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Suppose $h$ is continuously differentiable on $[0,L]$. For $0<x\le L$, the improper integral $g(x)={\int }_0^{x}h(t)(x-t{)}^{-1/2}dt$ exists, but substituting directly into the ordinary Leibniz integral rule produces an undefined endpoint term. First use integration by parts to obtainThe new endpoint term at $t=x$ vanishes, and the differentiated kernel is locally integrable. Differentiating gives the displayed formula. One rigorous justification rewrites the second term as ${\int }_0^x[{\int }_0^s{h}^{\prime }(t)(s-t{)}^{-1/2}dt]ds$ by Fubini's theorem; the inner integral is continuous for $s>0$, so the fundamental theorem of calculus applies. Moreover $g(x)=2h(0)\sqrt{x}+O(x^{3/2})$: if $h(0)\ne 0$, the right difference quotient at zero diverges rather than defining a finite derivative.

Suppose $q(t)=A/(t-c)+r(t)$, where $r$ is continuous, the pole $c$ is fixed, and differentiable endpoints obey $a(x)<c<b(x)$ locally. ThenApply the chain rule and the fundamental theorem of calculus to getThe same boundary expression as in the Leibniz integral rule thus applies to this explicitly specified Cauchy principal value. If $A\ne 0$, the two ordinary one-sided improper integrals diverge, so their sum is not an ordinary integral. A finite boundary derivative alone does not make that divergent integral meaningful.

Let $F(a)$ be the given integral. Its integrand extends continuously to $x=0$, with value $a$. The Leibniz integral rule includes both the moving upper endpoint and the derivative of the integrand. For $a\ne 0$ it givesConsequentlyThe zero case follows directly from $F(a)=a^3+O(a^9)$ near zero: expand $\sin (ax)/x=a+O(a^3{x}^2)$ and integrate up to $a^2$. Equivalently, the change of variables $t=ax$ gives $F(a)={\int }_0^{a^3}(\sin t)/tdt$, whose derivative is the same expression, interpreted continuously at zero. This substitution works for negative $a$ as an oriented integral too.

For $x>0$, the kernel is smooth between the two positive endpoints. The Leibniz integral rule gives both endpoint terms and the parameter derivative:The remaining integral equals $[\cos (x^3)-\cos (x^4)]/x$, soThe substitution $u=xt$ gives the same result directly from limits $x^3,x^4$. Negative $x$ would make the original interval cross the nonintegrable $1/t$ singularity, so it is not part of this ordinary-integral domain.

The kernel has an integrable singularity at the moving endpoint $t=x$. Direct substitution into the ordinary Leibniz integral rule would produce the undefined boundary value $h(x)/\sqrt{x-x}$ and the nonintegrable differentiated kernel $-h(t)/[2(x-t{)}^{3/2}]$. Those separately divergent expressions cannot be treated as ordinary finite terms. The derivative of an Abel integral is obtained by rewriting the integral before differentiating.

The Leibniz integral rule accounts for both the moving upper endpoint and the integrand's dependence on $a$:For the specified integrand this gives $I^{\prime }(a)=e^{-a^2}-{\int }_0^{a}x{e}^{-ax}dx$. When $a\ne 0$, integration by parts givessoFor the requested independent check, direct integration gives $I(a)=(1-e^{-a^2})/a$ at $a\ne 0$. Its derivative is $[2a^2{e}^{-a^2}-(1-e^{-a^2})]/a^2$, agreeing with the result above. At zero the apparent singularity is removable: $I(a)=a-a^3/2+O(a^5)$, so $\boxed{I(0)=0,I^{\prime }(0)=1}$.

Under the regularity and integrability conditions for the Leibniz integral rule,The first two terms account for the moving endpoints; the last accounts for the integrand's own parameter dependence.

The integrand and both limits depend on $x$, so use the Leibniz integral rule. The upper and lower endpoint contributions and the interior derivative giveFor $x\ne 0$, an antiderivative in $t$ is $t\sin (xt)/x+\cos (xt)/x^2$. Evaluating it and collecting terms givesAt $x=0$ the original Leibniz expression has zero endpoint terms and an integral over a zero-length interval, giving $\boxed{F^{\prime }(0)=0}$. The displayed quotient has a removable singularity and the same limiting value.

Apply the Leibniz integral rule, including both moving boundaries and the explicit parameter in the integrand. The upper bound has derivative $-\sin x$, the lower bound has derivative $\cos x$, and the integrand has partial derivative $-t^4{e}^{-x{t}^4}$ with respect to $x$. ThereforeThe formula remains valid when the bounds coincide or occur in reversed order, by the oriented definite integral convention. The integrand and its partial derivative are continuous on every finite integration interval.

Apply the Leibniz integral rule: differentiate both moving endpoints and the integrand's explicit $x$ dependence. The partial derivative is ${\partial }_x\sin (x{t}^2)=t^2\cos (x{t}^2)$. HenceThe signs distinguish upper and lower endpoint contributions. The formula also holds when the limits occur in reversed order, by the oriented-integral convention. If $\alpha =\beta$, the integral defining $f$ is identically zero and all displayed terms cancel, so

Under the usual regularity assumptions, the Leibniz integral rule isThere is a genuine domain problem in the following integral as printed in the PDF. PutThe point $t=2$ is a simple pole, withFor every $x>4$, the lower endpoint $\sin (1/x)$ is less than $2$ and the upper endpoint $\sqrt{x}$ is greater than $2$. The interval therefore crosses the pole. Since $q(t)=33/[7(t-2)]+O(1)$ near $2$, the two one-sided improper integrals diverge logarithmically, with opposite signs. Their divergences do not define an ordinary integral. Consequently the ordinary limit in the question is undefined as written: its underlying integral has no finite value for any $x\ge 4$.

A separately specified Cauchy principal value does exist for $x>4$. Under that interpretation, differentiating a principal-value integral across a fixed simple pole givesThe first term isand the second tends to zero because $q$ is bounded near $t=0$. Thus the principal-value version has limit $\boxed{1}$. The two conclusions are distinct: this regularised answer requires the additional principal-value convention, which the printed question does not state.

The Leibniz integral rule combines contributions from both moving endpoints with differentiation under the integral sign:The partial derivative in the last integral holds $t$ fixed. Suitable regularity and convergence are assumed as stipulated.

For the Leibniz integral rule, the upper endpoint has derivative $2\pi x$ and the lower endpoint has derivative $1$. The integrand has partial derivative $4xt+\sin t$. ThereforeThe last integral is $[2x{t}^2-\cos t{]}_x^{\pi x^2}=2{\pi }^2{x}^5-2x^3-\cos (\pi x^2)+\cos x$. Combining terms gives

Differentiate the integral formula using the Leibniz integral rule. Its two moving-endpoint terms cancel, leavingEach term before subtraction lies between zero and $m/(3x)$. ThereforeThe bounds are sharp at any fixed positive $x$: put $f=m$ below that point and zero above it for the upper bound, or reverse those choices for the lower bound. If continuity of the forcing is required, smooth approximations approach the same bounds.

For causal functions the convolution is $(f\ast g)(t)={\int }_0^{t}f(\tau )g(t-\tau )d\tau$. In a half-plane of absolute convergence, changing variables $u=t-\tau$ in its Laplace transform and using Fubini's theorem givesThis proves the convolution theorem for Laplace transforms. Transforming the differential equation gives $(s^2+1)Y(s)=sy(0)+y^{\prime }(0)+F(s)$. The direct transform pairs $\mathcal{L}(\cos t)=s/(s^2+1)$ and $\mathcal{L}(\sin t)=1/(s^2+1)$ then giveFor a continuous forcing, call the last term $w(t)$. The Leibniz integral rule gives $w^{\prime }={\int }_0^{t}f(\tau )\cos (t-\tau )d\tau$ and $w^{\prime \prime }=f(t)-w(t)$. Also $w(0)=w^{\prime }(0)=0$. Thus both the original equation and the initial values hold.

For an impulse at positive time $T$, the Dirac delta evaluates the convolution, yieldingHere $H$ is the Heaviside step function. The displacement remains continuous at $T$, while its derivative jumps upward by one, exactly as required by integrating the differential equation across the unit impulse.

Write $H(x^{\prime })=h^{\prime }(x^{\prime })$ and $p_x^{\prime }=d{p}^{\prime }/d{x}^{\prime }$ below, so the subscript $x$ in this shorthand denotes differentiation with respect to $x^{\prime }$. The vertical pressure gradient is zero, and integrating $u_{z^{\prime }{z}^{\prime }}^{\prime }=p_x^{\prime }$ twice with the wall boundary conditions gives the local Couette-Poiseuille flowThe first two terms are Couette flow due to the moving plate; the last is plane Poiseuille flow due to the local pressure gradient. Its dimensionless volume flux per unit width isUse the Leibniz integral rule and incompressibility to differentiate this integral:Thus the flux is independent of position, andThe normal velocity can then be recovered from incompressibility:It is zero on the lower wall by construction and zero on the upper wall by the constant-volume flux relation. This supplies the remaining component of the leading velocity field.

The endpoint pressure conditions are matching conditions, not an assertion that the lubrication approximation remains valid at arbitrarily large dimensional $x$. On an outer length scale $R$, a viscous pressure variation is of size $\mu U/R$, whose ratio to the inner scale isEven an outer inertial pressure scale $\rho U^2$ has ratioConsequently the gap pressure is much larger than either outer variation. Matching to fluid with a common ambient pressure at both ends, and taking that value as zero gauge pressure, givesThere is an overlap region $1\ll |x^{\prime }|\ll {\delta }^{-1}$ before the wall slope becomes order one. The infinite endpoints represent matching through that region. An externally imposed pressure difference comparable with $P$ would change these conditions and the resulting volume flux; equal outer pressure is the physical assumption being used here.

Quadratic energy loss from centred Gaussian gate timing error is the leading decrease in collected energy when a symmetric gate is displaced slightly from the centre of a Gaussian pulse. At the optimum the first-order change vanishes, but the second-order change is negative. The Gaussian pulse time-gate energy fraction gives the exact displaced-gate integral; here the local loss and its effect on jitter are derived explicitly.

Let $p(t)=e^{-t^2/(2{\sigma }^2)}/(\sqrt{2\pi }\sigma )$ be the centred energy-normalized pulse profile, let $a=T_g/2>0$, and setThe Leibniz integral rule gives $F^{\prime }(d)=p(d+a)-p(d-a)$. Since $p$ is even and $p^{\prime }(t)=-tp(t)/{\sigma }^2$,The Taylor expansion therefore yieldsThis is an absolute fraction of the full pulse energy. The relative loss compared with the centred gate is instead $[F(0)-F(d)]/F(0)$.

For a gate as wide as the power FWHM $\tau =2\sqrt{2\ln 2}\sigma$, $F(0)=0.760968$ and $C{\tau }^2=1.302337$. An offset $d=0.05\tau$ gives leading loss $0.00325584$, or about $0.428$ percent of the centred gate's collected energy. Neither the full pulse energy nor the gate width has changed.

A zero mean offset does not eliminate this loss. For a sufficiently narrow random offset $D$ with mean ${\mu }_D$ and variance ${\sigma }_D^2$, the leading expectation isIf $D$ has a centred normal distribution, the quadratic model also gives $\mathrm{SD}(F(D))\simeq \sqrt{2}C{\sigma }_D^2$. At ${\sigma }_D=0.05\tau$, the leading mean loss is $0.00325584$ and this standard deviation is $0.00460446$. These formulas neglect fourth-order effects; they are local approximations, not bounds for large timing excursions.

An apparatus check scans the gate delay about its optimum and compares the measured decrease with the model, while monitoring gate width and the pulse. A gate on the detector's electrical output need not act directly on optical power: the detector impulse response and readout convolution must be included. NIST's impulse-response measurements illustrate that distinction.^[ref]