h;z) are defined in the half-plane z>max(1-b,-c).

We are now ready to derive the asymptotic expansion of the integral I(x) in (2.5.3) as x. First we note that

2.5.29 I(x)=j,k=12Ijk(x),

where

2.5.30 Ijk(x)=0fj(t)hk(xt)t.

By direct computation

2.5.31 I21(x)=0,
for x1.

Next from Table 2.5.1 we observe that the integrals for the transform pair (fj;1-z) and (hk;z) are absolutely convergent in the domain Djk specified in Table 2.5.2, and these domains are nonempty as a consequence of (2.5.19) and (2.5.20).

Table 2.5.2: Domains of analyticity for Mellin transforms.
Transform Pair Domain Djk
(f1;1-z), (h1;z) -c<z<1+α0
(f1;1-z), (h2;z) z<min(1+α0,β0)
(f2;1-z), (h1;z) max(-c,1-b)<z
(f2;1-z), (h2;z) 1-b<z<β0

For simplicity, write

2.5.32 Gjk(z)=(fj;1-z)(hk;z).

From Table 2.5.2, we see that each Gjk(z) is analytic in the domain Djk. Furthermore, each Gjk(z) has an analytic or meromorphic extension to a half-plane containing Djk. Now suppose that there is a real number pjk in Djk such that the Parseval formula (2.5.5) applies and

2.5.33 Ijk(x)=12πpjk-pjk+x-zGjk(z)z.

If, in addition, there exists a number qjk>pjk such that

2.5.34 suppjkxqjk|Gjk(x+y)|0,
y±,

then

2.5.35 Ijk(x)=pjk<z<qjkres[-x-zGjk(z)]+Ejk(x),

where

2.5.36 Ejk(x)=12πqjk-qjk+x-zGjk(z)z=o(x-qjk)

as x+. (The last order estimate follows from the Riemann–Lebesgue lemma, §1.8(i).) The asymptotic expansion of I(x) is then obtained from (2.5.29).

For further discussion of this method and examples, see Wong (1989, Chapter 3), Paris and Kaminski (2001, Chapter 5), and Bleistein and Handelsman (1975, Chapters 4 and 6). The first reference also contains explicit expressions for the error terms, as do Soni (1980) and Carlson and Gustafson (1985).

The Mellin transform method can also be extended to derive asymptotic expansions of multidimensional integrals having algebraic or logarithmic singularities, or both; see Wong (1989, Chapter 3), Paris and Kaminski (2001, Chapter 7), and McClure and Wong (1987). See also Brüning (1984) for a different approach.

§2.5(iii) Laplace Transforms with Small Parameters

Let h(t) satisfy (2.5.18) and (2.5.20) with c>-1, and consider the Laplace transform

2.5.37 (h;ζ)=0h(t)-ζtt.

Put x=1/ζ and break the integration range at t=1, as in (2.5.23) and (2.5.24). Then

2.5.38 ζ(h;ζ)=I1(x)+I2(x),

where

2.5.39 Ij(x)=0-thj(xt)t,
j=1,2.

Since (-t;z)=Γ(z), by the Parseval formula (2.5.5), there are real numbers p1 and p2 such that -c<p1<1, p2<min(1,β0), and

2.5.40 Ij(x)=12πpj-pj+x-zΓ(1-z)(hj;z)z,
j=1,2.

Since (h;z) is analytic for z>-c, by (2.5.14),

2.5.41 I1(x)=(h1;1)x-1+12πρ-ρ+x-zΓ(1-z)(h1;z)z,

for any ρ satisfying 1<ρ<2. Similarly, since (h2;z) can be continued analytically to a meromorphic function (when κ=0) or to an entire function (when κ0), we can choose ρ so that (h2;z) has no poles in 1<zρ<2. Thus

2.5.42 I2(x)=β0z1res[-x-zΓ(1-z)(h2;z)]+12πρ-ρ+x-zΓ(1-z)(h2;z)z.

On substituting (2.5.41) and (2.5.42) into (2.5.38), we obtain

2.5.43 (h;ζ)=(h1;1)+β0z1res[-ζz-1Γ(1-z)(h2;z)]+1<z<lres[-ζz-1Γ(1-z)(h;z)]+12πl-δ-l-δ+ζz-1Γ(1-z)(h;z)z,

where l (2) is an arbitrary integer and δ is an arbitrary small positive constant. The last term is clearly O(ζl-δ-1) as ζ0+.

If κ=0 in (2.5.18) and c>-1 in (2.5.20), and if none of the exponents in (2.5.18) are positive integers, then the expansion (2.5.43) gives the following useful result:

2.5.44 (h;ζ)n=0bnΓ(1-βn)ζβn-1+n=0(-ζ)nn!(h;n+1),
ζ0+.

Example

2.5.45 (h;ζ)=0-ζt1+tt,
ζ>0.

With h(t)=1/(1+t), we have (h;z)=πcsc(πz) for 0<z<1. In the notation of (2.5.18) and (2.5.20), κ=0, βs=s+1, and c=0. Straightforward calculation gives

2.5.46 resz=k[-ζz-1Γ(1-z)πcsc(πz)]=(-lnζ+ψ(k))ζk-1(k-1)!,

where ψ(z)=Γ(z)/Γ(z). From (2.5.28)

2.5.47 resz=1[-ζz-1Γ(1-z)(h2;z)]=(-lnζ-γ)-(h1;1),

where γ is Euler’s constant (§5.2(ii)). Insertion of these results into (2.5.43) yields

2.5.48 (h;ζ)(-lnζ)k=0ζkk!+k=0ψ(k+1)ζkk!,
ζ0+.

To verify (2.5.48) we may use

2.5.49 (h;ζ)=ζE1(ζ);

compare (6.2.2) and (6.6.3).

For examples in which the integral defining the Mellin transform (h;z) does not exist for any value of z, see Wong (1989, Chapter 3), Bleistein and Handelsman (1975, Chapter 4), and Handelsman and Lew (1970).