# §1.14(i) Fourier Transform

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

 1.14.1 $F(x)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}f(t)e^{ixt}dt.$ Symbols: $dx$: differential of $x$, $e$: base of exponential function, $\int$: integral and $F(x)$: Fourier transform of $f(t)$ Referenced by: §1.14(v), §1.17(ii) Permalink: http://dlmf.nist.gov/1.14.E1 Encodings: TeX, pMML, png

(Some references replace $ixt$ by $-ixt$.)

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

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

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

# 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 $(f*g)(t)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}f(t-s)g(s)ds.$ Defines: $*$: $f*g$: convolution for Fourier transforms Symbols: $dx$: differential of $x$ and $\int$: integral Permalink: http://dlmf.nist.gov/1.14.E5 Encodings: TeX, pMML, png

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 $(-\infty,\infty)$, and $F(x)$ and $G(x)$ are their respective Fourier transforms. Then

 1.14.6 $(f*g)(t)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}F(x)G(x)e^{-itx}dx,$
 1.14.7 $\int^{\infty}_{-\infty}F(x)G(x)dx=\int^{\infty}_{-\infty}f(t)g(-t)dt,$
 1.14.8 $\int^{\infty}_{-\infty}|F(x)|^{2}dx=\int^{\infty}_{-\infty}|f(t)|^{2}dt.$

(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

These are defined respectively by

 1.14.9 $\displaystyle F_{c}(x)$ $\displaystyle=\sqrt{\frac{2}{\pi}}\int^{\infty}_{0}f(t)\mathop{\cos\/}% \nolimits\!\left(xt\right)dt,$ Symbols: $\mathop{\cos\/}\nolimits z$: cosine function, $dx$: differential of $x$, $\int$: integral and $F_{c}(x)$: cosine transformation of $f(t)$ Permalink: http://dlmf.nist.gov/1.14.E9 Encodings: TeX, pMML, png 1.14.10 $\displaystyle F_{s}(x)$ $\displaystyle=\sqrt{\frac{2}{\pi}}\int^{\infty}_{0}f(t)\mathop{\sin\/}% \nolimits\!\left(xt\right)dt.$ Symbols: $dx$: differential of $x$, $\int$: integral, $\mathop{\sin\/}\nolimits z$: sine function and $F_{s}(x)$: sine transformation of $f(t)$ Permalink: http://dlmf.nist.gov/1.14.E10 Encodings: TeX, pMML, png

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

# Parseval’s Formula

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

 1.14.13 $\displaystyle\int^{\infty}_{0}F_{c}(x)G_{c}(x)dx$ $\displaystyle=\int^{\infty}_{0}f(t)g(t)dt,$ 1.14.14 $\displaystyle\int^{\infty}_{0}F_{s}(x)G_{s}(x)dx$ $\displaystyle=\int^{\infty}_{0}f(t)g(t)dt,$
 1.14.15 $\displaystyle\int^{\infty}_{0}(F_{c}(x))^{2}dx$ $\displaystyle=\int^{\infty}_{0}(f(t))^{2}dt,$ 1.14.16 $\displaystyle\int^{\infty}_{0}(F_{s}(x))^{2}dx$ $\displaystyle=\int^{\infty}_{0}(f(t))^{2}dt,$

where $G_{c}(x)$ and $G_{s}(x)$ are respectively the cosine and sine transforms of $g(t)$.

# §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 $\mathop{\mathscr{L}\/}\nolimits\left(f(t);s\right)=\int^{\infty}_{0}e^{-st}f(t% )dt.$ Defines: $\mathop{\mathscr{L}\/}\nolimits\left(f;s\right)$: Laplace transform Symbols: $dx$: differential of $x$, $e$: 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

Alternative notations are $\mathop{\mathscr{L}\/}\nolimits(f(t))$, $\mathop{\mathscr{L}\/}\nolimits(f;s)$, or even $\mathop{\mathscr{L}\/}\nolimits(f)$, when it is not important to display all the variables.

# Convergence and Analyticity

Assume that on $[0,\infty)$ $f(t)$ is piecewise continuous 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: $e$: base of exponential function, $M$: constant and $\alpha$: constant Referenced by: §1.14(iii), §1.14(iii) Permalink: http://dlmf.nist.gov/1.14.E18 Encodings: TeX, pMML, png

Then $\mathop{\mathscr{L}\/}\nolimits\left(f(t);s\right)$ is an analytic function of $s$ for $\realpart{s}>\alpha$. Moreover,

 1.14.19 $\mathop{\mathscr{L}\/}\nolimits\left(f(t);s\right)\to 0,$ $\realpart{s}\to\infty$.

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

# Inversion

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

 1.14.20 $f(t)=\frac{1}{2\pi i}\lim_{T\to\infty}\int^{\sigma+iT}_{\sigma-iT}e^{ts}% \mathop{\mathscr{L}\/}\nolimits\left(f(t);s\right)ds,$ $\sigma>\alpha$.

Moreover, if $\mathop{\mathscr{L}\/}\nolimits\left(f(t);s\right)=\mathop{O\/}\nolimits\!% \left(s^{-K}\right)$ in some half-plane $\realpart{s}\geq\gamma$ and $K>1$, then (1.14.20) holds for $\sigma>\gamma$.

# Translation

If $\realpart{s}>\max(\realpart{(a+\alpha)},\alpha)$, then

 1.14.21 $\mathop{\mathscr{L}\/}\nolimits\left(f(t);s-a\right)=\mathop{\mathscr{L}\/}% \nolimits\left(e^{at}f(t);s\right).$

Also, if $a\geq 0$ then

 1.14.22 $\mathop{\mathscr{L}\/}\nolimits\left(\mathop{H\/}\nolimits\!\left(t-a\right)f(% t-a);s\right)=e^{-as}\mathop{\mathscr{L}\/}\nolimits\left(f(t);s\right),$

where $\mathop{H\/}\nolimits$ is the Heaviside function; see (1.16.13).

# Differentiation and Integration

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

 1.14.23 $\frac{{d}^{n}}{{ds}^{n}}\mathop{\mathscr{L}\/}\nolimits\left(f(t);s\right)=% \mathop{\mathscr{L}\/}\nolimits\left((-t)^{n}f(t);s\right),$ $n=1,2,3,\dots$.

If also $\lim_{t\to 0+}f(t)/t$ exists, then

 1.14.24 $\int^{\infty}_{s}\mathop{\mathscr{L}\/}\nolimits\left(f(t);u\right)du=\mathop{% \mathscr{L}\/}\nolimits\left(\frac{f(t)}{t};s\right).$

# Periodic Functions

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

 1.14.25 $\mathop{\mathscr{L}\/}\nolimits\left(f(t);s\right)=\frac{1}{1-e^{-as}}\int^{a}% _{0}e^{-st}f(t)dt.$

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

 1.14.26 $\mathop{\mathscr{L}\/}\nolimits\left(f(t);s\right)=\frac{1}{1+e^{-as}}\int^{a}% _{0}e^{-st}f(t)dt.$

# Derivatives

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

 1.14.27 $\mathop{\mathscr{L}\/}\nolimits\left(f^{\prime}(t);s\right)=s\mathop{\mathscr{% L}\/}\nolimits\left(f(t);s\right)-f(0+).$ Symbols: $\mathop{\mathscr{L}\/}\nolimits\left(f;s\right)$: Laplace transform A&S Ref: 29.2.4 Permalink: http://dlmf.nist.gov/1.14.E27 Encodings: TeX, pMML, png

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

 1.14.28 $\mathop{\mathscr{L}\/}\nolimits\left(f^{\prime}(t);s\right)=s\mathop{\mathscr{% L}\/}\nolimits\left(f(t);s\right)-f(0+)-\sum^{n}_{k=1}e^{-st_{k}}(f(t_{k}+)-f(% t_{k}-)).$

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 $\mathop{\mathscr{L}\/}\nolimits\left(f^{(n)}(t);s\right)=s^{n}\mathop{\mathscr% {L}\/}\nolimits\left(f(t);s\right)-s^{n-1}f(0+)-s^{n-2}f^{\prime}(0+)-\dots-f^% {(n-1)}(0+).$ Symbols: $\mathop{\mathscr{L}\/}\nolimits\left(f;s\right)$: Laplace transform and $n$: nonnegative integer A&S Ref: 29.2.5 Permalink: http://dlmf.nist.gov/1.14.E29 Encodings: TeX, pMML, png

# Convolution

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

 1.14.30 $(f*g)(t)=\int^{t}_{0}f(u)g(t-u)du.$

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

 1.14.31 $\mathop{\mathscr{L}\/}\nolimits(f*g)=\mathop{\mathscr{L}\/}\nolimits(f)\mathop% {\mathscr{L}\/}\nolimits(g).$ Symbols: $\mathop{\mathscr{L}\/}\nolimits\left(f;s\right)$: Laplace transform Permalink: http://dlmf.nist.gov/1.14.E31 Encodings: TeX, pMML, png

# Uniqueness

If $f(t)$ and $g(t)$ are continuous and $\mathop{\mathscr{L}\/}\nolimits(f)=\mathop{\mathscr{L}\/}\nolimits(g)$, 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 $\mathop{\mathscr{M}\/}\nolimits\left(f;s\right)=\int^{\infty}_{0}x^{s-1}f(x)dx.$ Defines: $\mathop{\mathscr{M}\/}\nolimits\left(f;s\right)$: Mellin transform Symbols: $dx$: 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

Alternative notations for $\mathop{\mathscr{M}\/}\nolimits\left(f;s\right)$ are $\mathop{\mathscr{M}\/}\nolimits\left(f(x);s\right)$ and $\mathop{\mathscr{M}\/}\nolimits(f)$.

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 $\mathop{\mathscr{M}\/}\nolimits\left(f;s\right)$ is an analytic function of $s$ in the vertical strip $a<\realpart{s}. Moreover, for $a<\sigma,

 1.14.33 $\lim_{t\to\pm\infty}\mathop{\mathscr{M}\/}\nolimits\left(f;\sigma+it\right)=0.$

Note: If $f(x)$ is continuous and $\alpha$ and $\beta$ are real numbers such that $f(x)=\mathop{O\/}\nolimits\!\left(x^{\alpha}\right)$ as $x\to 0+$ and $f(x)=\mathop{O\/}\nolimits\!\left(x^{\beta}\right)$ 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 $\realpart{s}=\sigma$ and $f(x)$ is of bounded variation in a neighborhood of $x=u$. Then

 1.14.34 $\tfrac{1}{2}(f(u+)+f(u-))=\frac{1}{2\pi i}\lim_{T\to\infty}\int^{\sigma+iT}_{% \sigma-iT}u^{-s}\mathop{\mathscr{M}\/}\nolimits\left(f;s\right)ds.$

If $f(x)$ is continuous on $(0,\infty)$ and $\mathop{\mathscr{M}\/}\nolimits\left(f;\sigma+it\right)$ is integrable on $(-\infty,\infty)$, then

 1.14.35 $f(x)=\frac{1}{2\pi i}\int^{\sigma+i\infty}_{\sigma-i\infty}x^{-s}\mathop{% \mathscr{M}\/}\nolimits\left(f;s\right)ds.$

# Parseval-type Formulas

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

 1.14.36 $\int^{\infty}_{0}f(x)g(yx)dx=\frac{1}{2\pi i}\*\int^{\sigma+i\infty}_{\sigma-i% \infty}y^{-s}\mathop{\mathscr{M}\/}\nolimits\left(f;1-s\right)\mathop{\mathscr% {M}\/}\nolimits\left(g;s\right)ds,$
 1.14.37 $\int^{\infty}_{0}f(x)g(x)dx=\frac{1}{2\pi i}\*\int^{\sigma+i\infty}_{\sigma-i% \infty}\mathop{\mathscr{M}\/}\nolimits\left(f;1-s\right)\mathop{\mathscr{M}\/}% \nolimits\left(g;s\right)ds.$

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

 1.14.38 $\int^{\infty}_{0}(f(x))^{2}dx=\frac{1}{2\pi}\int^{\infty}_{-\infty}\left|% \mathop{\mathscr{M}\/}\nolimits\left(f;\tfrac{1}{2}+it\right)\right|^{2}dt.$

# Convolution

Let

 1.14.39 $(f*g)(x)=\int^{\infty}_{0}f(y)g\left(\frac{x}{y}\right)\frac{dy}{y}.$

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 $\int^{\infty}_{0}x^{s-1}(f*g)(x)dx=\mathop{\mathscr{M}\/}\nolimits\left(f;s% \right)\mathop{\mathscr{M}\/}\nolimits\left(g;s\right).$

# §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 $\displaystyle\mathop{\mathcal{H}\/}\nolimits\left(f;x\right)$ $\displaystyle=\mathop{\mathcal{H}\/}\nolimits\left(f(t);x\right)$ $\displaystyle=\mathop{\mathcal{H}\/}\nolimits(f)$ $\displaystyle=\frac{1}{\pi}\pvint^{\infty}_{-\infty}\frac{f(t)}{t-x}dt,$ 1.14.42 $\displaystyle\mathop{\mathcal{H}\/}\nolimits\left(f;x\right)$ $\displaystyle=\lim_{y\to 0+}\frac{1}{\pi}\int^{\infty}_{-\infty}\frac{t-x}{(t-% x)^{2}+y^{2}}f(t)dt,$ 1.14.43 $\displaystyle\mathop{\mathcal{H}\/}\nolimits\left(f;x\right)$ $\displaystyle=\lim_{\epsilon\to 0+}\frac{1}{\pi}\int^{\infty}_{\epsilon}\frac{% f(x+t)-f(x-t)}{t}dt.$

# Inversion

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

 1.14.44 $f(x)=-\frac{1}{\pi}\pvint^{\infty}_{-\infty}\frac{\mathop{\mathcal{H}\/}% \nolimits\left(f;u\right)}{u-x}du.$

# Inequalities

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

 1.14.45 $\int^{\infty}_{-\infty}|\mathop{\mathcal{H}\/}\nolimits\left(f;x\right)|^{p}dx% \leq A_{p}\int^{\infty}_{-\infty}|f(t)|^{p}dt,$

where $A_{p}=\mathop{\tan\/}\nolimits\!\left(\tfrac{1}{2}\pi/p\right)$ when $1, or $\mathop{\cot\/}\nolimits\!\left(\tfrac{1}{2}\pi/p\right)$ 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 $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\mathop{\mathcal{H}\/}\nolimits% \left(f;t\right)e^{ixt}dt=-i(\mathop{\mathrm{sign}\/}\nolimits x)F(x),$

where $F(x)$ 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 $\mathop{\mathcal{S}\/}\nolimits\left(f;s\right)=\mathop{\mathcal{S}\/}% \nolimits\left(f(t);s\right)=\mathop{\mathcal{S}\/}\nolimits(f)=\int^{\infty}_% {0}\frac{f(t)}{s+t}dt.$ Defines: $\mathop{\mathcal{S}\/}\nolimits\left(f;s\right)$: Stieltjes transform Symbols: $dx$: 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

Sufficient conditions for the integral to converge are that $s$ is a positive real number, and $f(t)=\mathop{O\/}\nolimits\!\left(t^{-\delta}\right)$ 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, $\mathop{\mathcal{S}\/}\nolimits\left(f;s\right)$ represents an analytic function in the $s$-plane cut along the negative real axis, and

 1.14.48 $\frac{{d}^{m}}{{ds}^{m}}\mathop{\mathcal{S}\/}\nolimits\left(f;s\right)=(-1)^{% m}m!\int^{\infty}_{0}\frac{f(t)dt}{(s+t)^{m+1}},$ $m=0,1,2,\dots$.

# 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 $\lim_{t\to 0+}\frac{\mathop{\mathcal{S}\/}\nolimits\left(f;-\sigma-it\right)-% \mathop{\mathcal{S}\/}\nolimits\left(f;-\sigma+it\right)}{2\pi i}=\tfrac{1}{2}% (f(\sigma+)+f(\sigma-)),$

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 $\mathop{\mathcal{S}\/}\nolimits(f)=\mathop{\mathscr{L}\/}\nolimits(\mathop{% \mathscr{L}\/}\nolimits(f)).$

# §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).