# §2.5(i) Introduction

Let $f(t)$ be a locally integrable function on $(0,\infty)$, that is, $\int_{\rho}^{T}f(t)dt$ exists for all $\rho$ and $T$ satisfying $0<\rho. The Mellin transform of $f(t)$ is defined by

 2.5.1 $\mathop{\mathscr{M}\/}\nolimits\left(f;z\right)=\int_{0}^{\infty}t^{z-1}f(t)dt,$

when this integral converges. The domain of analyticity of $\mathop{\mathscr{M}\/}\nolimits\left(f;z\right)$ is usually an infinite strip $a<\realpart{z} parallel to the imaginary axis. The inversion formula is given by

 2.5.2 $f(t)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}t^{-z}\mathop{\mathscr{M}\/}% \nolimits\left(f;z\right)dz,$

with $a.

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

 2.5.3 $I(x)=\int_{0}^{\infty}f(t)\,h(xt)dt,$ $x>0$,

and

 2.5.4 $\mathop{\mathscr{M}\/}\nolimits\left(I;z\right)=\mathop{\mathscr{M}\/}% \nolimits\left(f;1-z\right)\mathop{\mathscr{M}\/}\nolimits\left(h;z\right).$

If $\mathop{\mathscr{M}\/}\nolimits\left(f;1-z\right)$ and $\mathop{\mathscr{M}\/}\nolimits\left(h;z\right)$ have a common strip of analyticity $a<\realpart{z}, then

 2.5.5 $I(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}x^{-z}\mathop{\mathscr{M}\/}% \nolimits\left(f;1-z\right)\mathop{\mathscr{M}\/}\nolimits\left(h;z\right)dz,$

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

If $\mathop{\mathscr{M}\/}\nolimits\left(f;1-z\right)$ and $\mathop{\mathscr{M}\/}\nolimits\left(h;z\right)$ can be continued analytically to meromorphic functions in a left half-plane, and if the contour $\realpart{z}=c$ can be translated to $\realpart{z}=d$ with $d, then

where

 2.5.7 $E(x)=\frac{1}{2\pi i}\int_{d-i\infty}^{d+i\infty}x^{-z}\mathop{\mathscr{M}\/}% \nolimits\left(f;1-z\right)\mathop{\mathscr{M}\/}\nolimits\left(h;z\right)dz.$

The sum in (2.5.6) is taken over all poles of $x^{-z}\mathop{\mathscr{M}\/}\nolimits\left(f;1-z\right)\mathop{\mathscr{M}\/}% \nolimits\left(h;z\right)$ in the strip $d<\realpart{z}, and it provides the asymptotic expansion of $I(x)$ for small values of $x$. Similarly, if $\mathop{\mathscr{M}\/}\nolimits\left(f;1-z\right)$ and $\mathop{\mathscr{M}\/}\nolimits\left(h;z\right)$ 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 $I(x)=\int_{0}^{\infty}\frac{{\mathop{J_{\nu}\/}\nolimits^{2}}\!\left(xt\right)% }{1+t}dt,$ $\nu>-\tfrac{1}{2}$,

where $\mathop{J_{\nu}\/}\nolimits$ denotes the Bessel function (§10.2(ii)), and $x$ is a large positive parameter. Let $h(t)={\mathop{J_{\nu}\/}\nolimits^{2}}\!\left(t\right)$ and $f(t)=1/(1+t)$. Then from Table 1.14.5 and Watson (1944, p. 403)

 2.5.9 $\mathop{\mathscr{M}\/}\nolimits\left(f;1-z\right)=\frac{\pi}{\mathop{\sin\/}% \nolimits\!\left(\pi z\right)},$ $0<\realpart{z}<1$,
 2.5.10 $\mathop{\mathscr{M}\/}\nolimits\left(h;z\right)=\frac{2^{z-1}\mathop{\Gamma\/}% \nolimits\!\left(\nu+\frac{1}{2}z\right)}{{\mathop{\Gamma\/}\nolimits^{2}}\!% \left(1-\frac{1}{2}z\right)\mathop{\Gamma\/}\nolimits\!\left(1+\nu-\frac{1}{2}% z\right)\mathop{\Gamma\/}\nolimits\!\left(z\right)}\frac{\pi}{\mathop{\sin\/}% \nolimits\!\left(\pi z\right)},$ $-2\nu<\realpart{z}<1$.

In the half-plane $\realpart{z}>\max(0,-2\nu)$, the product $\mathop{\mathscr{M}\/}\nolimits\left(f;1-z\right)\mathop{\mathscr{M}\/}% \nolimits\left(h;z\right)$ has a pole of order two at each positive integer, and

 2.5.11 $\Residue_{z=n}\left[x^{-z}\mathop{\mathscr{M}\/}\nolimits\left(f;1-z\right)% \mathop{\mathscr{M}\/}\nolimits\left(h;z\right)\right]=(a_{n}\mathop{\ln\/}% \nolimits x+b_{n})x^{-n},$

where

 2.5.12 $\displaystyle a_{n}$ $\displaystyle=\frac{2^{n-1}\mathop{\Gamma\/}\nolimits\!\left(\nu+\tfrac{1}{2}n% \right)}{{\mathop{\Gamma\/}\nolimits^{2}}\!\left(1-\tfrac{1}{2}n\right)\mathop% {\Gamma\/}\nolimits\!\left(1+\nu-\tfrac{1}{2}n\right)\mathop{\Gamma\/}% \nolimits\!\left(n\right)},$ Symbols: $\mathop{\Gamma\/}\nolimits\!\left(z\right)$: gamma function and $a_{n}$: coefficients Referenced by: ¶ ‣ §2.5(i) Permalink: http://dlmf.nist.gov/2.5.E12 Encodings: TeX, pMML, png 2.5.13 $\displaystyle b_{n}$ $\displaystyle=-a_{n}\left(\mathop{\ln\/}\nolimits 2+\tfrac{1}{2}\mathop{\psi\/% }\nolimits\!\left(\nu+\tfrac{1}{2}n\right)+\mathop{\psi\/}\nolimits\!\left(1-% \tfrac{1}{2}n\right)+\tfrac{1}{2}\mathop{\psi\/}\nolimits\!\left(1+\nu-\tfrac{% 1}{2}n\right)-\mathop{\psi\/}\nolimits\!\left(n\right)\right),$

and $\mathop{\psi\/}\nolimits$ is the logarithmic derivative of the gamma function (§5.2(i)).

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

 2.5.14 $|\mathop{\Gamma\/}\nolimits\!\left(x+iy\right)|=\sqrt{2\pi}e^{-\pi|y|/2}|y|^{x% -(1/2)}\left(1+\mathop{o\/}\nolimits\!\left(1\right)\right),$

as $y\to\pm\infty$, uniformly for bounded $|x|$; see (5.11.9). Then as in (2.5.6) and (2.5.7), with $d=2n+1-\epsilon$ $(0<\epsilon<1)$, we obtain

 2.5.15 $I(x)=-\sum_{s=0}^{2n}(a_{s}\mathop{\ln\/}\nolimits x+b_{s})x^{-s}+\mathop{O\/}% \nolimits\!\left(x^{-2n-1+\epsilon}\right),$ $n=0,1,2,\dots$.

From (2.5.12) and (2.5.13), it is seen that $a_{s}=b_{s}=0$ when $s$ is even. Hence

 2.5.16 $I(x)=\sum_{s=0}^{n-1}(c_{s}\mathop{\ln\/}\nolimits x+d_{s})x^{-2s-1}+\mathop{O% \/}\nolimits\!\left(x^{-2n-1+\epsilon}\right),$

where $c_{s}=-a_{2s+1}$, $d_{s}=-b_{2s+1}$.

# §2.5(ii) Extensions

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

 2.5.17 $f(t)\sim\sum_{s=0}^{\infty}a_{s}t^{\alpha_{s}},$ $t\to 0+$, Symbols: $\sim$: asymptotic equality, $f(x)$: locally integrable function and $a_{s}$: coefficients Referenced by: §2.5(ii) Permalink: http://dlmf.nist.gov/2.5.E17 Encodings: TeX, pMML, png

where $\realpart{\alpha_{s}}>\realpart{\alpha_{s^{\prime}}}$ for $s>s^{\prime}$, and $\realpart{\alpha_{s}}\to+\infty$ as $s\to\infty$. Also, let

 2.5.18 $h(t)\sim\mathop{\exp\/}\nolimits\!\left(i\kappa t^{p}\right)\sum_{s=0}^{\infty% }b_{s}t^{-\beta_{s}},$ $t\to+\infty$,

where $\kappa$ is real, $p>0$, $\realpart{\beta_{s}}>\realpart{\beta_{s^{\prime}}}$ for $s>s^{\prime}$, and $\realpart{\beta_{s}}\to+\infty$ as $s\to\infty$. To ensure that the integral (2.5.3) converges we assume that

 2.5.19 $f(t)=\mathop{O\/}\nolimits\!\left(t^{-b}\right),$ $t\to+\infty$,

with $b+\realpart{\beta_{0}}>1$, and

 2.5.20 $h(t)=\mathop{O\/}\nolimits\!\left(t^{c}\right),$ $t\to 0+$,

with $c+\realpart{\alpha_{0}}>-1$. To apply the Mellin transform method outlined in §2.5(i), we require the transforms $\mathop{\mathscr{M}\/}\nolimits\left(f;1-z\right)$ and $\mathop{\mathscr{M}\/}\nolimits\left(h;z\right)$ to have a common strip of analyticity. This, in turn, requires $-b<\realpart{\alpha_{0}}$, $-c<\realpart{\beta_{0}}$, and either $-c<\realpart{\alpha_{0}}+1$ or $1-b<\realpart{\beta_{0}}$. 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 $f_{1}(t)$ and $f_{2}(t)$ defined by

 2.5.21 $\displaystyle f_{1}(t)$ $\displaystyle=\begin{cases}f(t),&0 Symbols: $f(x)$: locally integrable function and $f_{j}(t)$: truncated functions Permalink: http://dlmf.nist.gov/2.5.E21 Encodings: TeX, pMML, png 2.5.22 $\displaystyle f_{2}(t)$ $\displaystyle=f(t)-f_{1}(t).$ Symbols: $f(x)$: locally integrable function and $f_{j}(t)$: truncated functions Permalink: http://dlmf.nist.gov/2.5.E22 Encodings: TeX, pMML, png

Similarly,

 2.5.23 $\displaystyle h_{1}(t)$ $\displaystyle=\begin{cases}h(t),&0 Symbols: $h(x)$: locally integrable function Referenced by: §2.5(iii) Permalink: http://dlmf.nist.gov/2.5.E23 Encodings: TeX, pMML, png 2.5.24 $\displaystyle h_{2}(t)$ $\displaystyle=h(t)-h_{1}(t).$ Symbols: $h(x)$: locally integrable function Referenced by: §2.5(iii) Permalink: http://dlmf.nist.gov/2.5.E24 Encodings: TeX, pMML, png

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.

Furthermore, $\mathop{\mathscr{M}\/}\nolimits\left(f_{1};z\right)$ can be continued analytically to a meromorphic function on the entire $z$-plane, whose singularities are simple poles at $-\alpha_{s}$, $s=0,1,2,\dots$, with principal part

 2.5.25 $a_{s}/\left(z+\alpha_{s}\right).$ Symbols: $a_{s}$: coefficients Permalink: http://dlmf.nist.gov/2.5.E25 Encodings: TeX, pMML, png

By Table 2.5.1, $\mathop{\mathscr{M}\/}\nolimits\left(f_{2};z\right)$ is an analytic function in the half-plane $\realpart{z}. Hence we can extend the definition of the Mellin transform of $f$ by setting

 2.5.26 $\mathop{\mathscr{M}\/}\nolimits\left(f;z\right)=\mathop{\mathscr{M}\/}% \nolimits\left(f_{1};z\right)+\mathop{\mathscr{M}\/}\nolimits\left(f_{2};z\right)$

for $\realpart{z}. The extended transform $\mathop{\mathscr{M}\/}\nolimits\left(f;z\right)$ has the same properties as $\mathop{\mathscr{M}\/}\nolimits\left(f_{1};z\right)$ in the half-plane $\realpart{z}.

Similarly, if $\kappa=0$ in (2.5.18), then $\mathop{\mathscr{M}\/}\nolimits\left(h_{2};z\right)$ can be continued analytically to a meromorphic function on the entire $z$-plane with simple poles at $\beta_{s}$, $s=0,1,2,\dots$, with principal part

 2.5.27 $-b_{s}/\left(z-\beta_{s}\right).$ Symbols: $b$: right endpoint Permalink: http://dlmf.nist.gov/2.5.E27 Encodings: TeX, pMML, png

Alternatively, if $\kappa\neq 0$ in (2.5.18), then $\mathop{\mathscr{M}\/}\nolimits\left(h_{2};z\right)$ can be continued analytically to an entire function.

Since $\mathop{\mathscr{M}\/}\nolimits\left(h_{1};z\right)$ is analytic for $\realpart{z}>-c$ by Table 2.5.1, the analytically-continued $\mathop{\mathscr{M}\/}\nolimits\left(h_{2};z\right)$ allows us to extend the Mellin transform of $h$ via

 2.5.28 $\mathop{\mathscr{M}\/}\nolimits\left(h;z\right)=\mathop{\mathscr{M}\/}% \nolimits\left(h_{1};z\right)+\mathop{\mathscr{M}\/}\nolimits\left(h_{2};z\right)$ Symbols: $\mathop{\mathscr{M}\/}\nolimits\left(f;s\right)$: Mellin transform and $h(x)$: locally integrable function Referenced by: ¶ ‣ §2.5(iii), §2.5(ii) Permalink: http://dlmf.nist.gov/2.5.E28 Encodings: TeX, pMML, png

in the same half-plane. From (2.5.26) and (2.5.28), it follows that both $\mathop{\mathscr{M}\/}\nolimits\left(f;1-z\right)$ and $\mathop{\mathscr{M}\/}\nolimits\left(h;z\right)$ are defined in the half-plane $\realpart{z}>\max(1-b,-c)$.

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

 2.5.29 $I(x)=\sum\limits_{j,k=1}^{2}I_{jk}(x),$ Symbols: $I(x)$: convolution integral Referenced by: §2.5(ii) Permalink: http://dlmf.nist.gov/2.5.E29 Encodings: TeX, pMML, png

where

 2.5.30 $I_{jk}(x)=\int_{0}^{\infty}f_{j}(t)h_{k}(xt)dt.$

By direct computation

 2.5.31 $I_{21}(x)=0,$ for $x\geq 1$. Symbols: $I(x)$: convolution integral Permalink: http://dlmf.nist.gov/2.5.E31 Encodings: TeX, pMML, png

Next from Table 2.5.1 we observe that the integrals for the transform pair $\mathop{\mathscr{M}\/}\nolimits\left(f_{j};1-z\right)$ and $\mathop{\mathscr{M}\/}\nolimits\left(h_{k};z\right)$ are absolutely convergent in the domain $D_{jk}$ specified in Table 2.5.2, and these domains are nonempty as a consequence of (2.5.19) and (2.5.20).

For simplicity, write

 2.5.32 $G_{jk}(z)=\mathop{\mathscr{M}\/}\nolimits\left(f_{j};1-z\right)\mathop{% \mathscr{M}\/}\nolimits\left(h_{k};z\right).$

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

 2.5.33 $I_{jk}(x)=\frac{1}{2\pi i}\int_{p_{jk}-i\infty}^{p_{jk}+i\infty}x^{-z}G_{jk}(z% )dz.$

If, in addition, there exists a number $q_{jk}>p_{jk}$ such that

 2.5.34 $\sup_{p_{jk}\leq x\leq q_{jk}}\left|G_{jk}(x+iy)\right|\to 0,$ $y\to\pm\infty$,

then

 2.5.35 $I_{jk}(x)=\sum_{p_{jk}<\realpart{z}

where

 2.5.36 $E_{jk}(x)=\frac{1}{2\pi i}\int_{q_{jk}-i\infty}^{q_{jk}+i\infty}x^{-z}G_{jk}(z% )dz=\mathop{o\/}\nolimits\!\left(x^{-q_{jk}}\right)$

as $x\to+\infty$. (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 $\mathop{\mathscr{L}\/}\nolimits\left(h;\zeta\right)=\int_{0}^{\infty}h(t)e^{-% \zeta t}dt.$

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

 2.5.38 $\zeta\mathop{\mathscr{L}\/}\nolimits\left(h;\zeta\right)=I_{1}(x)+I_{2}(x),$

where

 2.5.39 $I_{j}(x)=\int_{0}^{\infty}e^{-t}h_{j}(xt)dt,$ $j=1,2$.

Since $\mathop{\mathscr{M}\/}\nolimits\left(e^{-t};z\right)=\mathop{\Gamma\/}% \nolimits\!\left(z\right)$, by the Parseval formula (2.5.5), there are real numbers $p_{1}$ and $p_{2}$ such that $-c, $p_{2}<\min(1,\realpart{\beta_{0}})$, and

 2.5.40 $I_{j}(x)=\frac{1}{2\pi i}\int_{p_{j}-i\infty}^{p_{j}+i\infty}x^{-z}\mathop{% \Gamma\/}\nolimits\!\left(1-z\right)\mathop{\mathscr{M}\/}\nolimits\left(h_{j}% ;z\right)dz,$ $j=1,2$.

Since $\mathop{\mathscr{M}\/}\nolimits\left(h;z\right)$ is analytic for $\realpart{z}>-c$, by (2.5.14),

 2.5.41 $I_{1}(x)=\mathop{\mathscr{M}\/}\nolimits\left(h_{1};1\right)x^{-1}+\frac{1}{2% \pi i}\int_{\rho-i\infty}^{\rho+i\infty}x^{-z}\mathop{\Gamma\/}\nolimits\!% \left(1-z\right)\mathop{\mathscr{M}\/}\nolimits\left(h_{1};z\right)dz,$

for any $\rho$ satisfying $1<\rho<2$. Similarly, since $\mathop{\mathscr{M}\/}\nolimits\left(h_{2};z\right)$ can be continued analytically to a meromorphic function (when $\kappa=0$) or to an entire function (when $\kappa\neq 0$), we can choose $\rho$ so that $\mathop{\mathscr{M}\/}\nolimits\left(h_{2};z\right)$ has no poles in $1<\realpart{z}\leq\rho<2$. Thus

 2.5.42 $I_{2}(x)=\sum_{\realpart{\beta_{0}}\leq\realpart{z}\leq 1}\Residue\left[-x^{-z% }\mathop{\Gamma\/}\nolimits\!\left(1-z\right)\mathop{\mathscr{M}\/}\nolimits% \left(h_{2};z\right)\right]+\frac{1}{2\pi i}\int_{\rho-i\infty}^{\rho+i\infty}% x^{-z}\mathop{\Gamma\/}\nolimits\!\left(1-z\right)\mathop{\mathscr{M}\/}% \nolimits\left(h_{2};z\right)dz.$

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

 2.5.43 $\mathop{\mathscr{L}\/}\nolimits\left(h;\zeta\right)=\mathop{\mathscr{M}\/}% \nolimits\left(h_{1};1\right)+\sum_{\realpart{\beta_{0}}\leq\realpart{z}\leq 1% }\Residue\left[-\zeta^{z-1}\mathop{\Gamma\/}\nolimits\!\left(1-z\right)\mathop% {\mathscr{M}\/}\nolimits\left(h_{2};z\right)\right]+\sum\limits_{1<\realpart{z% }

where $l$ ($\geq 2$) is an arbitrary integer and $\delta$ is an arbitrary small positive constant. The last term is clearly $\mathop{O\/}\nolimits\!\left(\zeta^{l-\delta-1}\right)$ as $\zeta\to 0+$.

If $\kappa=0$ 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 $\mathop{\mathscr{L}\/}\nolimits\left(h;\zeta\right)\sim\sum_{n=0}^{\infty}b_{n% }\mathop{\Gamma\/}\nolimits\!\left(1-\beta_{n}\right)\zeta^{\beta_{n}-1}+\sum% \limits_{n=0}^{\infty}\frac{(-\zeta)^{n}}{n!}\mathop{\mathscr{M}\/}\nolimits% \left(h;n+1\right),$ $\zeta\to 0+$.

# ¶ Example

 2.5.45 $\mathop{\mathscr{L}\/}\nolimits\left(h;\zeta\right)=\int_{0}^{\infty}\frac{e^{% -\zeta t}}{1+t}dt,$ $\realpart{\zeta}>0$.

With $h(t)=1/(1+t)$, we have $\mathop{\mathscr{M}\/}\nolimits\left(h;z\right)=\pi\mathop{\csc\/}\nolimits\!% \left(\pi z\right)$ for $0<\realpart{z}<1$. In the notation of (2.5.18) and (2.5.20), $\kappa=0$, $\beta_{s}=s+1$, and $c=0$. Straightforward calculation gives

 2.5.46 $\Residue_{z=k}\left[-\zeta^{z-1}\mathop{\Gamma\/}\nolimits\!\left(1-z\right)% \pi\mathop{\csc\/}\nolimits\!\left(\pi z\right)\right]=\left(-\mathop{\ln\/}% \nolimits\zeta+\mathop{\psi\/}\nolimits\!\left(k\right)\right)\dfrac{\zeta^{k-% 1}}{(k-1)!},$

where $\mathop{\psi\/}\nolimits\!\left(z\right)={\mathop{\Gamma\/}\nolimits^{\prime}}% \!\left(z\right)/\mathop{\Gamma\/}\nolimits\!\left(z\right)$. From (2.5.28)

 2.5.47 $\Residue_{z=1}\left[-\zeta^{z-1}\mathop{\Gamma\/}\nolimits\!\left(1-z\right)% \mathop{\mathscr{M}\/}\nolimits\left(h_{2};z\right)\right]=\left(-\mathop{\ln% \/}\nolimits\zeta-\EulerConstant\right)-\mathop{\mathscr{M}\/}\nolimits\left(h% _{1};1\right),$

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

 2.5.48 $\mathop{\mathscr{L}\/}\nolimits\left(h;\zeta\right)\sim(-\mathop{\ln\/}% \nolimits\zeta)\sum_{k=0}^{\infty}\frac{\zeta^{k}}{k!}+\sum_{k=0}^{\infty}% \mathop{\psi\/}\nolimits\!\left(k+1\right)\frac{\zeta^{k}}{k!},$ $\zeta\to 0+$.

To verify (2.5.48) we may use

 2.5.49 $\mathop{\mathscr{L}\/}\nolimits\left(h;\zeta\right)=e^{\zeta}\mathop{E_{1}\/}% \nolimits\!\left(\zeta\right);$

compare (6.2.2) and (6.6.3).

For examples in which the integral defining the Mellin transform $\mathop{\mathscr{M}\/}\nolimits\left(h;z\right)$ does not exist for any value of $z$, see Wong (1989, Chapter 3), Bleistein and Handelsman (1975, Chapter 4), and Handelsman and Lew (1970).