§1.14 Integral Transforms

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

1.14.3		$$\frac{1}{2}(f(u+)+f(u-))=\frac{1}{\sqrt{2\pi }}{⨍}_{-\mathrm{\infty }}^{\mathrm{\infty }}F(x){\mathrm{e}}^{-\mathrm{i}xu}dx,$$
ⓘ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $dx$: differential of $x$, $\mathrm{e}$: base of natural logarithm, $\mathrm{i}$: imaginary unit, ${⨍}_a^b$: Cauchy principal value and $F(x)$: Fourier transform of $f(t)$ Permalink: http://dlmf.nist.gov/1.14.E3 Encodings: TeX, pMML, png See also: Annotations for §1.14(i), §1.14(i), §1.14 and Ch.1

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 $(-\mathrm{\infty },\mathrm{\infty })$. Then

1.14.4		$$f(t)=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}F(x){\mathrm{e}}^{-\mathrm{i}xt}dx.$$
ⓘ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $dx$: differential of $x$, $\mathrm{e}$: base of natural logarithm, $\mathrm{i}$: imaginary unit, $\int$: integral and $F(x)$: Fourier transform of $f(t)$ Referenced by: §1.17(ii) Permalink: http://dlmf.nist.gov/1.14.E4 Encodings: TeX, pMML, png See also: Annotations for §1.14(i), §1.14(i), §1.14 and Ch.1

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

1.14.5		$$(f\mathrm{*}g)(t)=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t-s)g(s)ds.$$
ⓘ Defines: $\mathrm{*}$: convolution (Fourier) (locally) Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $dx$: 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 Ch.1

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

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

1.14.6		$$(f\mathrm{*}g)(t)=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}F(x)G(x){\mathrm{e}}^{-\mathrm{i}tx}dx,$$
ⓘ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $dx$: differential of $x$, $\mathrm{e}$: base of natural logarithm, $\mathrm{i}$: imaginary unit, $\int$: integral, $F(x)$: Fourier transform of $f(t)$, $\mathrm{*}$: convolution (Fourier) and $G(x)$: Fourier transform of $g(t)$ Permalink: http://dlmf.nist.gov/1.14.E6 Encodings: TeX, pMML, png See also: Annotations for §1.14(i), §1.14(i), §1.14 and Ch.1

1.14.7		$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}F(x)G(x)dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t)g(-t)dt,$$
ⓘ Symbols: $dx$: differential of $x$, $\int$: integral, $F(x)$: Fourier transform of $f(t)$ and $G(x)$: Fourier transform of $g(t)$ Permalink: http://dlmf.nist.gov/1.14.E7 Encodings: TeX, pMML, png See also: Annotations for §1.14(i), §1.14(i), §1.14 and Ch.1

1.14.8		$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|F(x)|}^{2}dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|f(t)|}^{2}dt.$$
ⓘ Symbols: $dx$: 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 Ch.1

(1.14.8) is Parseval’s formula.

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

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

1.14.11		$\frac{1}{2}(f(u+)+f(u-))$	$=\sqrt{{\displaystyle \frac{2}{\pi }}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{F}_c(x)\cos \left(ux\right)dx,$
ⓘ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $\cos z$: cosine function, $dx$: differential of $x$, $\int$: integral and $F_c(x)$: cosine transformation of $f(t)$ Permalink: http://dlmf.nist.gov/1.14.E11 Encodings: TeX, pMML, png See also: Annotations for §1.14(ii), §1.14(ii), §1.14 and Ch.1
1.14.12		$\frac{1}{2}(f(u+)+f(u-))$	$=\sqrt{{\displaystyle \frac{2}{\pi }}}{\displaystyle {\int }_0^{\mathrm{\infty }}}{F}_s(x)\sin \left(ux\right)dx.$
ⓘ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $dx$: differential of $x$, $\int$: integral, $\sin z$: sine function and $F_s(x)$: sine transformation of $f(t)$ Permalink: http://dlmf.nist.gov/1.14.E12 Encodings: TeX, pMML, png See also: Annotations for §1.14(ii), §1.14(ii), §1.14 and Ch.1

If , $g(t)$ is of bounded variation on $(0,\mathrm{\infty })$ and $g(t)\to 0$ as $t\to \mathrm{\infty }$, then

1.14.13		${\displaystyle {\int }_0^{\mathrm{\infty }}}{F}_c(x)G_c(x)dx$	$={\displaystyle {\int }_0^{\mathrm{\infty }}}f(t)g(t)dt,$
ⓘ Symbols: $dx$: differential of $x$, $\int$: integral, $F_c(x)$: cosine transformation of $f(t)$ and $G_c(x)$: cosine transformation of $g(t)$ Permalink: http://dlmf.nist.gov/1.14.E13 Encodings: TeX, pMML, png See also: Annotations for §1.14(ii), §1.14(ii), §1.14 and Ch.1
1.14.14		${\displaystyle {\int }_0^{\mathrm{\infty }}}{F}_s(x)G_s(x)dx$	$={\displaystyle {\int }_0^{\mathrm{\infty }}}f(t)g(t)dt,$
ⓘ Symbols: $dx$: differential of $x$, $\int$: integral, $F_s(x)$: sine transformation of $f(t)$ and $G_s(x)$: sine transformation of $g(t)$ Permalink: http://dlmf.nist.gov/1.14.E14 Encodings: TeX, pMML, png See also: Annotations for §1.14(ii), §1.14(ii), §1.14 and Ch.1

1.14.15		${\displaystyle {\int }_0^{\mathrm{\infty }}}{(F_c(x))}^{2}dx$	$={\displaystyle {\int }_0^{\mathrm{\infty }}}{(f(t))}^{2}dt,$
ⓘ Symbols: $dx$: 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 Ch.1
1.14.16		${\displaystyle {\int }_0^{\mathrm{\infty }}}{(F_s(x))}^{2}dx$	$={\displaystyle {\int }_0^{\mathrm{\infty }}}{(f(t))}^{2}dt.$
ⓘ Symbols: $dx$: 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 Ch.1

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

1.14.18		$$|f(t)|\le M{\mathrm{e}}^{\alpha t},$$
.
ⓘ Symbols: $\mathrm{e}$: base of natural logarithm, $M$: 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 Ch.1

Then $ℒf\left(s\right)$ is an analytic function of $s$ for $\mathrm{\Re }s>\alpha$. Moreover,

1.14.19		$$ℒf\left(s\right)\to 0,$$
$\mathrm{\Re }s\to \mathrm{\infty }$.
ⓘ Symbols: $ℒ\left(f\right)\left(s\right)$: Laplace transform and $\mathrm{\Re }$: real part Keywords: Laplace transform 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 $ℒf\left(s\right)$ from $ℒ(f(t);s)$. See also: Annotations for §1.14(iii), §1.14(iii), §1.14 and Ch.1

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

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

1.14.20		$$f(t)=\frac{1}{2\pi \mathrm{i}}\underset{T\to \mathrm{\infty }}{lim}{\int }_{\sigma -\mathrm{i}T}^{\sigma +\mathrm{i}T}{\mathrm{e}}^{ts}ℒf\left(s\right)ds,$$
$\sigma >\alpha$.
ⓘ Symbols: $ℒ\left(f\right)\left(s\right)$: Laplace transform, $\pi$: the ratio of the circumference of a circle to its diameter, $dx$: differential of $x$, $\mathrm{e}$: base of natural logarithm, $\mathrm{i}$: imaginary unit, $\int$: integral, $\sigma \in (a,b)$: parameter and $\alpha$: constant Keywords: Laplace transform 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 $ℒf\left(s\right)$ from $ℒ(f(t);s)$. See also: Annotations for §1.14(iii), §1.14(iii), §1.14 and Ch.1

Moreover, if $ℒf\left(s\right)=O\left(s^{-K}\right)$ in some half-plane $\mathrm{\Re }s\ge \gamma$ and $K>1$, then (1.14.20) holds for $\sigma >\gamma$.

If $\mathrm{\Re }s>\max (\mathrm{\Re }\left(a+\alpha \right),\alpha )$, then

1.14.21		$$ℒf\left(s-a\right)=ℒf_a\left(s\right),$$
ⓘ Symbols: $ℒ\left(f\right)\left(s\right)$: Laplace transform and $\mathrm{e}$: base of natural logarithm Keywords: Laplace transform 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 $ℒf\left(s\right)$ from $ℒ(f(t);s)$. In addition, the auxiliary function $f_a(t)={\mathrm{e}}^{at}f(t)$ was introduced. See also: Annotations for §1.14(iii), §1.14(iii), §1.14 and Ch.1

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

1.14.22		$$ℒf_a^{+}\left(s\right)={\mathrm{e}}^{-as}ℒf\left(s\right),$$
ⓘ Symbols: $H\left(x\right)$: Heaviside function, $ℒ\left(f\right)\left(s\right)$: Laplace transform and $\mathrm{e}$: base of natural logarithm Keywords: Laplace transform 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 $ℒ\left(f\right)\left(s\right)$ from $ℒ(f(t);s)$. In addition, the auxiliary function $f_a^{+}(t)=H\left(t-a\right)f(t-a)$ was introduced. See also: Annotations for §1.14(iii), §1.14(iii), §1.14 and Ch.1

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

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

1.14.23		$${(-1)}^n\frac{d^n}{{ds}^n}ℒf\left(s\right)=ℒf_n\left(s\right),$$
$n=1,2,3,\mathrm{\dots }$,
ⓘ Symbols: $ℒ\left(f\right)\left(s\right)$: Laplace transform, $\frac{df}{dx}$: derivative of $f$ with respect to $x$ and $n$: nonnegative integer Keywords: Laplace transform 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 $ℒf\left(s\right)$ from $ℒ(f(t);s)$. In addition, the auxiliary function $f_n(t)=t^{n}f(t)$ was introduced. See also: Annotations for §1.14(iii), §1.14(iii), §1.14 and Ch.1

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

1.14.24		$${\int }_s^{\mathrm{\infty }}ℒf\left(u\right)du=ℒf_{-1}\left(s\right),$$
ⓘ Symbols: $ℒ\left(f\right)\left(s\right)$: Laplace transform, $dx$: differential of $x$ and $\int$: integral Keywords: Laplace transform 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 $ℒf\left(s\right)$ from $ℒ(f(t);s)$. In addition, the auxiliary function $f_{-1}(t)=\frac{f(t)}{t}$ was introduced. See also: Annotations for §1.14(iii), §1.14(iii), §1.14 and Ch.1

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

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

1.14.25		$$ℒf\left(s\right)=\frac{1}{1-{\mathrm{e}}^{-as}}{\int }_0^a{\mathrm{e}}^{-st}f(t)dt.$$
ⓘ Symbols: $ℒ\left(f\right)\left(s\right)$: Laplace transform, $dx$: differential of $x$, $\mathrm{e}$: base of natural logarithm and $\int$: integral Keywords: Laplace transform 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 $ℒf\left(s\right)$ from $ℒ(f(t);s)$. See also: Annotations for §1.14(iii), §1.14(iii), §1.14 and Ch.1

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

1.14.26		$$ℒf\left(s\right)=\frac{1}{1+{\mathrm{e}}^{-as}}{\int }_0^a{\mathrm{e}}^{-st}f(t)dt.$$
ⓘ Symbols: $ℒ\left(f\right)\left(s\right)$: Laplace transform, $dx$: differential of $x$, $\mathrm{e}$: base of natural logarithm and $\int$: integral Keywords: Laplace transform 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 $ℒf\left(s\right)$ from $ℒ(f(t);s)$. See also: Annotations for §1.14(iii), §1.14(iii), §1.14 and Ch.1

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

1.14.27		$$ℒ\left(f^{\prime }\right)\left(s\right)=sℒ\left(f\right)\left(s\right)-f(0+).$$
ⓘ Symbols: $ℒ\left(f\right)\left(s\right)$: Laplace transform Keywords: 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 $ℒ\left(f^{\prime }\right)\left(s\right)$ and $ℒ\left(f\right)\left(s\right)$ from $ℒ(f^{\prime }(t);s)$ and $ℒ(f(t);s)$. See also: Annotations for §1.14(iii), §1.14(iii), §1.14 and Ch.1

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

1.14.28		$$ℒ\left(f^{\prime }\right)\left(s\right)=sℒ\left(f\right)\left(s\right)-f(0+)-\sum _{k=1}^n{\mathrm{e}}^{-s{t}_k}(f(t_k+)-f(t_k-)).$$
ⓘ Symbols: $ℒ\left(f\right)\left(s\right)$: Laplace transform, $\mathrm{e}$: base of natural logarithm, $k$: integer and $n$: nonnegative integer Keywords: Laplace transform 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 $ℒ\left(f^{\prime }\right)\left(s\right)$ and $ℒ\left(f\right)\left(s\right)$ from $ℒ(f^{\prime }(t);s)$ and $ℒ(f(t);s)$. See also: Annotations for §1.14(iii), §1.14(iii), §1.14 and Ch.1

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

1.14.29		$$ℒ\left(f^{(n)}\right)\left(s\right)=s^nℒ\left(f\right)\left(s\right)-s^{n-1}f(0+)-s^{n-2}{f}^{\prime }(0+)-\mathrm{\cdots }-f^{(n-1)}(0+).$$
ⓘ Symbols: $ℒ\left(f\right)\left(s\right)$: Laplace transform and $n$: nonnegative integer Keywords: Laplace transform 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 $ℒ\left(f^{(n)}\right)\left(s\right)$ and $ℒ\left(f\right)\left(s\right)$ from $ℒ(f^{(n)}(t);s)$ and $ℒ(f(t);s)$. See also: Annotations for §1.14(iii), §1.14(iii), §1.14 and Ch.1

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

1.14.30		$$(f\mathrm{*}g)(t)={\int }_0^{t}f(u)g(t-u)du.$$
ⓘ Defines: $\mathrm{*}$: convolution (Laplace) (locally) Symbols: $dx$: 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 Ch.1

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

1.14.31		$$ℒ\left(f\mathrm{*}g\right)=ℒ\left(f\right)ℒ\left(g\right).$$
ⓘ Symbols: $ℒ\left(f\right)\left(s\right)$: Laplace transform and $\mathrm{*}$: convolution (Laplace) Keywords: Laplace transform 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 Ch.1

If $f(t)$ and $g(t)$ are continuous and $ℒf\left(s\right)=ℒg\left(s\right)$, then $f(t)=g(t)$.

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

1.14.34		$$\frac{1}{2}(f(u+)+f(u-))=\frac{1}{2\pi \mathrm{i}}\underset{T\to \mathrm{\infty }}{lim}{\int }_{\sigma -\mathrm{i}T}^{\sigma +\mathrm{i}T}{u}^{-s}ℳf\left(s\right)ds.$$
ⓘ Symbols: $ℳ\left(f\right)\left(s\right)$: Mellin transform, $\pi$: the ratio of the circumference of a circle to its diameter, $dx$: differential of $x$, $\mathrm{i}$: imaginary unit, $\int$: integral and $\sigma \in (a,b)$: parameter Keywords: Mellin transform 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 $ℳf\left(s\right)$ from $ℳ(f;s)$. See also: Annotations for §1.14(iv), §1.14(iv), §1.14 and Ch.1

If $f(x)$ is continuous on $(0,\mathrm{\infty })$ and $ℳf\left(\sigma +\mathrm{i}t\right)$ is integrable on $(-\mathrm{\infty },\mathrm{\infty })$, then

1.14.35		$$f(x)=\frac{1}{2\pi \mathrm{i}}{\int }_{\sigma -\mathrm{i}\mathrm{\infty }}^{\sigma +\mathrm{i}\mathrm{\infty }}{x}^{-s}ℳf\left(s\right)ds.$$
ⓘ Symbols: $ℳ\left(f\right)\left(s\right)$: Mellin transform, $\pi$: the ratio of the circumference of a circle to its diameter, $dx$: differential of $x$, $\mathrm{i}$: imaginary unit, $\int$: integral and $\sigma \in (a,b)$: parameter Keywords: Mellin transform 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 $ℳf\left(s\right)$ from $ℳ(f;s)$. See also: Annotations for §1.14(iv), §1.14(iv), §1.14 and Ch.1

Suppose $x^{-\sigma }f(x)$ and $x^{\sigma -1}g(x)$ are absolutely integrable on $(0,\mathrm{\infty })$ and either $ℳg\left(\sigma +\mathrm{i}t\right)$ or $ℳf\left(1-\sigma -\mathrm{i}t\right)$ is absolutely integrable on $(-\mathrm{\infty },\mathrm{\infty })$. Then for $y>0$,

1.14.36		$${\int }_0^{\mathrm{\infty }}f(x)g(yx)dx=\frac{1}{2\pi \mathrm{i}}{\int }_{\sigma -\mathrm{i}\mathrm{\infty }}^{\sigma +\mathrm{i}\mathrm{\infty }}{y}^{-s}ℳf\left(1-s\right)ℳg\left(s\right)ds,$$
ⓘ Symbols: $ℳ\left(f\right)\left(s\right)$: Mellin transform, $\pi$: the ratio of the circumference of a circle to its diameter, $dx$: differential of $x$, $\mathrm{i}$: imaginary unit, $\int$: integral and $\sigma \in (a,b)$: parameter Keywords: Mellin transform 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 $ℳf\left(s\right)$ from $ℳ(f;s)$. See also: Annotations for §1.14(iv), §1.14(iv), §1.14 and Ch.1

1.14.37		$${\int }_0^{\mathrm{\infty }}f(x)g(x)dx=\frac{1}{2\pi \mathrm{i}}{\int }_{\sigma -\mathrm{i}\mathrm{\infty }}^{\sigma +\mathrm{i}\mathrm{\infty }}ℳf\left(1-s\right)ℳg\left(s\right)ds.$$
ⓘ Symbols: $ℳ\left(f\right)\left(s\right)$: Mellin transform, $\pi$: the ratio of the circumference of a circle to its diameter, $dx$: differential of $x$, $\mathrm{i}$: imaginary unit, $\int$: integral and $\sigma \in (a,b)$: parameter Keywords: Mellin transform 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 $ℳf\left(s\right)$ from $ℳ(f;s)$. See also: Annotations for §1.14(iv), §1.14(iv), §1.14 and Ch.1