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§1.14 Integral Transforms

Contents
  1. §1.14(i) Fourier Transform
  2. §1.14(ii) Fourier Cosine and Sine Transforms
  3. §1.14(iii) Laplace Transform
  4. §1.14(iv) Mellin Transform
  5. §1.14(v) Hilbert Transform
  6. §1.14(vi) Stieltjes Transform
  7. §1.14(vii) Tables
  8. §1.14(viii) Compendia

§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-3.0muf\mskip 3.0mu%
\left(x\right)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}f(t){\mathrm{e}}^{%
\mathrm{i}xt}\,\mathrm{d}t.

(Some references replace \mathrm{i}xt by -\mathrm{i}xt). 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

Inversion

Suppose that f(t) is absolutely integrable on (-\infty,\infty) and of bounded variation in a neighborhood of t=u1.4(v)). Then

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

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.

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 and square integrable on (-\infty,\infty), then

1.14.7_5 \int^{\infty}_{-\infty}F(x)\overline{G(x)}\,\mathrm{d}x=\int^{\infty}_{-\infty%
}f(t)\overline{g(t)}\,\mathrm{d}t,
1.14.8 \int^{\infty}_{-\infty}{\left|F(x)\right|}^{2}\,\mathrm{d}x=\int^{\infty}_{-%
\infty}{\left|f(t)\right|}^{2}\,\mathrm{d}t.

(1.14.7_5) and (1.14.8) are Parseval’s formulas.

Poisson’s Summation Formula

See §1.8(iv) and especially (1.8.14).

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 \mathscr{F}_{\mkern-3.0muc}\left(f\right)\left(x\right)=\mathscr{F}_{\mkern-3.%
0muc}\mskip-1.0muf\mskip 3.0mu\left(x\right)=\sqrt{\frac{2}{\pi}}\int^{\infty}%
_{0}f(t)\cos\left(xt\right)\,\mathrm{d}t,
1.14.10 \mathscr{F}_{\mkern-2.0mus}\left(f\right)\left(x\right)=\mathscr{F}_{\mkern-2.%
0mus}\mskip-1.0muf\mskip 3.0mu\left(x\right)=\sqrt{\frac{2}{\pi}}\int^{\infty}%
_{0}f(t)\sin\left(xt\right)\,\mathrm{d}t.

In this subsection we let F_{c}(x)=\mathscr{F}_{\mkern-3.0muc}\mskip-1.0muf\mskip 3.0mu\left(x\right), F_{s}(x)=\mathscr{F}_{\mkern-2.0mus}\mskip-1.0muf\mskip 3.0mu\left(x\right), G_{c}(x)=\mathscr{F}_{\mkern-3.0muc}\mskip-1.0mug\mskip 3.0mu\left(x\right), and G_{s}(x)=\mathscr{F}_{\mkern-2.0mus}\mskip-1.0mug\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

Parseval’s Formula

Suppose f(t) and g(t) are absolutely and square integrable on [0,\infty), then

1.14.15 \int^{\infty}_{0}(F_{c}(x))^{2}\,\mathrm{d}x=\int^{\infty}_{0}(f(t))^{2}\,%
\mathrm{d}t,
1.14.16 \int^{\infty}_{0}(F_{s}(x))^{2}\,\mathrm{d}x=\int^{\infty}_{0}(f(t))^{2}\,%
\mathrm{d}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 \mathscr{L}\left(f\right)\left(s\right)=\mathscr{L}\mskip-3.0muf\mskip 3.0mu%
\left(s\right)=\int^{\infty}_{0}{\mathrm{e}}^{-st}f(t)\,\mathrm{d}t.

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

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

1.14.19 \mathscr{L}\mskip-3.0muf\mskip 3.0mu\left(s\right)\to 0,\Re s\to\infty.

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\mathrm{i}}\lim_{T\to\infty}\int^{\sigma+\mathrm{i}T}_{%
\sigma-\mathrm{i}T}{\mathrm{e}}^{ts}\mathscr{L}\mskip-3.0muf\mskip 3.0mu\left(%
s\right)\,\mathrm{d}s,\sigma>\alpha.

Moreover, if \mathscr{L}\mskip-3.0muf\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-3.0muf\mskip 3.0mu\left(s-a\right)=\mathscr{L}\mskip-3.0muf_%
{a}\mskip 3.0mu\left(s\right),

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

1.14.22 \mathscr{L}\mskip-3.0muf_{a}^{+}\mskip 3.0mu\left(s\right)={\mathrm{e}}^{-as}%
\mathscr{L}\mskip-3.0muf\mskip 3.0mu\left(s\right),

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-3.0muf%
\mskip 3.0mu\left(s\right)=\mathscr{L}\mskip-3.0muf_{n}\mskip 3.0mu\left(s%
\right),n=1,2,3,\dots,

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-3.0muf\mskip 3.0mu\left(u\right)\,\mathrm{d%
}u=\mathscr{L}\mskip-3.0muf_{-1}\mskip 3.0mu\left(s\right),

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-3.0muf\mskip 3.0mu\left(s\right)=\frac{1}{1-{\mathrm{e}}^{-%
as}}\int^{a}_{0}{\mathrm{e}}^{-st}f(t)\,\mathrm{d}t.

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

1.14.26 \mathscr{L}\mskip-3.0muf\mskip 3.0mu\left(s\right)=\frac{1}{1+{\mathrm{e}}^{-%
as}}\int^{a}_{0}{\mathrm{e}}^{-st}f(t)\,\mathrm{d}t.

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+).

If f(t) and f^{\prime}(t) are piecewise continuous on [0,\infty) with discontinuities at (0=) t_{0}<t_{1}<\cdots<t_{n}, 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}{\mathrm{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 \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+).

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.

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).

Uniqueness

If f(t) and g(t) are continuous and \mathscr{L}\mskip-3.0muf\mskip 3.0mu\left(s\right)=\mathscr{L}\mskip-3.0mug%
\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-3.0muf\mskip 3.0mu%
\left(s\right)=\int^{\infty}_{0}x^{s-1}f(x)\,\mathrm{d}x.

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

1.14.33 \lim_{t\to\pm\infty}\mathscr{M}\mskip-3.0muf\mskip 3.0mu\left(\sigma+\mathrm{i%
}t\right)=0.

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\mathrm{i}}\lim_{T\to\infty}\int^{%
\sigma+\mathrm{i}T}_{\sigma-\mathrm{i}T}u^{-s}\mathscr{M}\mskip-3.0muf\mskip 3%
.0mu\left(s\right)\,\mathrm{d}s.

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

1.14.35 f(x)=\frac{1}{2\pi\mathrm{i}}\int^{\sigma+\mathrm{i}\infty}_{\sigma-\mathrm{i}%
\infty}x^{-s}\mathscr{M}\mskip-3.0muf\mskip 3.0mu\left(s\right)\,\mathrm{d}s.

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-3.0mug\mskip 3.0mu\left(\sigma+\mathrm{i}t\right) or \mathscr{M}\mskip-3.0muf\mskip 3.0mu\left(1-\sigma-\mathrm{i}t\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\mathrm{i}}\*\int^{\sigma%
+\mathrm{i}\infty}_{\sigma-\mathrm{i}\infty}y^{-s}\mathscr{M}\mskip-3.0muf%
\mskip 3.0mu\left(1-s\right)\mathscr{M}\mskip-3.0mug\mskip 3.0mu\left(s\right)%
\,\mathrm{d}s,
1.14.37 \int^{\infty}_{0}f(x)g(x)\,\mathrm{d}x=\frac{1}{2\pi\mathrm{i}}\*\int^{\sigma+%
\mathrm{i}\infty}_{\sigma-\mathrm{i}\infty}\mathscr{M}\mskip-3.0muf\mskip 3.0%
mu\left(1-s\right)\mathscr{M}\mskip-3.0mug\mskip 3.0mu\left(s\right)\,\mathrm{%
d}s.

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-3.0muf\mskip 3.0mu\left(\tfrac{1}{2}+\mathrm{i}t%
\right)\right|}^{2}\,\mathrm{d}t.

Convolution

Let

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

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

1.14.40 \int^{\infty}_{0}x^{s-1}(f*g)(x)\,\mathrm{d}x=\mathscr{M}\mskip-3.0muf\mskip 3%
.0mu\left(s\right)\mathscr{M}\mskip-3.0mug\mskip 3.0mu\left(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 \mathcal{H}\left(f\right)\left(x\right)=\mathcal{H}\mskip-3.0muf\mskip 3.0mu%
\left(x\right)=\frac{1}{\pi}\pvint^{\infty}_{-\infty}\frac{f(t)}{t-x}\,\mathrm%
{d}t,
1.14.42 \mathcal{H}\mskip-3.0muf\mskip 3.0mu\left(x\right)=\lim_{y\to 0+}\frac{1}{\pi}%
\int^{\infty}_{-\infty}\frac{t-x}{(t-x)^{2}+y^{2}}f(t)\,\mathrm{d}t,
1.14.43 \mathcal{H}\mskip-3.0muf\mskip 3.0mu\left(x\right)=\lim_{\epsilon\to 0+}\frac{%
1}{\pi}\int^{\infty}_{\epsilon}\frac{f(x+t)-f(x-t)}{t}\,\mathrm{d}t.

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-3.0muf%
\mskip 3.0mu\left(u\right)}{u-x}\,\mathrm{d}u.

Inequalities

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

1.14.45 \int^{\infty}_{-\infty}{\left|\mathcal{H}\mskip-3.0muf\mskip 3.0mu\left(x%
\right)\right|}^{p}\,\mathrm{d}x\leq A_{p}\int^{\infty}_{-\infty}{\left|f(t)%
\right|}^{p}\,\mathrm{d}t,

where A_{p}=\tan\left(\tfrac{1}{2}\pi/p\right) when 1<p\leq 2, 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-3.0muf\mskip 3.0%
mu\left(u\right){\mathrm{e}}^{\mathrm{i}ux}\,\mathrm{d}u=-\mathrm{i}(%
\operatorname{sign}x)\mathscr{F}\mskip-3.0muf\mskip 3.0mu\left(x\right),

where \mathscr{F}\mskip-3.0muf\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-3.0muf\mskip 3.0mu%
\left(s\right)=\int^{\infty}_{0}\frac{f(t)}{s+t}\,\mathrm{d}t.

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-3.0muf\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-3.0muf\mskip 3.0mu%
\left(s\right)=(-1)^{m}m!\int^{\infty}_{0}\frac{f(t)\,\mathrm{d}t}{(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{\mathcal{S}\mskip-3.0muf\mskip 3.0mu\left(-\sigma-\mathrm{%
i}t\right)-\mathcal{S}\mskip-3.0muf\mskip 3.0mu\left(-\sigma+\mathrm{i}t\right%
)}{2\pi\mathrm{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 \mathcal{S}\left(f\right)=\mathscr{L}\left(\mathscr{L}\left(f\right)\right).

§1.14(vii) Tables

Table 1.14.4: Laplace transforms.
f(t) \displaystyle\int^{\infty}_{0}{\mathrm{e}}^{-st}f(t)\,\mathrm{d}t
1 \dfrac{1}{s}, \Re s>0
\dfrac{t^{n}}{n!} \dfrac{1}{s^{n+1}}, \Re s>0
\dfrac{1}{\sqrt{\pi t}} \dfrac{1}{\sqrt{s}}, \Re s>0
{\mathrm{e}}^{-at} \dfrac{1}{s+a}, \Re(s+a)>0
\dfrac{t^{n}{\mathrm{e}}^{-at}}{n!} \dfrac{1}{(s+a)^{n+1}}, \Re(s+a)>0
\dfrac{{\mathrm{e}}^{-at}-{\mathrm{e}}^{-bt}}{b-a} \dfrac{1}{(s+a)(s+b)}, a\not=b, \Re s>-\Re a, \Re s>-\Re b
\sin\left(at\right) \dfrac{a}{s^{2}+a^{2}}, \Re s>\left|\Im a\right|
\cos\left(at\right) \dfrac{s}{s^{2}+a^{2}}, \Re s>\left|\Im a\right|
\sinh\left(at\right) \dfrac{a}{s^{2}-a^{2}}, \Re s>\left|\Re a\right|
\cosh\left(at\right) \dfrac{s}{s^{2}-a^{2}}, \Re s>\left|\Re a\right|
t\sin\left(at\right) \dfrac{2as}{(s^{2}+a^{2})^{2}}, \Re s>\left|\Im a\right|
t\cos\left(at\right) \dfrac{s^{2}-a^{2}}{(s^{2}+a^{2})^{2}}, \Re s>\left|\Im a\right|
\dfrac{{\mathrm{e}}^{-bt}-{\mathrm{e}}^{-at}}{t} \displaystyle{\ln\left(\frac{s+a}{s+b}\right)}, \Re s>-\Re a, \Re s>-\Re b
\dfrac{2(1-\cosh\left(at\right))}{t} \displaystyle{\ln\left(1-\frac{a^{2}}{s^{2}}\right)}, \Re(s+a)>0
\dfrac{2(1-\cos\left(at\right))}{t} \displaystyle{\ln\left(1+\frac{a^{2}}{s^{2}}\right)}, \Re s>0
\dfrac{\sin\left(at\right)}{t} \operatorname{arctan}\left(\dfrac{a}{s}\right), \Re s>0

§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 (2015), Marichev (1983), Oberhettinger (1972, 1974, 1990), Oberhettinger and Badii (1973), Oberhettinger and Higgins (1961), Prudnikov et al. (1986a, b, 1990, 1992a, 1992b).