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If $m\ge 1$ is an integer and $\mathbb{E}[|X{|}^m]<\infty$, the characteristic function has continuous derivatives through order $m$, withThus finite absolute moments justify recovering moments from derivatives.^[ref]

For a short justification of differentiation under the integral sign, the difference quotient for the $(k-1)$st derivative has integrand $(iX{)}^{k-1}{e}^{itX}(e^{ihX}-1)/h$. Its absolute value is at most $|X{|}^k$, since $|e^{iu}-1|\le |u|$ for real $u$. Lower absolute moments are finite because $|X{|}^k\le 1+|X{|}^m$. Apply the dominated convergence theorem successively, then use the same integrable bound to obtain continuity of each derivative.

For the Bernoulli distribution with parameter $1/3$, direct differentiation gives ${\phi }_B^{\prime }(0)=i/3$ and ${\phi }_B^{\prime \prime }(0)=-1/3$, agreeing with $\mathbb{E}[B]=\mathbb{E}[B^2]=1/3$. Its variance is therefore $1/3-(1/3{)}^2=2/9$. Existence and continuity of a characteristic function alone do not authorize these moment calculations: the standard Cauchy distribution, whose expected value is undefined, still has a characteristic function. The stated finite absolute moment hypothesis is a sufficient condition for the derivative formula, not an assumption built into the definition of a characteristic function.

Under fixed-support regularity permitting differentiation under the integral sign, apply the product rule to $\int s_{\lambda }{W}_{\lambda }$. The first derivative term is $\int W_{\lambda }\nabla s_{\lambda }=\mathbb{E}[W_{\lambda }\nabla \log s_{\lambda }]$ and the second is $\mathbb{E}[\nabla W_{\lambda }]$. This proves the displayed identity. For an evidence lower bound, take $W_{\lambda }=\log p(t_{\lambda }(\epsilon ),x)-\log q_{\lambda }(t_{\lambda }(\epsilon ))$, holding noise $\epsilon$ fixed in the inner derivatives. Parameter-independent continuous noise contributes a reparameterization gradient; untransformed discrete choices contribute a score-function gradient estimator. A mixed base law cannot be called parameter independent if its discrete factors retain variational parameters.^[ref]

For the angular-frequency Fourier transform and Schwartz seminorms $p_{m,n}(\varphi )={\sup }_x|x^m{\varphi }^{(n)}(x)|$, every output seminorm is bounded by finitely many input seminorms. To prove this, differentiation under the integral sign and integration by parts giveThe generalized Leibniz rule, $\Vert \hat{g}{\Vert }_{\infty }\le \Vert g{\Vert }_1$, and $\Vert x^a{\varphi }^{(b)}{\Vert }_1\le \pi (p_{a,b}+p_{a+2,b})$ implyThis proves continuity of the Fourier transform on the Schwartz space and rapid decay of every transformed derivative.

The fundamental strip of a Mellin transform is the interior of the set of real parts $\sigma$ for whichViewed in the complex plane, it is a vertical strip $\alpha <\mathrm{Re}s<\beta$; either bound may be infinite, and the interior may be empty. Both endpoints of the improper integral must be checked. Convergence on a boundary line, whether absolute convergence or conditional convergence, requires a separate test. The Mellin transform is holomorphic throughout a nonempty fundamental strip.^[ref]

A practical sufficient test uses Big O notation: if $f(x)=O(x^p)$ near zero and $f(x)=O(x^{-q})$ near infinity, with real $p,q$, then comparison with elementary improper integrals givesThus $-p<\sigma <q$ guarantees absolute convergence. These upper bounds on $|f|$ need not locate the maximal Mellin fundamental strip; faster actual decay can widen it. The strip property itself follows by bounding the weight at any intermediate $\sigma$ by the sum of two weights already known to be integrable. On a closed narrower strip, neighboring integrable weights also dominate the natural logarithm factors introduced by differentiation under the integral sign, making the transform a holomorphic function.

An exact two-endpoint example isHere both bounds are sharp: at $\sigma =-2$ the zero-end absolute convergence test contains ${\int }_0^{1}dx/x$, and at $\sigma =3$ the infinity-end test contains ${\int }_1^{\infty }dx/x$. The displayed rational function continues meromorphically outside the Mellin fundamental strip and has simple poles at $-2$ and $3$. Its other finite continued values outside the strip are values of a meromorphic continuation, not of a convergent defining integral. More general endpoint information leads to meromorphic continuation of a Mellin transform from an asymptotic expansion.

On a star-shaped neighbourhood of $a$, a smooth function satisfiesThe fundamental theorem of calculus along the segment proves the identity, and differentiation under the integral sign proves smoothness of the coefficients.

Suppose $\phi$ is continuous at $x$ and, for some $T>0$,For $0<t<T$, its convolution with the heat kernel $K_t(s)=(4\pi t{)}^{-1/2}{e}^{-s^2/(4t)}$ is smooth and solves the heat equation. On every compact set in $\mathbb{R}\times (0,T)$, spatial or time derivatives of the heat kernel are bounded by the displayed integrable weight times a constant: their polynomial factors are absorbed by the strictly narrower Gaussian. This justifies differentiation under the integral sign.

Moreover, $\int K_t=1$, soInside $|\xi -x|<\delta$, continuity makes the difference uniformly small. Outside that interval, for $t\le T/4$, the inequality ${\xi }^2\le 2(\xi -x{)}^2+2x^2$ givesThe right side tends to zero, proving that the weighted tail contribution vanishes; the constant term $\phi (x)$ is controlled by the vanishing Gaussian tail mass. Thus $(K_t\ast \phi )(x)\to \phi (x)$ as $t\downarrow 0$. In particular, continuous data of polynomial growth satisfy this hypothesis, even when they are unbounded. This extends the Gaussian approximate identity argument beyond bounded data.

Let $\phi$ be continuous and suppose that the displayed improper integral converges separately at both infinite endpoints for some $T>0$. Absolute convergence is not required. Then its heat-kernel solution exists, is smooth and solves the heat equation for $0<t<T$, and tends to $\phi (x)$ as $t\downarrow 0$ at each fixed $x$.

To prove the smoothing claim, take a primitive $H$ of $\phi (\xi )e^{-{\xi }^2/(4T)}$. The assumed convergence makes $H$ bounded. For $q(x,\xi ,t)=K_t(x-\xi )e^{{\xi }^2/(4T)}$, integration by parts givesThe endpoint term vanishes. The coefficient of ${\xi }^2$ in the exponent of $q$ is $-(T-t)/(4tT)<0$. Its derivatives therefore have Gaussian decay on every compact subset of $\mathbb{R}\times (0,T)$, justifying differentiation under the integral sign in the absolutely convergent right-hand expression. Since $q_t=q_{xx}$, it satisfies the heat equation.

For the initial limit, apply the same construction to $\phi (\xi )-\phi (x)$ and split at $|\xi -x|=\delta$. Continuity controls the local part by its small oscillation times the unit kernel mass. On each remaining tail, $q$ is monotone for sufficiently small $t$: its maximum lies at $\xi =xT/(T-t)$, approaching $x$. If the relevant primitive has absolute bound $M$, integration by parts bounds that tail by $2M$ times the value of $q$ at its finite endpoint. For either endpoint $\xi =x\pm \delta$, that value tends to zero like $t^{-1/2}{e}^{-{\delta }^2/(4t)}$ times a fixed constant. Both tails vanish, proving the limit. Separate endpoint convergence matters here; merely assuming a symmetric Cauchy principal value does not provide the bounded primitive used in this proof.

For differentiable endpoints $a,b$ and a function $f$ with suitable continuous partial derivative in the parameter, the moving-endpoint rule isTo prove it locally, put $H(x,u)={\int }_c^{u}f(x,t)dt$ for a fixed reference endpoint $c$. The original integral is $H(x,b(x))-H(x,a(x))$. Apply the chain rule, use the fundamental theorem of calculus for $H_u=f$, and use differentiation under the integral sign for $H_x$. The two boundary terms account for the moving endpoints, while the remaining integral accounts for variation of the integrand with the parameter. For improper integrals, convergence and a suitable domination condition must be verified before interchanging the derivative and integral.

The natural logarithm has the parameter-integral representationFor $0<t<1$, the identity $(t^x-1)/\ln t={\int }_0^x{t}^{s}ds$ shows convergence: the absolute value is at most $|x|t^{\min (x,0)}$, which is integrable when $x>-1$. At $t=1$, the apparent singularity is removable, with limiting value $x$.

Write the left side as $I(x)$. On a compact parameter interval $c\le x\le d$ with $c>-1$, its differentiated integrand $t^x$ is bounded by the integrable function $t^c$. The dominated convergence theorem justifies differentiation under the integral sign, giving $I^{\prime }(x)={\int }_0^1{t}^{x}dt=1/(1+x)$. Since $I(0)=0$, the fundamental theorem of calculus gives $I(x)=\ln (1+x)$. At $x=-1$, the original improper integral diverges as $-{\int }_{0}dt/[t|\ln t|]$, and it also diverges for $x<-1$.

For real $m>-1$ and integer $n\ge 0$, start from ${\int }_0^1{x}^{m}dx=1/(m+1)$ and differentiate $n$ times with respect to $m$. Each derivative adds one factor $\log x$, givingTo justify differentiation under the integral sign, restrict $m$ to a compact interval with lower endpoint $m_0>-1$. The derivatives are dominated by $x^{m_0}|\log x{|}^n$, which is integrable: $x=e^{-t}$ turns its integral into ${\int }_0^{\infty }{e}^{-(m_0+1)t}{t}^{n}dt$. A power of $t$ is integrable against this decaying exponential. This also explains why $m=-1$ is the convergence threshold.

For real $\alpha$, the displayed improper integral converges and is an even function of $\alpha$. For $\alpha >0$, differentiation under the integral sign and partial fraction decomposition giveFor $\alpha \ne 1$, use $x^2/[(1+{\alpha }^2{x}^2)(1+x^2)]=[1/(1+x^2)-1/(1+{\alpha }^2{x}^2)]/({\alpha }^2-1)$. Integrate the resulting rational functions and extend to $\alpha =1$ by continuity. Integrating in $\alpha$ and using $E(0)=0$ gives $E(\alpha )=\pi \ln (1+\alpha )$ for $\alpha \ge 0$; evenness gives the formula for negative parameters. Thus $E^{\prime }(\alpha )=\pi \mathrm{sgn}(\alpha )/(1+|\alpha |)$ away from zero. At zero the one-sided derivatives are opposite, so $E$ is continuous but not differentiable. Substituting $\alpha =0$ into the pointwise differentiated integrand would therefore be an invalid interchange.

For every real parameter $a$,where $F(0)=0$. For $a>0$, differentiation under the integral sign and a partial fraction decomposition giveSolving this first-order linear differential equation yields $F(a)=a[\pi \ln (1+a)+C]$. To fix the constant, dominated convergence gives ${\lim }_{a\to 0^{+}}F(a)/a={\int }_0^{\infty }\ln (x^2)/(1+x^2)dx=0$; the last absolutely convergent integral changes sign under $x=1/t$. Thus $C=0$. Oddness in $a$ supplies the result for negative parameters. The initial value $F(0)=0$ alone does not determine $C$, since every solution $a[\pi \ln (1+a)+C]$ tends to zero as $a\to 0^{+}$.

For every real $a$, define $I(a)={\int }_0^1\sin (a\log x)/\log xdx$. The apparent endpoint singularity at $x=1$ is removable, with limit $a$, and the bound $|\sin (a\log x)/\log x|\le |a|$ ensures integrability. Differentiation under the integral sign is justified by the integrable bound $|\cos (a\log x)|\le 1$ on the parameter derivative. Thuswhere $x=e^{-t}$ and the last integral is a Laplace transform. Since $I(0)=0$, integrating with respect to $a$ gives $I(a)=\arctan a$. This obtains an integral with a logarithmic denominator by differentiating away that denominator, then restoring the parameter dependence with one elementary antiderivative.

Suppose $\varphi (\lambda )=\mathbb{E}{e}^{\lambda X}<\infty$ for all $\lambda \ge 0$. Under the exponential tilting $d{\mu }^{\lambda }=e^{\lambda x}d\mu /\varphi (\lambda )$, $X$ is an integrable random variable whenever $\lambda >0$: its negative exponential tail is bounded, and its positive contribution is dominated by a larger positive exponential moment. Differentiation under the integral sign gives the displayed mean. At the endpoint $\lambda =0$, the positive part of $X$ is integrable, but the negative part need not be. The right derivative ${\varphi }_{+}^{\prime }(0)=\mathbb{E}X$ is valid as an extended value, possibly $-\infty$; a finite mean at zero requires an extra negative-tail integrability assumption.

For real $|k|<1$, define $E(\varphi ,k)={\int }_0^{\varphi }\sqrt{1-k^2{\sin }^{2}x}dx$ and $F(\varphi ,k)={\int }_0^{\varphi }(1-k^2{\sin }^{2}x{)}^{-1/2}dx$. Smoothness of the integrands permits differentiation under the integral sign. Subtracting the two integrands gives $E-F=-k^2{\int }_0^{\varphi }{\sin }^{2}x/\sqrt{1-k^2{\sin }^{2}x}dx$, which is $k{\partial }_{k}E$. This proves the displayed identity for $k\ne 0$. At $k=0$, both sides extend continuously with value zero. The variable $k$ is the modulus; derivatives with respect to the parameter $m=k^2$ instead obey ${\partial }_{m}E=(E-F)/(2m)$, with removable value $-\frac{1}{2}{\int }_0^{\varphi }{\sin }^{2}xdx$ at zero.

In the Laplace integral method for a differential equation, differentiate the integral three times with respect to $z$, and replace $z{e}^{z\zeta }$ by ${\partial }_{\zeta }{e}^{z\zeta }$. Integration by parts givesThus the weight equation for this construction and its boundary condition arewhich giveThese are the conditions obtained by eliminating the transformed interior term and the endpoint term separately. As usual for this method, differentiation and integration by parts need convergence assumptions; the contours constructed next make every such step valid locally uniformly for all complex $z$. A literal necessity claim for arbitrary weights and contours would additionally require uniqueness of the chosen representation by a Laplace integral; a null weight in a contour Laplace representation can otherwise be added without changing the solution.

LetThe quartic decay sectors for a contour integral contain the four rays $\zeta =r{u}_j$. Along each ray, the quartic term is ${\zeta }^4/4=-r^4/4$. For $j=1,2,3$, choose ${\Gamma }_j$ to travel from infinity on ray $u_0$ to zero, and then from zero to infinity on ray $u_j$. Set $A=1$ and defineOn any compact set of $z$ values, the integrand and each of its $z$ derivatives are bounded by a constant times $r^m{e}^{-r^4/4+Mr}$ for some $M$. These bounds are integrable at infinity, so differentiation under the integral sign is justified and the functions are entire. The endpoint term vanishes at both infinite ends. There is no extra term at zero because the two pieces join with the same integrand value and opposite endpoint orientations. Therefore all three functions solve the differential equation.

To prove independence, evaluate their first three initial data. For $m=0,1,2$,If ${\sum }_{j=1}^3{c}_j{w}_j=0$, these initial data give ${\sum }_{j=1}^3{c}_j(u_j^{\ell }-u_0^{\ell })=0$ for $\ell =1,2,3$. Define $c_0=-{\sum }_{j=1}^3{c}_j$. ThenThe coefficient matrix is a Vandermonde matrix at four distinct fourth roots of $-1$, and hence is invertible. All $c_j$ vanish. There is also a concrete reason for the earlier qualification about necessity. Deform the incoming ray of ${\Gamma }_1$ to angle $\pi /3$ and its outgoing ray to angle $2\pi /3$, within their respective quartic decay sectors. The quartic integral is unchanged and remains nonzero. On this deformed contour, the extra entire weight $e^{{\zeta }^2}$ contributes zero for every $z$: throughout the closing wedge, $\mathrm{Re}({\zeta }^2)\le -r^2/2$, so the large closing arc tends to zero and the Cauchy integral theorem gives a zero integral. Thus $g=e^{{\zeta }^4/4}+e^{{\zeta }^2}$ represents the same entire solution and has vanishing endpoint terms, although it does not satisfy $g^{\prime }={\zeta }^{3}g$. The canonical weight equation is the standard sufficient construction, rather than an unconditional uniqueness assertion about contour representations.

Thus the three contours give linearly independent nonzero solutions, and they form a fundamental system of solutions for this third-order linear equation. This proves independence directly instead of inferring it merely from the presence of three different contours.

For $|z|<r$, the Schwarz integral on the unit disk, rescaled to radius $r$, isOn each smaller closed disk $|z|\le s<r$, its kernel and all its derivatives in $z$ are uniformly bounded. Since the boundary values of the real harmonic function $u$ are continuous on this circle, differentiation under the integral sign proves that $g$ is holomorphic. Taking the real part gives the Poisson integral of those boundary values. It agrees with $u$ by uniqueness of the Dirichlet problem on the radius-$r$ disk. Hence $g$ is holomorphic and $\mathrm{Re}g=u$ on $|z|<r$. Its imaginary part is the harmonic conjugate normalized by $\mathrm{Im}g(0)=0$.

Use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$, with complex-linear distribution pairings throughout. The Riemann-Lebesgue lemma says that, for an integrable function $f\in L^1(\mathbb{R})$, its Fourier transform is continuous and tends to zero as $|\xi |\to \infty$. It also satisfies $\Vert \hat{f}{\Vert }_{\infty }\le \Vert f{\Vert }_1$. The product rule, or convolution theorem, isHere $(f\ast g)(x)=\int f(y)g(x-y)dy$ belongs to $L^1(\mathbb{R})$ by Tonelli's theorem, and Fubini's theorem proves the displayed identity. The companion Fourier product formula, for Schwartz functions, is $\widehat{fg}=(2\pi {)}^{-1}\hat{f}\ast \hat{g}$.

A Schwartz function is a smooth function $\varphi :\mathbb{R}\to \mathbb{C}$ for which every Schwartz seminormis finite. These seminorms define the topology of the Schwartz space $\mathcal{S}(\mathbb{R})$. In particular,Thus every Schwartz function is an integrable function.

The Fourier bounds for Schwartz functions assert that each Schwartz seminorm of $\hat{\varphi }$ is bounded by finitely many Schwartz seminorms of $\varphi$. Explicitly, differentiation under the integral sign and integration by parts giveso the generalized Leibniz rule and the preceding integrability estimate giveIn particular the Fourier transform maps the Schwartz space continuously into itself.

The distributional derivative is defined by $⟨u^{\prime },\varphi ⟩=-⟨u,{\varphi }^{\prime }⟩$. Since differentiation under the integral sign gives $(\hat{\varphi }{)}^{\prime }=\widehat{-i\xi \varphi }$, we obtainThis proves the Fourier transform of a derivative identity $\boxed{\widehat{u^{\prime }}=i\xi \hat{u}}$ for all tempered distributions.

If $u,u^{\prime }\in H^s(\mathbb{R})$, their Fourier transforms are represented by locally square-integrable functions, and the identity identifies $\widehat{u^{\prime }}$ with $i\xi \hat{u}$ almost everywhere. Hence the Sobolev norms satisfyso $\boxed{u\in H^{s+1}(\mathbb{R}),\Vert u{\Vert }_{H^{s+1}}^2=\Vert u{\Vert }_{H^s}^2+\Vert u^{\prime }{\Vert }_{H^s}^2}$ for every real $s$.

We prove the required differentiation under the integral sign directly. Fix $t_0>0$, and take a compact parameter interval $J\subseteq [0,\infty )$ containing $t_0$ in its interior. The jointly continuous partial derivative ${\partial }_{t}f$ is bounded on the compact rectangle $J\times [0,1]$ by some $M_1$.

For nonzero $h$ sufficiently close to zero, defineFor each $x$, the mean value theorem in the parameter gives $|D_h(x)|\le M_1$. The definition of the partial derivative also gives $D_h(x)\to {\partial }_{t}f(t_0,x)$. Since $M_1$ is integrable on $[0,1]$, dominated convergence applied along any sequence $h_n\to 0$ yieldsThusAt $t=0$, repeat the argument on $J=[0,\delta ]$ with $h>0$. It proves that the right derivative exists and satisfies the same formula at zero. This is the appropriate meaning of differentiability on the closed half-line. In fact $F^{\prime }$ is continuous there: apply part (i) to the continuous function ${\partial }_{t}f$.

The proof establishes differentiation of parameter integrals on compact rectangles under the exact regularity assumptions used here. In particular, it supplies the theorem's proof instead of assuming an interchange of a limit and an integral.