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The geometric series gives $Q_r(t)=2{\sum }_{n\ge 1}{r}^n\sin nt$. Its convolution multiplier is $-i\mathrm{sgn}(n)r^{|n|}$ for nonzero frequencies and zero at frequency zero. Since the series is uniformly absolutely convergent for $r<1$, these identities hold for every Lebesgue integrable function by termwise integration. If all negative Fourier coefficients vanish, $Q_r\ast f=-i(P_r\ast f-\hat{f}(0))$.

Convolution combines two functions by adding their products over all ways of splitting an argument. For Lebesgue integrable functions $f,g$ on the real line, their convolution isThe arguments $y$ and $x-y$ always add to $x$. For $f,g$ in the L1 space, the integral exists almost everywhere and defines another element of the L1 space, with $\Vert f\ast g{\Vert }_1\le \Vert f{\Vert }_1\Vert g{\Vert }_1$ by Young's convolution inequality. Particular examples below have a value at every $x$.

To see what the integral counts, take the indicator function $q=1_{[0,1]}$. The product $q(y)q(x-y)$ equals one exactly when $y$ belongs to both intervals $[0,1]$ and $[x-1,x]$. Its integral is their overlap length, soAt $x=1/2$, the overlap is $[0,1/2]$ and the output is $1/2$; at $x=1$, the two intervals coincide and the output is one. The result is the shifted triangular function $\mathrm{tri}(x-1)$. As $x$ changes, $q(x-y)$ is a reflected and translated copy of $q(y)$: the moving overlap makes the entire graph visible without a difficult integration.

The same calculation gives the probability distribution of a sum of two independent random variables. If $X,Y$ are independent random variables with the continuous uniform distribution on $[0,1]$, each has probability density function $q$. Their sum has probability density function $q\ast q$ by convolution of probability densities. The probability density function at one is larger than near zero or two because there are more combinations of the two inputs giving that sum. Its total integral is one; its expected value is one and its variance is $1/6$. These follow equally from integrating the triangular function or from linearity of expectation and variance additivity for independent random variables.

Convolution also gives a useful averaging operation. For $\epsilon >0$, let $k_{\epsilon }(y)=1_{[-\epsilon ,\epsilon ]}(y)/(2\epsilon )$. Thenthe mean of $f$ in a neighborhood of $x$. A smaller $\epsilon$ examines a smaller neighborhood. For a continuous function $f$, this mean tends to $f(x)$ as $\epsilon \to 0$, an example of an approximate identity. Integrating a sharp input against a distributed convolution kernel can produce a smoother output; the triangular example turns two discontinuous inputs into a continuous output.

In signal processing, a linear time-invariant system combines shifted copies of its impulse response by the same rule. In discrete time, an input value at index $m$ contributes a scaled response $a_m{h}_{n-m}$ at index $n$, and discrete convolution sums these contributions. This is the superposition interpretation developed in Oppenheim, MIT Signals and Systems, Lecture 4, convolution sums and integrals. The convolution theorem then gives another description: transform both inputs with a Fourier transform and multiply their transforms. Combining shifts, combining independent quantities and multiplying frequency descriptions are three faces of the same operation.

For Lebesgue integrable functions $f,g$ on $\mathbb{R}$, use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-2\pi ix\xi }dx$. ThenThus the integral combining all translated products becomes an ordinary multiplication at each frequency. Different Fourier transform normalizations can place a constant in this identity; fixing the convention prevents that ambiguity. Its use in filtering is illustrated by ^[ref].

For the indicator function $q=1_{[0,1]}$,with $\mathrm{sinc}(0)=1$. Therefore the triangular function $q\ast q$ has Fourier transform $e^{-2\pi i\xi }{\mathrm{sinc}}^2(\xi )$. The extra factor of the sinc function improves the large-frequency bound from a constant times $1/|\xi |$ to one times $1/|\xi {|}^2$, matching the smoother output in the overlap picture. Fourier decay of repeated interval convolutions extends this calculation to further repetitions.

For a proof outline, Fubini's theorem permits exchanging the two integrals because the integral of $|f(y)g(x-y)|$ over $(x,y)$ is $\Vert f{\Vert }_1\Vert g{\Vert }_1<\infty$. Set $u=x-y$ in the inner integral:The factorization of the complex exponential is the mechanism behind the simplification. The discrete convolution version similarly turns coefficient sums into products of generating functions.

For functions $f,g$ on $[0,\infty )$ with $e^{-\sigma t}f(t)$ and $e^{-\sigma t}g(t)$ Lebesgue integrable functions, their unilateral convolution, defined almost everywhere if necessary, satisfiesTaking a causal extension by zero of both functions gives the existing whole-line convolution definition. Absolute convergence and Fubini's theorem justify the product rule: put $t=\tau +u$ in the double integral, so $e^{-st}=e^{-s\tau }{e}^{-su}$. For piecewise continuous functions of exponential order $a$ and $b$, the hypotheses hold for $\mathrm{Re}s>\max (a,b)$. The Laplace transform article works out the concrete example $h(t)=e^{-2t}$ and the Heaviside step function input, whose output is $(1-e^{-2t})/2$. ^[ref].

If $g$ is locally Lebesgue integrable andthen $G^{\prime }(x)=g(x)$ at every Lebesgue point of $g$, and hence almost everywhere.

Let $t_0$ be interior to a parameter interval. Suppose $h(t,\cdot )$ is a measurable function for each $t$, $h(t_0,\cdot )$ is a Lebesgue integrable function, and $h(\cdot ,x)$ is differentiable near $t_0$ for $x$ almost everywhere. If one integrable nonnegative function $D$ satisfies $|{\partial }_{t}h(t,x)|\le D(x)$ throughout that neighbourhood off one fixed null set, then $F(t)={\int }_{X}h(t,x)d\mu (x)$ is defined there and satisfies the displayed derivative formula at $t_0$.

The mean value theorem gives $|h(t,x)-h(t_0,x)|\le |t-t_0|D(x)$, proving local integrability. It also bounds the difference quotient by $D$. That quotient converges almost everywhere to ${\partial }_{t}h(t_0,x)$, so the dominated convergence theorem permits passing the limit through the Lebesgue integral.

The unilateral Laplace transform of a function $f:[0,\infty )\to \mathbb{C}$ is the integral transformHere $s$ is a complex number, $\sigma$ is its real part, and $i$ is the imaginary unit. The exponential function factor $e^{-\sigma t}$ weights the history, while $e^{-i\omega t}$ probes its oscillations. The integral starts at zero, so it uses only the history from that starting time onward. This section concerns ordinary functions; their value at the single endpoint does not change the integral. See MIT 18.03SC, definition and elementary examples.

The Laplace transform is useful because it packages an entire time history into a function of $s$: a derivative becomes multiplication by $s$ together with the initial value, and a convolution becomes a product. Thus a linear differential equation can become an algebraic equation that already remembers its initial value problem. The worked example below shows both descriptions of the same response.

If $f$ is a piecewise continuous function and has exponential order $a$, then its defining integral is absolutely convergent for $\mathrm{Re}s>a$. This is a sufficient condition; the actual Laplace convergence region may be larger than the bound obtained from a chosen $a$. ^[ref].

The bound has a short explanation. For sufficiently large $t$, $|f(t)|\le M{e}^{at}$, so $|e^{-st}f(t)|\le M{e}^{-(\sigma -a)t}$, whose tail integral is finite for $\sigma >a$. The finite initial interval is covered by piecewise continuity.

Three calculations make the convergence condition concrete. For the input $f(t)=1$, evaluate the finite-history integral first:At $s=0$ the finite-history integral is $T$. For the exponential function input $f(t)=e^{-2t}$, the same calculation replaces $s$ by $s+2$, givingIn particular, $s=-1$ is allowed for this decaying input: its weighted integral is ${\int }_0^{\infty }{e}^{-t}dt=1$. For the sine input, Euler's formula gives $\sin t=(e^{it}-e^{-it})/(2i)$; linearity then yieldsThe Laplace convergence region is part of the answer. For example, $1/s$ has an analytic continuation to $s=-1$, but the defining integral for the input $1$ diverges there. The sine calculation is also developed in MIT 18.03SC, further elementary transforms.

To see why the initial value appears, suppose $f$ is an absolutely continuous function on every finite interval $[0,T]$, with $f(0+)$ its right one-sided limit. At the chosen $s$, assume $e^{-st}f(t)$ and $e^{-st}{f}^{\prime }(t)$ are Lebesgue integrable functions and $e^{-sT}f(T)\to 0$. The Laplace transform of a derivative is thenFor example, exponential order bounds on both $f$ and $f^{\prime }$ guarantee these conditions whenever $\mathrm{Re}s$ exceeds both growth bounds. The initial value is the lower-end boundary contribution of integration by parts.

Consider the initial value problemWrite $Y=\mathcal{L}\{y\}$. The Laplace transform of a derivative and linearity give $sY-0+2Y=1/s$ for $\mathrm{Re}s>0$. A partial fraction decomposition therefore givesWe recognize the two elementary Laplace transforms already computed. Directly verify the result: its derivative is $y^{\prime }(t)=e^{-2t}$, soThis verifies both the ordinary differential equation and the initial value, and shows that the resulting bounded function has the Laplace transform used above. The response rises from zero toward $1/2$: at that value the input $1$ balances the term $2y$. The general method is described in MIT 18.03SC, solving initial value problems.

The same solution is a convolution. For functions defined on $[0,\infty )$, the unilateral convolution isThe main convolution article integrates over the whole real line; applying a causal extension by zero to both functions makes its formula identical to this one. Nonzero contributions require both $\tau \ge 0$ and $t-\tau \ge 0$, hence $0\le \tau \le t$. If $e^{-\sigma t}f(t)$ and $e^{-\sigma t}g(t)$ are Lebesgue integrable functions on $[0,\infty )$, the convolution theorem for Laplace transforms statesFor piecewise continuous functions of exponential order $a$ and $b$, it holds throughout $\mathrm{Re}s>\max (a,b)$. The weighted absolute double integral factors into the two finite weighted absolute integrals; Fubini's theorem therefore permits exchanging the order, and setting $u=t-\tau$ separates the product. ^[ref].

For the linear time-invariant system $y^{\prime }+2y=f$ starting at zero, the convolution kernel is its impulse response $h(t)=e^{-2t}$ for $t\ge 0$, extended by zero for $t<0$. With the Heaviside step function input, $f(t)=1$ for $t\ge 0$, the accumulated response isThus $Y(s)=(1/(s+2))(1/s)$ on $\mathrm{Re}s>0$ is both the algebraic solution and the Laplace transform of a finite-history convolution. Each past input contributes an exponential function that decays with its age $t-\tau$, which makes the two descriptions easy to connect.

Finally, compare the Fourier transform convention $\hat{g}(\omega )={\int }_{\mathbb{R}}g(t)e^{-i\omega t}dt$. If $f_0$ is the causal extension by zero of $f$, define $g_{\sigma }(t)=e^{-\sigma t}{f}_0(t)$. Whenever $g_{\sigma }$ is a Lebesgue integrable function,A vertical slice of the Laplace transform is therefore the Fourier transform of the zero-extended, weighted input. Positive $\sigma$ damps its positive-time history. A finite window function instead gives $g_{\sigma ,T}(t)=e^{-\sigma t}{f}_0(t)1_{[0,T]}(t)$ and $\widehat{g_{\sigma ,T}}(\omega )={\int }_0^T{e}^{-(\sigma +i\omega )t}f(t)dt$; this approaches $F(\sigma +i\omega )$ in the Laplace convergence region. Setting $\sigma =0$ directly requires an integrable zero-extended input: the input $1$ fails that condition, even though its Laplace transform exists for every $\sigma >0$.

The history of the Laplace transform spans several stages. Schiff's historical summary notes that Laplace's 1812 treatise acknowledged earlier work on integral transforms and describes later development of the modern Laplace transform method. ^[ref].

Suppose $f$ is an absolutely continuous function on every finite interval $[0,T]$, both $e^{-st}f(t)$ and $e^{-st}{f}^{\prime }(t)$ are Lebesgue integrable functions, and $e^{-sT}f(T)\to 0$. Then integration by parts on $[0,T]$ followed by a limit givesHere $f(0+)$ is the right one-sided limit. Exponential order bounds on $f$ and $f^{\prime }$ suffice when $\mathrm{Re}s$ exceeds both bounds. For $f(t)=e^{-2t}$, the identity reads $-2/(s+2)=s/(s+2)-1$ on $\mathrm{Re}s>-2$. This is the ordinary derivative on nonnegative time; differentiating a causal extension by zero at its jump instead introduces a Dirac delta function. ^[ref].

If $f$ is locally Lebesgue integrable on ${\mathbb{R}}^n$, then for Lebesgue almost every $x$,

A point $x\in {\mathbb{R}}^n$ is a Lebesgue point of a locally Lebesgue integrable function $f$ whenThe Lebesgue differentiation theorem says that almost every point is a Lebesgue point.

Suppose $u$ and $\log (1+x)u$ are Lebesgue integrable functions on $(0,\infty )$. For integers $n\ge 1$, $x+n\le n(1+x)$ gives the displayed L1 norm bound. It tends to zero because $\log n/n\to 0$, proving L1 convergence of the logarithmically weighted functions to zero. Alternatively, $|\log (n+x)u(x)/n|\le (1+\log (1+x))|u(x)|$ and pointwise convergence permit the dominated convergence theorem.

For $u(x)=\sin x/x^q$ with $1<q<2$, both integrability conditions hold. Near zero, $|\sin x|\le x$ and $\log (1+x)\le x$ give integrable bounds by powers. At infinity, $|\sin x|\le 1$ and $\log (1+x)\le \log 2+\log x$, with ${\int }_1^{\infty }{x}^{-q}\log xdx=(q-1{)}^{-2}$. The L1 norm consequently decays as $O(\log n/n)$ in Big O notation.

The change of variables $x=e^u$, using the exponential function, turns the Mellin transform intoThe factor $dx=e^{u}du$ changes $x^{s-1}$ to $e^{su}$; missing this factor would shift every exponent by one. With $h_{\sigma }(u)=e^{\sigma u}f(e^u)$ and the Fourier transform convention $\hat{h}(\omega )={\int }_{\mathbb{R}}h(u)e^{-i\omega u}du$, the identity isChoosing $\sigma$ inside the Mellin fundamental strip makes $h_{\sigma }$ a Lebesgue integrable function. The minus sign in the Fourier transform argument follows from the stated convention. For the Mellin transform of an interval indicator with $[b,c]=[1,4]$ and $\sigma =0$, $h_0=1_{[0,\log 4]}$: the Fourier transform of an interval indicator is literally the same computation in a different coordinate.

Splitting the real line at $u=0$ also connects the Mellin transform to two unilateral Laplace transforms. If $k_{+}(v)=f(e^v)$ and $k_{-}(v)=f(e^{-v})$ for $v\ge 0$, thenwhere both integrals converge absolutely. Their Laplace convergence regions impose opposite endpoint conditions, explaining the two sides of the Mellin fundamental strip. Finally, $f(a{e}^u)=f(e^{u+\log a})$ shows that a change of scale is a translation in $u$. Weighted Fourier transforms of this translation give exactly the factor $a^{-s}$ in the scaling property of the Mellin transform. For a smooth function with compact support inside $(0,\infty )$, the Mellin inversion theorem follows from Fourier inversion; the existing formula states the corresponding hypotheses.

DefineThis is a Lebesgue integrable function: on $(0,1]$ it is bounded by $x^{-1/2}$, whose integral is two, while on $[1,\infty )$ it is bounded by $e^{-x}$, whose integral is $e^{-1}$. Since $|\sin (xy)|\le x|y|$,Thus $h_y$ is Lebesgue integrable for every $y\ge 0$, including $h_0=0$; the same argument actually works for all real $y$.

The mean value theorem for the sine gives $|\sin u-\sin v|\le |u-v|$. Henceand the triangle inequality for an integral givesThus $F$ is a globally Lipschitz function, in particular continuous on $[0,\infty )$, including the endpoint zero.

For the derivative, fix $y>0$ and take nonzero $t$ sufficiently small that $y+t>0$. The difference quotient has the integrandFor every $x>0$, $D_t(x)\to H(x)\cos (xy)$ as $t\to 0$. The same mean value theorem gives the uniform-in-$t$ bound $|D_t(x)|\le H(x)$.

The dominated convergence theorem states that if measurable functions $u_n$ converge almost everywhere to $u$ and $|u_n|\le G$ for a single Lebesgue integrable function $G$, then $u$ is Lebesgue integrable and $\int |u_n-u|\to 0$, hence $\int u_n\to \int u$. Apply it to the difference quotients along every sequence $t_n\to 0$, using the common dominator $H$. The resulting limit is independent of the sequence, soThe right-hand integral is absolutely convergent because its integrand has absolute value at most $H$. This is dominated differentiation under the integral sign, justified by an explicit integrable bound rather than a formal interchange.

For every fixed $x>0$, $\log (x+n)/n\to 0$, so $f_n(x)\to 0$. To exchange the limit with the Lebesgue integral, we need a common integrable bound over the whole half-line.

For $n\ge 1$, $x+n\le n(1+x)$, soTogether with $|\sin x|\le \min (x,1)$, this givesFor $0<x\le 1$, $\log (1+x)\le x\le 1$, hence $G(x)\le 2x^{1-q}$, which is integrable because $q<2$. For $x\ge 1$,Both terms are integrable because $q>1$:The second equality follows from integration by parts, or the substitution $x=e^t$. Consequently $G$ is a Lebesgue integrable function, and every $f_n$ is Lebesgue integrable.

The dominated convergence theorem states that almost everywhere convergence of measurable functions under one integrable absolute-value bound implies L1 convergence, and in particular convergence of their integrals. Here the pointwise limit is zero and the dominator is $G$, soIn fact we obtain the stronger conclusion $\Vert f_n{\Vert }_1\to 0$. A quantitative version follows by retaining the factors $1/n$ in the bound:The preceding endpoint estimates show $A,B<\infty$, so the L1 norm is $O(\log n/n)$ as $n\to \infty$, with $O$ denoting Big O notation. This is logarithmic reweighting under dominated convergence.

The dominated convergence theorem says that if measurable $f_n$ converge almost everywhere to $f$ and $|f_n|\le H$ almost everywhere for a single Lebesgue integrable function $H\ge 0$, then $f$ is integrable andHere, for $x>0$, put $H_0(x)=x^{-1/2}{e}^{-x}$ and $f_n(x)=H_0(x)(1-(-1{)}^n{e}^{-nx})$. The exponential function gives $e^{-nx}\to 0$ at every such $x$, so $f_n\to H_0$. Also $|f_n(x)|\le 2H_0(x)$. This dominator is integrable: on $(0,1)$, $H_0(x)\le x^{-1/2}$, whose integral is two; on $[1,\infty )$, $H_0(x)\le e^{-x}$, whose integral is finite. DCT therefore proves the required limit.

The limiting value can also be evaluated. Substitute $x=t^2$ and use the Gaussian integral:In fact, scaling the second exponential term gives the exact integral $\sqrt{\pi }[1-(-1{)}^n/\sqrt{n+1}]$. The integral error decays as $n^{-1/2}$ despite exponential convergence at each fixed positive $x$; the slow part lies near the endpoint zero.

Use the power series of the exponential function:Each term is a Lebesgue integrable function, andThe sum of these nonnegative integrals is at most $2{\sum }_{n\ge 0}1/n!=2e<\infty$. Applying integration of a series with summable L1 norms, just proved in part (ii), therefore justifies exchanging the series and the integral:The absolute-integral bound is the justification; pointwise convergence of the exponential series alone would not suffice.

The monotone convergence theorem states that, for nonnegative measurable $f_n$ on a Lebesgue measurable set $E$ with $f_n\uparrow f$ almost everywhere,allowing infinity. Fatou's lemma states that, for nonnegative measurable $f_n$,Now write $C={\sup }_n{\int }_E|g_n|<\infty$. A pointwise limit superior of measurable functions is measurable, and it agrees with $g$ off the exceptional null set. By completeness of Lebesgue measure, assigning any values on that null set preserves measurability. Therefore $g$ is measurable. Since $|g_n|\to |g|$ almost everywhere, Fatou's lemma givesThus $g$ is finite almost everywhere and is a Lebesgue integrable function. This is an almost-everywhere limit under a uniform L1 bound. The argument proves integrability of the limit, not convergence in L1 space.

A sufficient theorem is dominated differentiation under the integral sign. Let $t_0$ be interior to a parameter interval, suppose $u(\cdot ,t)$ is measurable for each $t$, and suppose $u(\cdot ,t_0)$ is integrable. Assume $u(x,\cdot )$ is differentiable near $t_0$ for $x$ outside one fixed null set, and that throughout this neighborhoodfor one nonnegative Lebesgue integrable function $H$. Then $F(t)={\int }_{E}u(x,t)dx$ is defined near $t_0$, is differentiable at $t_0$, andIndeed the mean value theorem bounds both $|u(x,t)-u(x,t_0)|$ by $|t-t_0|H(x)$ and the difference quotient by $H(x)$. The first bound proves local integrability; the second lets the dominated convergence theorem pass the limit of difference quotients through the integral.

Again the function is continuous on its domain and therefore a measurable function. For $0<x<1$, $|(\log x)e^{-x}|\le -\log x$, andThis standard improper integral follows from the antiderivative $x-x\log x$ and $x\log x\to 0$ at zero. For $x\ge 1$, $0\le \log x\le x$, and integration by parts gives ${\int }_1^{\infty }x{e}^{-x}dx=2/e$. ThusThe logarithmic singularity is integrable and the exponential factor controls the tail, so this too is a Lebesgue integrable function.

The function is continuous on $(0,\infty )$, hence a measurable function. On $(0,1)$ its absolute value is at most one, while on $[1,\infty )$ the elementary estimate $|\sin u|\le |u|$ gives $|\sin (1/x^3)|\le x^{-3}$. By comparison for the Lebesgue integral,It is therefore a Lebesgue integrable function. The oscillations near zero do not obstruct absolute integrability because their amplitude stays bounded on an interval of finite measure.