# §1.14 Integral Transforms

## §1.14(i) Fourier Transform

The Fourier transform of a real- or complex-valued function $f(t)$ is defined by

 1.14.1 $\mathscr{F}\left(f\right)\left(x\right)=\mathscr{F}\mskip-3.0mu f\mskip 3.0mu % \left(x\right)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}f(t)e^{ixt}\mathrm{% d}t.$ ⓘ Defines: $\mathscr{F}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Fourier transform Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$, $\mathrm{e}$: base of exponential function and $\int$: integral Referenced by: §1.14(v), §1.14(i), (i), §1.17(ii), (i) Permalink: http://dlmf.nist.gov/1.14.E1 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): This equation defines $\mathscr{F}\left(f\right)\left(x\right)$ or $\mathscr{F}\mskip-3.0mu f\mskip 3.0mu \left(x\right)$ as the Fourier transform of functions of a single variable. An analogous notation is defined for the Fourier transform of tempered distributions in (1.16.29) and the Fourier transform of special distributions in (1.16.38). Reported 2017-03-07 See also: Annotations for 1.14(i), 1.14 and 1

(Some references replace $ixt$ by $-ixt$). The same notation $\mathscr{F}$ is used for Fourier transforms of functions of several variables and for Fourier transforms of distributions; see §1.16(vii).

In this subsection we let $F(x)=\mathscr{F}\left(f\right)\left(x\right)$.

If $f(t)$ is absolutely integrable on $(-\infty,\infty)$, then $F(x)$ is continuous, $F(x)\to 0$ as $x\to\pm\infty$, and

 1.14.2 $|F(x)|\leq\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}|f(t)|\mathrm{d}t.$

### Inversion

Suppose that $f(t)$ is absolutely integrable on $(-\infty,\infty)$ and of bounded variation in a neighborhood of $t=u$1.4(v)). Then

 1.14.3 $\tfrac{1}{2}(f(u+)+f(u-))=\frac{1}{\sqrt{2\pi}}\pvint^{\infty}_{-\infty}F(x)e^% {-ixu}\mathrm{d}x,$

where the last integral denotes the Cauchy principal value (1.4.25).

In many applications $f(t)$ is absolutely integrable and $f^{\prime}(t)$ is continuous on $(-\infty,\infty)$. Then

 1.14.4 $f(t)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}F(x)e^{-ixt}\mathrm{d}x.$

### Convolution

For Fourier transforms, the convolution $(f*g)(t)$ of two functions $f(t)$ and $g(t)$ defined on $(-\infty,\infty)$ is given by

 1.14.5 $(f*g)(t)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}f(t-s)g(s)\mathrm{d}s.$ ⓘ Defines: $*$: convolution (Fourier) (locally) Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$ and $\int$: integral Permalink: http://dlmf.nist.gov/1.14.E5 Encodings: TeX, pMML, png See also: Annotations for 1.14(i), 1.14(i), 1.14 and 1

If $f(t)$ and $g(t)$ are absolutely integrable on $(-\infty,\infty)$, then so is $(f*g)(t)$, and its Fourier transform is $F(x)G(x)$, where $G(x)$ is the Fourier transform of $g(t)$.

### Parseval’s Formula

Suppose $f(t)$ and $g(t)$ are absolutely integrable on $(-\infty,\infty)$, then

 1.14.6 $(f*g)(t)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}F(x)G(x)e^{-itx}\mathrm{d% }x,$
 1.14.7 $\int^{\infty}_{-\infty}F(x)G(x)\mathrm{d}x=\int^{\infty}_{-\infty}f(t)g(-t)% \mathrm{d}t,$
 1.14.8 $\int^{\infty}_{-\infty}|F(x)|^{2}\mathrm{d}x=\int^{\infty}_{-\infty}|f(t)|^{2}% \mathrm{d}t.$ ⓘ Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $F(x)$: Fourier transform of $f(t)$ Referenced by: §1.14(i) Permalink: http://dlmf.nist.gov/1.14.E8 Encodings: TeX, pMML, png See also: Annotations for 1.14(i), 1.14(i), 1.14 and 1

(1.14.8) is Parseval’s formula.

### Uniqueness

If $f(t)$ and $g(t)$ are continuous and absolutely integrable on $(-\infty,\infty)$, and $F(x)=G(x)$ for all $x$, then $f(t)=g(t)$ for all $t$.

## §1.14(ii) Fourier Cosine and Sine Transforms

The Fourier cosine transform and Fourier sine transform are defined respectively by

 1.14.9 $\displaystyle\mathscr{F}_{\mskip-3.0mu c}\left(f\right)\left(x\right)$ $\displaystyle=\mathscr{F}_{\mskip-3.0mu c}\mskip-1.0mu f\mskip 3.0mu \left(x% \right)=\sqrt{\frac{2}{\pi}}\int^{\infty}_{0}f(t)\cos\left(xt\right)\mathrm{d}t,$ ⓘ Defines: $\mathscr{F}_{\mskip-3.0mu c}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Fourier cosine transform Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $\cos\NVar{z}$: cosine function, $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $F_{c}(x)$: cosine transformation of $f(t)$ Permalink: http://dlmf.nist.gov/1.14.E9 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): This equation defines the general notation $\mathscr{F}_{\mskip-3.0mu c}\left(f\right)\left(x\right)$ or $\mathscr{F}_{\mskip-3.0mu c}\mskip-1.0mu f\mskip 3.0mu \left(x\right)$. Previously it defined the local notation $F_{c}(x)$ used in this subsection. Reported 2017-03-07 See also: Annotations for 1.14(ii), 1.14 and 1 1.14.10 $\displaystyle\mathscr{F}_{\mskip-2.0mu s}\left(f\right)\left(x\right)$ $\displaystyle=\mathscr{F}_{\mskip-2.0mu s}\mskip-1.0mu f\mskip 3.0mu \left(x% \right)=\sqrt{\frac{2}{\pi}}\int^{\infty}_{0}f(t)\sin\left(xt\right)\mathrm{d}t.$ ⓘ Defines: $\mathscr{F}_{\mskip-2.0mu s}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Fourier sine transform Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral, $\sin\NVar{z}$: sine function and $F_{s}(x)$: sine transformation of $f(t)$ Permalink: http://dlmf.nist.gov/1.14.E10 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): This equation defines the general notation $\mathscr{F}_{\mskip-2.0mu s}\left(f\right)\left(x\right)$ or $\mathscr{F}_{\mskip-2.0mu s}\mskip-1.0mu f\mskip 3.0mu \left(x\right)$. Previously it defined the local notation $F_{s}(x)$ used in this subsection. Reported 2017-03-07 See also: Annotations for 1.14(ii), 1.14 and 1

In this subsection we let $F_{c}(x)=\mathscr{F}_{\mskip-3.0mu c}\mskip-1.0mu f\mskip 3.0mu \left(x\right)$, $F_{s}(x)=\mathscr{F}_{\mskip-2.0mu s}\mskip-1.0mu f\mskip 3.0mu \left(x\right)$, $G_{c}(x)=\mathscr{F}_{\mskip-3.0mu c}\mskip-1.0mu g\mskip 3.0mu \left(x\right)$, and $G_{s}(x)=\mathscr{F}_{\mskip-2.0mu s}\mskip-1.0mu g\mskip 3.0mu \left(x\right)$.

### Inversion

If $f(t)$ is absolutely integrable on $[0,\infty)$ and of bounded variation (§1.4(v)) in a neighborhood of $t=u$, then

 1.14.11 $\displaystyle\tfrac{1}{2}(f(u+)+f(u-))$ $\displaystyle=\sqrt{\frac{2}{\pi}}\int^{\infty}_{0}F_{c}(x)\cos\left(ux\right)% \mathrm{d}x,$ 1.14.12 $\displaystyle\tfrac{1}{2}(f(u+)+f(u-))$ $\displaystyle=\sqrt{\frac{2}{\pi}}\int^{\infty}_{0}F_{s}(x)\sin\left(ux\right)% \mathrm{d}x.$

### Parseval’s Formula

If $\int^{\infty}_{0}|f(t)|\mathrm{d}t<\infty$, $g(t)$ is of bounded variation on $(0,\infty)$ and $g(t)\to 0$ as $t\to\infty$, then

 1.14.13 $\displaystyle\int^{\infty}_{0}F_{c}(x)G_{c}(x)\mathrm{d}x$ $\displaystyle=\int^{\infty}_{0}f(t)g(t)\mathrm{d}t,$ 1.14.14 $\displaystyle\int^{\infty}_{0}F_{s}(x)G_{s}(x)\mathrm{d}x$ $\displaystyle=\int^{\infty}_{0}f(t)g(t)\mathrm{d}t,$
 1.14.15 $\displaystyle\int^{\infty}_{0}(F_{c}(x))^{2}\mathrm{d}x$ $\displaystyle=\int^{\infty}_{0}(f(t))^{2}\mathrm{d}t,$ ⓘ Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $F_{c}(x)$: cosine transformation of $f(t)$ Permalink: http://dlmf.nist.gov/1.14.E15 Encodings: TeX, pMML, png See also: Annotations for 1.14(ii), 1.14(ii), 1.14 and 1 1.14.16 $\displaystyle\int^{\infty}_{0}(F_{s}(x))^{2}\mathrm{d}x$ $\displaystyle=\int^{\infty}_{0}(f(t))^{2}\mathrm{d}t.$ ⓘ Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $F_{s}(x)$: sine transformation of $f(t)$ Permalink: http://dlmf.nist.gov/1.14.E16 Encodings: TeX, pMML, png See also: Annotations for 1.14(ii), 1.14(ii), 1.14 and 1

## §1.14(iii) Laplace Transform

Suppose $f(t)$ is a real- or complex-valued function and $s$ is a real or complex parameter. The Laplace transform of $f$ is defined by

 1.14.17 $\mathscr{L}\left(f\right)\left(s\right)=\mathscr{L}\mskip-3.0mu f\mskip 3.0mu % \left(s\right)=\int^{\infty}_{0}e^{-st}f(t)\mathrm{d}t.$ ⓘ Defines: $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\mathrm{e}$: base of exponential function and $\int$: integral A&S Ref: 29.1.1 Permalink: http://dlmf.nist.gov/1.14.E17 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Laplace transform was changed to $\mathscr{L}\left(f\right)\left(s\right)$ or $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{L}(f(t);s)$. Reported 2017-03-07 See also: Annotations for 1.14(iii), 1.14 and 1

### Convergence and Analyticity

Assume that $f(t)$ is piecewise continuous on $[0,\infty)$ and of exponential growth, that is, constants $M$ and $\alpha$ exist such that

 1.14.18 $|f(t)|\leq Me^{\alpha t},$ $0\leq t<\infty$. ⓘ Symbols: $\mathrm{e}$: base of exponential function, $M$: constant and $\alpha$: constant Referenced by: §1.14(iii), §1.14(iii), §1.14(iii) Permalink: http://dlmf.nist.gov/1.14.E18 Encodings: TeX, pMML, png See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

Then $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ is an analytic function of $s$ for $\Re s>\alpha$. Moreover,

 1.14.19 $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)\to 0,$ $\Re s\to\infty$. ⓘ Symbols: $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform and $\Re$: real part Permalink: http://dlmf.nist.gov/1.14.E19 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Laplace transform was changed to $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{L}(f(t);s)$. Reported 2017-03-07 See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

Throughout the remainder of this subsection we assume (1.14.18) is satisfied and $\Re s>\alpha$.

### Inversion

If $f(t)$ is continuous and $f^{\prime}(t)$ is piecewise continuous on $[0,\infty)$, then

 1.14.20 $f(t)=\frac{1}{2\pi i}\lim_{T\to\infty}\int^{\sigma+iT}_{\sigma-iT}e^{ts}% \mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)\mathrm{d}s,$ $\sigma>\alpha$. ⓘ Symbols: $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$, $\mathrm{e}$: base of exponential function, $\int$: integral, $\sigma\in(a,b)$: parameter and $\alpha$: constant A&S Ref: 29.2.2 Referenced by: §1.14(iii) Permalink: http://dlmf.nist.gov/1.14.E20 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Laplace transform was changed to $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{L}(f(t);s)$. Reported 2017-03-07 See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

Moreover, if $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)=O\left(s^{-K}\right)$ in some half-plane $\Re s\geq\gamma$ and $K>1$, then (1.14.20) holds for $\sigma>\gamma$.

### Translation

If $\Re s>\max(\Re(a+\alpha),\alpha)$, then

 1.14.21 $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s-a\right)=\mathscr{L}\mskip-3.0mu% f_{a}\mskip 3.0mu \left(s\right),$ ⓘ Symbols: $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform and $\mathrm{e}$: base of exponential function Permalink: http://dlmf.nist.gov/1.14.E21 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Laplace transform was changed to $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{L}(f(t);s)$. In addition, the auxiliary function $f_{a}(t)=e^{at}f(t)$ was introduced. Reported 2017-03-07 See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

where $f_{a}(t)=e^{at}f(t)$. Also, if $a\geq 0$ then

 1.14.22 $\mathscr{L}\mskip-3.0mu f_{a}^{+}\mskip 3.0mu \left(s\right)=e^{-as}\mathscr{L% }\mskip-3.0mu f\mskip 3.0mu \left(s\right),$ ⓘ Symbols: $H\left(\NVar{x}\right)$: Heaviside function, $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform and $\mathrm{e}$: base of exponential function Permalink: http://dlmf.nist.gov/1.14.E22 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Laplace transform was changed to $\mathscr{L}\left(f\right)\left(s\right)$ from $\mathscr{L}(f(t);s)$. In addition, the auxiliary function $f^{+}_{a}(t)=H\left(t-a\right)f(t-a)$ was introduced. Reported 2017-03-07 See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

where $f^{+}_{a}(t)=H\left(t-a\right)f(t-a)$ and $H$ is the Heaviside function; see (1.16.13).

### Differentiation and Integration

If $f(t)$ is piecewise continuous, then

 1.14.23 ${(-1)}^{n}\frac{{\mathrm{d}}^{n}}{{\mathrm{d}s}^{n}}\mathscr{L}\mskip-3.0mu f% \mskip 3.0mu \left(s\right)=\mathscr{L}\mskip-3.0mu f_{n}\mskip 3.0mu \left(s% \right),$ $n=1,2,3,\dots$, ⓘ Symbols: $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform, $\frac{\mathrm{d}\NVar{f}}{\mathrm{d}\NVar{x}}$: derivative of $f$ with respect to $x$ and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.14.E23 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Laplace transform was changed to $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{L}(f(t);s)$. In addition, the auxiliary function $f_{n}(t)=t^{n}f(t)$ was introduced. Reported 2017-03-07 See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

where $f_{n}(t)=t^{n}f(t)$. If also $\lim_{t\to 0+}f(t)/t$ exists, then

 1.14.24 $\int^{\infty}_{s}\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(u\right)\mathrm{d% }u=\mathscr{L}\mskip-3.0mu f_{-1}\mskip 3.0mu \left(s\right),$ ⓘ Symbols: $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform, $\mathrm{d}\NVar{x}$: differential of $x$ and $\int$: integral Permalink: http://dlmf.nist.gov/1.14.E24 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Laplace transform was changed to $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{L}(f(t);s)$. In addition, the auxiliary function $f_{-1}(t)=\frac{f(t)}{t}$ was introduced. Reported 2017-03-07 See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

where $f_{-1}(t)=\frac{f(t)}{t}$.

### Periodic Functions

If $a>0$ and $f(t+a)=f(t)$ for $t>0$, then

 1.14.25 $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)=\frac{1}{1-e^{-as}}\int^{% a}_{0}e^{-st}f(t)\mathrm{d}t.$ ⓘ Symbols: $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform, $\mathrm{d}\NVar{x}$: differential of $x$, $\mathrm{e}$: base of exponential function and $\int$: integral Permalink: http://dlmf.nist.gov/1.14.E25 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Laplace transform was changed to $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{L}(f(t);s)$. Reported 2017-03-07 See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

Alternatively if $f(t+a)=-f(t)$ for $t>0$, then

 1.14.26 $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)=\frac{1}{1+e^{-as}}\int^{% a}_{0}e^{-st}f(t)\mathrm{d}t.$ ⓘ Symbols: $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform, $\mathrm{d}\NVar{x}$: differential of $x$, $\mathrm{e}$: base of exponential function and $\int$: integral Permalink: http://dlmf.nist.gov/1.14.E26 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Laplace transform was changed to $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{L}(f(t);s)$. Reported 2017-03-07 See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

### Derivatives

If $f(t)$ is continuous on $[0,\infty)$ and $f^{\prime}(t)$ is piecewise continuous on $(0,\infty)$, then

 1.14.27 $\mathscr{L}\left(f^{\prime}\right)\left(s\right)=s\mathscr{L}\left(f\right)% \left(s\right)-f(0+).$ ⓘ Symbols: $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform A&S Ref: 29.2.4 Permalink: http://dlmf.nist.gov/1.14.E27 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Laplace transform was changed to $\mathscr{L}\left(f^{\prime}\right)\left(s\right)$ and $\mathscr{L}\left(f\right)\left(s\right)$ from $\mathscr{L}(f^{\prime}(t);s)$ and $\mathscr{L}(f(t);s)$. Reported 2017-03-07 See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

If $f(t)$ and $f^{\prime}(t)$ are piecewise continuous on $[0,\infty)$ with discontinuities at ($0=$) $t_{0}, then

 1.14.28 $\mathscr{L}\left(f^{\prime}\right)\left(s\right)=s\mathscr{L}\left(f\right)% \left(s\right)-f(0+)-\sum^{n}_{k=1}e^{-st_{k}}(f(t_{k}+)-f(t_{k}-)).$ ⓘ Symbols: $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform, $\mathrm{e}$: base of exponential function, $k$: integer and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.14.E28 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Laplace transforms was changed to $\mathscr{L}\left(f^{\prime}\right)\left(s\right)$ and $\mathscr{L}\left(f\right)\left(s\right)$ from $\mathscr{L}(f^{\prime}(t);s)$ and $\mathscr{L}(f(t);s)$. Reported 2017-03-07 See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

Next, assume $f(t)$, $f^{\prime}(t)$, $\dots$, $f^{(n-1)}(t)$ are continuous and each satisfies (1.14.18). Also assume that $f^{(n)}(t)$ is piecewise continuous on $[0,\infty)$. Then

 1.14.29 $\mathscr{L}\left(f^{(n)}\right)\left(s\right)=s^{n}\mathscr{L}\left(f\right)% \left(s\right)-s^{n-1}f(0+)-s^{n-2}f^{\prime}(0+)-\dots-f^{(n-1)}(0+).$ ⓘ Symbols: $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform and $n$: nonnegative integer A&S Ref: 29.2.5 Permalink: http://dlmf.nist.gov/1.14.E29 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Laplace transforms was changed to $\mathscr{L}\left(f^{(n)}\right)\left(s\right)$ and $\mathscr{L}\left(f\right)\left(s\right)$ from $\mathscr{L}(f^{(n)}(t);s)$ and $\mathscr{L}(f(t);s)$. Reported 2017-03-07 See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

### Convolution

For Laplace transforms, the convolution of two functions $f(t)$ and $g(t)$, defined on $[0,\infty)$, is

 1.14.30 $(f*g)(t)=\int^{t}_{0}f(u)g(t-u)\mathrm{d}u.$ ⓘ Defines: $*$: convolution (Laplace) (locally) Symbols: $\mathrm{d}\NVar{x}$: differential of $x$ and $\int$: integral Permalink: http://dlmf.nist.gov/1.14.E30 Encodings: TeX, pMML, png See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

If $f(t)$ and $g(t)$ are piecewise continuous, then

 1.14.31 $\mathscr{L}\left(f*g\right)=\mathscr{L}\left(f\right)\mathscr{L}\left(g\right).$ ⓘ Symbols: $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform and $*$: convolution (Laplace) Permalink: http://dlmf.nist.gov/1.14.E31 Encodings: TeX, pMML, png See also: Annotations for 1.14(iii), 1.14(iii), 1.14 and 1

### Uniqueness

If $f(t)$ and $g(t)$ are continuous and $\mathscr{L}\mskip-3.0mu f\mskip 3.0mu \left(s\right)=\mathscr{L}\mskip-3.0mu g% \mskip 3.0mu \left(s\right)$, then $f(t)=g(t)$.

## §1.14(iv) Mellin Transform

The Mellin transform of a real- or complex-valued function $f(x)$ is defined by

 1.14.32 $\mathscr{M}\left(f\right)\left(s\right)=\mathscr{M}\mskip-3.0mu f\mskip 3.0mu % \left(s\right)=\int^{\infty}_{0}x^{s-1}f(x)\mathrm{d}x.$ ⓘ Defines: $\mathscr{M}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Mellin transform Symbols: $\mathrm{d}\NVar{x}$: differential of $x$ and $\int$: integral Referenced by: §1.14(iv), §1.14(iv) Permalink: http://dlmf.nist.gov/1.14.E32 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Mellin transform was changed to $\mathscr{M}\left(f\right)\left(s\right)$ or $\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{M}(f;s)$. Reported 2017-03-07 See also: Annotations for 1.14(iv), 1.14 and 1

If $x^{\sigma-1}f(x)$ is integrable on $(0,\infty)$ for all $\sigma$ in $a<\sigma, then the integral (1.14.32) converges and $\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ is an analytic function of $s$ in the vertical strip $a<\Re s. Moreover, for $a<\sigma,

 1.14.33 $\lim_{t\to\pm\infty}\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(\sigma+it% \right)=0.$ ⓘ Symbols: $\mathscr{M}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Mellin transform and $\sigma\in(a,b)$: parameter Permalink: http://dlmf.nist.gov/1.14.E33 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Mellin transform was changed to $\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{M}(f;s)$. Reported 2017-03-07 See also: Annotations for 1.14(iv), 1.14 and 1

Note: If $f(x)$ is continuous and $\alpha$ and $\beta$ are real numbers such that $f(x)=O\left(x^{\alpha}\right)$ as $x\to 0+$ and $f(x)=O\left(x^{\beta}\right)$ as $x\to\infty$, then $x^{\sigma-1}f(x)$ is integrable on $(0,\infty)$ for all $\sigma\in(-\alpha,-\beta)$.

### Inversion

Suppose the integral (1.14.32) is absolutely convergent on the line $\Re s=\sigma$ and $f(x)$ is of bounded variation in a neighborhood of $x=u$. Then

 1.14.34 $\tfrac{1}{2}(f(u+)+f(u-))=\frac{1}{2\pi i}\lim_{T\to\infty}\int^{\sigma+iT}_{% \sigma-iT}u^{-s}\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(s\right)\mathrm{d}s.$ ⓘ Symbols: $\mathscr{M}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Mellin transform, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $\sigma\in(a,b)$: parameter Permalink: http://dlmf.nist.gov/1.14.E34 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Mellin transform was changed to $\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{M}(f;s)$. Reported 2017-03-07 See also: Annotations for 1.14(iv), 1.14(iv), 1.14 and 1

If $f(x)$ is continuous on $(0,\infty)$ and $\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(\sigma+it\right)$ is integrable on $(-\infty,\infty)$, then

 1.14.35 $f(x)=\frac{1}{2\pi i}\int^{\sigma+i\infty}_{\sigma-i\infty}x^{-s}\mathscr{M}% \mskip-3.0mu f\mskip 3.0mu \left(s\right)\mathrm{d}s.$ ⓘ Symbols: $\mathscr{M}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Mellin transform, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $\sigma\in(a,b)$: parameter Permalink: http://dlmf.nist.gov/1.14.E35 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Mellin transform was changed to $\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{M}(f;s)$. Reported 2017-03-07 See also: Annotations for 1.14(iv), 1.14(iv), 1.14 and 1

### Parseval-type Formulas

Suppose $x^{-\sigma}f(x)$ and $x^{\sigma-1}g(x)$ are absolutely integrable on $(0,\infty)$ and either $\mathscr{M}\mskip-3.0mu g\mskip 3.0mu \left(\sigma+it\right)$ or $\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(1-\sigma-it\right)$ is absolutely integrable on $(-\infty,\infty)$. Then for $y>0$,

 1.14.36 $\int^{\infty}_{0}f(x)g(yx)\mathrm{d}x=\frac{1}{2\pi i}\*\int^{\sigma+i\infty}_% {\sigma-i\infty}y^{-s}\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(1-s\right)% \mathscr{M}\mskip-3.0mu g\mskip 3.0mu \left(s\right)\mathrm{d}s,$ ⓘ Symbols: $\mathscr{M}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Mellin transform, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $\sigma\in(a,b)$: parameter Permalink: http://dlmf.nist.gov/1.14.E36 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Mellin transform was changed to $\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{M}(f;s)$. Reported 2017-03-07 See also: Annotations for 1.14(iv), 1.14(iv), 1.14 and 1
 1.14.37 $\int^{\infty}_{0}f(x)g(x)\mathrm{d}x=\frac{1}{2\pi i}\*\int^{\sigma+i\infty}_{% \sigma-i\infty}\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(1-s\right)\mathscr{% M}\mskip-3.0mu g\mskip 3.0mu \left(s\right)\mathrm{d}s.$ ⓘ Symbols: $\mathscr{M}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Mellin transform, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $\sigma\in(a,b)$: parameter Permalink: http://dlmf.nist.gov/1.14.E37 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Mellin transform was changed to $\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{M}(f;s)$. Reported 2017-03-07 See also: Annotations for 1.14(iv), 1.14(iv), 1.14 and 1

When $f$ is real and $\sigma=\tfrac{1}{2}$,

 1.14.38 $\int^{\infty}_{0}(f(x))^{2}\mathrm{d}x=\frac{1}{2\pi}\int^{\infty}_{-\infty}{% \left|\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(\tfrac{1}{2}+it\right)\right% |^{2}}\mathrm{d}t.$ ⓘ Symbols: $\mathscr{M}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Mellin transform, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$ and $\int$: integral Permalink: http://dlmf.nist.gov/1.14.E38 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Mellin transform was changed to $\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{M}(f;s)$. Reported 2017-03-07 See also: Annotations for 1.14(iv), 1.14(iv), 1.14 and 1

### Convolution

Let

 1.14.39 $(f*g)(x)=\int^{\infty}_{0}f(y)g\left(\frac{x}{y}\right)\frac{\mathrm{d}y}{y}.$ ⓘ Defines: $*$: convolution (Mellin) (locally) Symbols: $\mathrm{d}\NVar{x}$: differential of $x$ and $\int$: integral Permalink: http://dlmf.nist.gov/1.14.E39 Encodings: TeX, pMML, png See also: Annotations for 1.14(iv), 1.14(iv), 1.14 and 1

If $x^{\sigma-1}f(x)$ and $x^{\sigma-1}g(x)$ are absolutely integrable on $(0,\infty)$, then for $s=\sigma+it$,

 1.14.40 $\int^{\infty}_{0}x^{s-1}(f*g)(x)\mathrm{d}x=\mathscr{M}\mskip-3.0mu f\mskip 3.% 0mu \left(s\right)\mathscr{M}\mskip-3.0mu g\mskip 3.0mu \left(s\right).$ ⓘ Symbols: $\mathscr{M}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Mellin transform, $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $*$: convolution (Mellin) Permalink: http://dlmf.nist.gov/1.14.E40 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Mellin transform was changed to $\mathscr{M}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathscr{M}(f;s)$. Reported 2017-03-07 See also: Annotations for 1.14(iv), 1.14(iv), 1.14 and 1

## §1.14(v) Hilbert Transform

The Hilbert transform of a real-valued function $f(t)$ is defined in the following equivalent ways:

 1.14.41 $\displaystyle\mathcal{H}\left(f\right)\left(x\right)$ $\displaystyle=\mathcal{H}\mskip-3.0mu f\mskip 3.0mu \left(x\right)=\frac{1}{% \pi}\pvint^{\infty}_{-\infty}\frac{f(t)}{t-x}\mathrm{d}t,$ ⓘ Symbols: $\mathcal{H}\left(\NVar{f}\right)\left(\NVar{x}\right)$: Hilbert transform, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$ and $\pvint_{\NVar{a}}^{\NVar{b}}$: Cauchy principal value Permalink: http://dlmf.nist.gov/1.14.E41 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Hilbert transform was changed to $\mathcal{H}\left(f\right)\left(x\right)$ or $\mathcal{H}\mskip-3.0mu f\mskip 3.0mu \left(x\right)$ from $\mathcal{H}(f;x)$. Reported 2017-03-07 See also: Annotations for 1.14(v), 1.14 and 1 1.14.42 $\displaystyle\mathcal{H}\mskip-3.0mu f\mskip 3.0mu \left(x\right)$ $\displaystyle=\lim_{y\to 0+}\frac{1}{\pi}\int^{\infty}_{-\infty}\frac{t-x}{(t-% x)^{2}+y^{2}}f(t)\mathrm{d}t,$ ⓘ Symbols: $\mathcal{H}\left(\NVar{f}\right)\left(\NVar{x}\right)$: Hilbert transform, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$ and $\int$: integral Permalink: http://dlmf.nist.gov/1.14.E42 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Hilbert transform was changed to $\mathcal{H}\mskip-3.0mu f\mskip 3.0mu \left(x\right)$ from $\mathcal{H}(f;x)$. Reported 2017-03-07 See also: Annotations for 1.14(v), 1.14 and 1 1.14.43 $\displaystyle\mathcal{H}\mskip-3.0mu f\mskip 3.0mu \left(x\right)$ $\displaystyle=\lim_{\epsilon\to 0+}\frac{1}{\pi}\int^{\infty}_{\epsilon}\frac{% f(x+t)-f(x-t)}{t}\mathrm{d}t.$ ⓘ Symbols: $\mathcal{H}\left(\NVar{f}\right)\left(\NVar{x}\right)$: Hilbert transform, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$ and $\int$: integral Permalink: http://dlmf.nist.gov/1.14.E43 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Hilbert transform was changed to $\mathcal{H}\mskip-3.0mu f\mskip 3.0mu \left(x\right)$ from $\mathcal{H}(f;x)$. Reported 2017-03-07 See also: Annotations for 1.14(v), 1.14 and 1

### Inversion

Suppose $f(t)$ is continuously differentiable on $(-\infty,\infty)$ and vanishes outside a bounded interval. Then

 1.14.44 $f(x)=-\frac{1}{\pi}\pvint^{\infty}_{-\infty}\frac{\mathcal{H}\mskip-3.0mu f% \mskip 3.0mu \left(u\right)}{u-x}\mathrm{d}u.$ ⓘ Symbols: $\mathcal{H}\left(\NVar{f}\right)\left(\NVar{x}\right)$: Hilbert transform, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$ and $\pvint_{\NVar{a}}^{\NVar{b}}$: Cauchy principal value Referenced by: §1.14(v) Permalink: http://dlmf.nist.gov/1.14.E44 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Hilbert transform was changed to $\mathcal{H}\mskip-3.0mu f\mskip 3.0mu \left(u\right)$ from $\mathcal{H}(f;u)$. Reported 2017-03-07 See also: Annotations for 1.14(v), 1.14(v), 1.14 and 1

### Inequalities

If $|f(t)|^{p}$, $p>1$, is integrable on $(-\infty,\infty)$, then so is $|\mathcal{H}\mskip-3.0mu f\mskip 3.0mu \left(x\right)|^{p}$ and

 1.14.45 $\int^{\infty}_{-\infty}|\mathcal{H}\mskip-3.0mu f\mskip 3.0mu \left(x\right)|^% {p}\mathrm{d}x\leq A_{p}\int^{\infty}_{-\infty}|f(t)|^{p}\mathrm{d}t,$ ⓘ Symbols: $\mathcal{H}\left(\NVar{f}\right)\left(\NVar{x}\right)$: Hilbert transform, $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $A_{p}$: coefficient Permalink: http://dlmf.nist.gov/1.14.E45 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Hilbert transform was changed to $\mathcal{H}\mskip-3.0mu f\mskip 3.0mu \left(x\right)$ from $\mathcal{H}(f;x)$. Reported 2017-03-07 See also: Annotations for 1.14(v), 1.14(v), 1.14 and 1

where $A_{p}=\tan\left(\tfrac{1}{2}\pi/p\right)$ when $1, or $\cot\left(\tfrac{1}{2}\pi/p\right)$ when $p\geq 2$. These bounds are sharp, and equality holds when $p=2$.

### Fourier Transform

When $f(t)$ satisfies the same conditions as those for (1.14.44),

 1.14.46 $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\mathcal{H}\mskip-3.0mu f\mskip 3.% 0mu \left(u\right)e^{iux}\mathrm{d}u=-\mathrm{i}(\operatorname{sign}x)\mathscr% {F}\mskip-3.0mu f\mskip 3.0mu \left(x\right),$ ⓘ Symbols: $\mathscr{F}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Fourier transform, $\mathcal{H}\left(\NVar{f}\right)\left(\NVar{x}\right)$: Hilbert transform, $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$, $\mathrm{e}$: base of exponential function, $\int$: integral and $\operatorname{sign}\NVar{x}$: sign of $x$ Referenced by: §1.14(v) Permalink: http://dlmf.nist.gov/1.14.E46 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Hilbert transform was changed to $\mathcal{H}\mskip-3.0mu f\mskip 3.0mu \left(t\right)$ from $\mathcal{H}(f;t)$, and the notation for the Fourier transform was changed to $\mathscr{F}\mskip-3.0mu f\mskip 3.0mu \left(x\right)$ from $F(x)$. In addition, the integration variable was changed from $t$ to $u$. Reported 2017-03-07 See also: Annotations for 1.14(v), 1.14(v), 1.14 and 1

where $\mathscr{F}\mskip-3.0mu f\mskip 3.0mu \left(x\right)$ is given by (1.14.1).

## §1.14(vi) Stieltjes Transform

The Stieltjes transform of a real-valued function $f(t)$ is defined by

 1.14.47 $\mathcal{S}\left(f\right)\left(s\right)=\mathcal{S}\mskip-3.0mu f\mskip 3.0mu % \left(s\right)=\int^{\infty}_{0}\frac{f(t)}{s+t}\mathrm{d}t.$ ⓘ Defines: $\mathcal{S}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Stieltjes transform Symbols: $\mathrm{d}\NVar{x}$: differential of $x$ and $\int$: integral Referenced by: §1.14(vi), §1.14(vi) Permalink: http://dlmf.nist.gov/1.14.E47 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Stieltjes transform was changed to $\mathcal{S}\left(f\right)\left(s\right)$ or $\mathcal{S}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathcal{S}(f;s)$. Reported 2017-03-07 See also: Annotations for 1.14(vi), 1.14 and 1

Sufficient conditions for the integral to converge are that $s$ is a positive real number, and $f(t)=O\left(t^{-\delta}\right)$ as $t\to\infty$, where $\delta>0$.

If the integral converges, then it converges uniformly in any compact domain in the complex $s$-plane not containing any point of the interval $(-\infty,0]$. In this case, $\mathcal{S}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ represents an analytic function in the $s$-plane cut along the negative real axis, and

 1.14.48 $\frac{{\mathrm{d}}^{m}}{{\mathrm{d}s}^{m}}\mathcal{S}\mskip-3.0mu f\mskip 3.0% mu \left(s\right)=(-1)^{m}m!\int^{\infty}_{0}\frac{f(t)\mathrm{d}t}{(s+t)^{m+1% }},$ $m=0,1,2,\dots$. ⓘ Symbols: $\mathcal{S}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Stieltjes transform, $\frac{\mathrm{d}\NVar{f}}{\mathrm{d}\NVar{x}}$: derivative of $f$ with respect to $x$, $\mathrm{d}\NVar{x}$: differential of $x$, $!$: factorial (as in $n!$), $\int$: integral and $m$: nonnegative integer Permalink: http://dlmf.nist.gov/1.14.E48 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Stieltjes transform was changed to $\mathcal{S}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathcal{S}(f;s)$. Reported 2017-03-07 See also: Annotations for 1.14(vi), 1.14 and 1

### Inversion

If $f(t)$ is absolutely integrable on $[0,R]$ for every finite $R$, and the integral (1.14.47) converges, then

 1.14.49 $\lim_{t\to 0+}\frac{\mathcal{S}\mskip-3.0mu f\mskip 3.0mu \left(-\sigma-it% \right)-\mathcal{S}\mskip-3.0mu f\mskip 3.0mu \left(-\sigma+it\right)}{2\pi i}% =\tfrac{1}{2}(f(\sigma+)+f(\sigma-)),$ ⓘ Symbols: $\mathcal{S}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Stieltjes transform, $\pi$: the ratio of the circumference of a circle to its diameter and $\sigma\in(a,b)$: parameter Permalink: http://dlmf.nist.gov/1.14.E49 Encodings: TeX, pMML, png Notational Change (effective with 1.0.15): The notation for the Stieltjes transform was changed to $\mathcal{S}\mskip-3.0mu f\mskip 3.0mu \left(s\right)$ from $\mathcal{S}(f;s)$. Reported 2017-03-07 See also: Annotations for 1.14(vi), 1.14(vi), 1.14 and 1

for all values of the positive constant $\sigma$ for which the right-hand side exists.

### Laplace Transform

If $f(t)$ is piecewise continuous on $[0,\infty)$ and the integral (1.14.47) converges, then

 1.14.50 $\mathcal{S}\left(f\right)=\mathscr{L}\left(\mathscr{L}\left(f\right)\right).$ ⓘ Symbols: $\mathscr{L}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Laplace transform and $\mathcal{S}\left(\NVar{f}\right)\left(\NVar{s}\right)$: Stieltjes transform Permalink: http://dlmf.nist.gov/1.14.E50 Encodings: TeX, pMML, png See also: Annotations for 1.14(vi), 1.14(vi), 1.14 and 1

## §1.14(viii) Compendia

For more extensive tables of the integral transforms of this section and tables of other integral transforms, see Erdélyi et al. (1954a, b), Gradshteyn and Ryzhik (2000), Marichev (1983), Oberhettinger (1972, 1974, 1990), Oberhettinger and Badii (1973), Oberhettinger and Higgins (1961), Prudnikov et al. (1986a, b, 1990, 1992a, 1992b).