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The theorem givesIntegration by parts has zero boundary term: the sine vanishes at zero and the Gaussian tends to zero at infinity. Hence $F^{\prime }(t)=-2t{\int }_0^{\infty }{e}^{-x^2}\cos (2xt)dx=-2tF(t)$.

The same Gaussian majorant and dominated convergence give continuity at zero, where the allowed Gaussian integral gives $F(0)=\sqrt{\pi }/2$. The differential equation implies $(e^{t^2}F(t){)}^{\prime }=0$; continuity at zero fixes that constant. Thus

Write $K[y]={\int }_a^{b}G(x,y,y^{\prime })dx$. At a regular constrained extremal, the first variation of $K$ is not identically zero. Choose a variation $\zeta$ for which $\delta K[\zeta ]\ne 0$; the implicit function theorem supplies a correction in the $\zeta$ direction to each other variation so that $K$ remains one. Stationarity of $J$ on these corrected variations implies $\delta J=\lambda \delta K$ for one scalar $\lambda$.

Put $L=F-\lambda G$. Integration by parts gives, for an arbitrary twice continuously differentiable variation $\eta$,Taking variations with zero endpoint values and using the stated fundamental lemma of the calculus of variations gives the Euler-Lagrange equation. The remaining endpoint values are free and independent, giving the natural boundary conditions:If the constraint is not regular, the general multiplier statement permits an abnormal pair ${\lambda }_{0}F-{\lambda }_{1}G$ with ${\lambda }_0=0$; the source's general normal-form derivation implicitly excludes that case. In its quadratic example, regularity follows automatically from normalization: $\delta K[y]=2\int w{y}^2=2$.

For that example $L=p{y}^{\prime 2}-q{y}^2-\lambda w{y}^2$, so the equation and endpoint conditions reduce toMultiply the equation by $y$ and integrate. The boundary term $[p{y}^{\prime }y{]}_a^b$ vanishes, givingThe last integral equals one, so $\lambda =J[y]$, the functional value at the constrained extremal.

Finally multiply the equation for $y_1$ by $y_2$ and the equation for $y_2$ by $y_1$, then subtract and integrate. The derivative terms give only the vanishing boundary term $[p(y_2{y}_1^{\prime }-y_1{y}_2^{\prime }){]}_a^b$, while the $q$ terms cancel. Therefore $({\lambda }_1-{\lambda }_2){\int }_a^{b}w{y}_1{y}_{2}dx=0$. For distinct multipliers this proves

The central limit theorem states that for independent identically distributed variables of finite mean $\mu$ and positive finite variance ${\sigma }^2$,For a continuous strictly increasing cumulative distribution function, its inverse is defined on $(0,1)$. Thus $\mathbb{P}(F(X)\le y)=\mathbb{P}(X\le F^{-1}(y))=y$ for $0<y<1$, proving the uniform law.

If $T=-\log (1-Y)$, then for $t\ge 0$, $\mathbb{P}(T\le t)=\mathbb{P}(Y\le 1-e^{-t})=1-e^{-t}$. Hence $f_T(t)=e^{-t}{1}_{(0,\infty )}(t)$, an exponential density of rate one. Its first two moments are one and two, by integration by parts, so $\mathbb{E}T=1$ and $\mathrm{Var}T=1$.

Although the original variables may have different distributions, their separate transforms $Y_k=F_k(X_k)$ are independent and all uniform. Consequently $T_k=-\log (1-Y_k)$ are independent identically distributed exponentials of rate one, and the statistic is exactlyThe stated central limit theorem therefore yields $\boxed{Z_n\stackrel{d}{\to }N(0,1)}$.

Substitute $t=\log x$, with $dx=e^{t}dt$. The nonnegative integral becomesby scaling and integration by parts. The exponential decays because $\alpha >1$, with zero boundary term at infinity, proving integrability and

Substitute $t=\log x$, with $dx=e^{t}dt$. The nonnegative integral becomesby scaling and integration by parts. The exponential decays because $\alpha >1$, with zero boundary term at infinity, proving integrability and

The function is measurable and negative on this interval. Its absolute value obeysThe final integral follows by integration by parts and $x^2\log x\to 0$ at zero. The comparison test for integrals proves integrability.

Take a smooth variation $u+\epsilon v$ with $v=0$ on the boundary. Differentiating under the integral gives the first variationIntegration by parts removes the derivatives of $v$, and the boundary contribution vanishes. The fundamental lemma of the calculus of variations therefore givesFor a regular constraint $J[u]=c$, the Lagrange multiplier rule says that an extremal satisfies $\delta I=\lambda \delta J$ for a constant $\lambda$. Equivalently, apply the preceding Euler–Lagrange equation to $F-\lambda G$. Regularity means that $\delta J$ is not identically zero as a linear functional on the admissible variations. For the quadratic constraint below this holds at every nonzero admissible $u$; a zero constraint leaves only $u=0$.

In polar coordinates the constrained integrand isIts Euler–Lagrange equation is $-2\lambda ru-{\partial }_r(2r{u}_r)-{\partial }_{\theta }(2u_{\theta }/r)=0$, orThe field is single-valued and periodic in $\theta$. Substitution $u=f(r)\cos (n\theta )$ with $n\in \{0,1,2,\dots \}$ givesThe centre is part of the disk, so one must also select the regular radial solution: for $n\ge 1$, smoothness gives $f(r)=O(r^n)$ and the regular derivative behaviour; for $n=0$, $f$ is bounded and $f^{\prime }(0)=0$. A singular solution at zero does not define an admissible smooth field on the disk.

For fixed $n$, let $c_n={\int }_0^{2\pi }{\cos }^2(n\theta )d\theta$, so $c_0=2\pi$ and $c_n=\pi$ for $n\ge 1$. The angular derivative term has integral $n^2{c}_n$, including zero when $n=0$. ConsequentlyThe one-dimensional Euler–Lagrange equation for $\hat{I}-\lambda \hat{J}$ isexactly the same radial equation. The common factor $c_n$ cancels.

For two regular solutions with the same $n$ and different multipliers, multiply the equation for $f_1$ by $f_2$, multiply that for $f_2$ by $f_1$, subtract and integrate. The result isAt $a$, both functions vanish; at zero, regularity makes the boundary term tend to zero. Since the multipliers are distinct,This is weighted Sturm–Liouville orthogonality with weight $r$.

Finally, multiply one radial equation by $f$ and integrate by parts. The same endpoint conditions yieldMultiplication by $c_n$ provesFor any nonzero admissible solution the energy on the left before subtraction is positive, so its multiplier is the positive Rayleigh quotient $\lambda =\hat{I}/\hat{J}$.

For nonnegative integers $j,k$, repeated integration by parts gives the beta integral ${\int }_0^1{y}^j(1-y{)}^{k}dy=j!k!/(j+k+1)!$. Multiplying the stated density by $y$ and using this formula yieldsCancellation gives

For $\alpha >0$, integration by parts on $[0,R]$ givesThe exponential remainder tends to zero. Since the integrand is nonnegative, the monotone convergence theorem applied to truncations identifies the improper integral with the Lebesgue integral. ThereforeIts finite value proves integrability as well as evaluating it.

For fixed endpoint variations, the Euler-Lagrange equations areIndeed the first variation is $\int {\sum }_i(L_{q_i}\delta q_i+L_{{\dot{q}}_i}\delta {\dot{q}}_i)dt$. Integration by parts removes the derivatives from the variations; their zero endpoint values remove the boundary term. The fundamental lemma then makes every remaining coefficient vanish.

For $\alpha >0$, integration by parts on $[0,R]$ givesThe exponential remainder tends to zero. Since the integrand is nonnegative, the monotone convergence theorem applied to truncations identifies the improper integral with the Lebesgue integral. ThereforeIts finite value proves integrability as well as evaluating it.

The first weak derivative $D_{j}u=v$ is characterized by ${\int }_{\Omega }u{\partial }_j\varphi =-{\int }_{\Omega }v\varphi$ for every $\varphi \in C_c^{\infty }(\Omega )$. The fundamental lemma of the calculus of variations in the locally integrable form says that a function pairing to zero with every such test is zero almost everywhere. Applied to the difference of two candidate derivatives, it proves uniqueness. The Sobolev space consists of $u\in L^p$ with all first weak derivatives in $L^p$, with normFor the one-dimensional tent, integration by parts on each side of one has cancelling junction terms because the function is continuous. Thus $u^{\prime }=1$ on $(0,1)$ and $u^{\prime }=-1$ on $(1,2)$, almost everywhere. Both it and its derivative are bounded, so it belongs to every stated Sobolev space; its displayed norm to the power $p$ is $2/(p+1)+2$.

For $u=x_1/r$ in the punctured ball, ordinary differentiation givesThe function is bounded and $|D_{j}u|\le 2/r$, so polar integration gives finite derivative $L^p$ norms when ${\int }_0^1{r}^{n-1-p}dr<\infty$, namely $p<n$. To verify these are weak derivatives across the origin, integrate by parts outside a ball of radius $\epsilon$ against a compactly supported smooth test. The inner boundary term is at most a fixed constant times ${\epsilon }^{n-1}$, since $u$ is bounded. It tends to zero for $n>1$. The volume terms converge by local integrability of the displayed derivative. Thus $u\in W^{1,p}(B)$ for $1\le p<n$ with exactly these derivatives. Its value at the origin can be chosen arbitrarily.

Choose a smooth compactly supported mollifier of integral one. On every compact subset of $\Omega$, and for sufficiently small $\epsilon$, each translated mollifier is an allowed test, so the assumption gives $w\ast {\rho }_{\epsilon }=0$. The standard mollifier approximation theorem gives $w\ast {\rho }_{\epsilon }\to w$ in local $L^1$. Thus $w=0$ almost everywhere.

For a multi-index $\alpha$, the weak derivative $v=D^{\alpha }u$ is the locally integrable function satisfyingTwo such functions have zero pairing difference against every test, so the first result proves uniqueness. A compactly supported smooth test on an open subset $E$ extends by zero to one on $\Omega$, proving restriction of the derivative to $E$.

Put $w=D^{\beta }u$. For every test $\varphi$ the defining identity applied to $D^{\alpha }\varphi$ givesThus these two distributions are identical; one is represented by a locally integrable function if and only if the other is. Their weak derivatives then agree, proving $D^{\alpha }(D^{\beta }u)=D^{\alpha +\beta }u$ with exactly the asserted existence equivalence.

On $(0,2)$ the displayed piecewise affine function is continuous at one. Integration by parts separately on the two intervals shows its first weak derivative is one on $(0,1)$ and minus one on $(1,2)$; the first boundary terms cancel by continuity. The derivative of that step function is $-2{\delta }_1$. This distribution is not represented by a locally integrable function: tests of height one supported in shrinking neighborhoods of one have pairing $-2$ with it, whereas their integrals against any fixed local $L^1$ function tend to zero. Hence there is a first weak derivative but no locally integrable second weak derivative.

For fixed nonzero $x$, expand the denominator geometrically:The effective expansion parameter is ${ϵ}^2/x^2$, so the expansion fails when $x=O(ϵ)$. In the inner variable $X=x/ϵ$,for fixed $X$. Expanding the leading outer term toward $x=0$ gives $x^{-2}$ at leading order; expanding the leading inner term for large $X$ and returning to $x$ gives ${ϵ}^{-2}{X}^{-2}=x^{-2}$. Thus the two one-term re-expansions match in $ϵ\ll x\ll 1$. The next common contribution is $-x^{-1}$ when the corresponding next inner order is retained.

On $x\ge \delta$, the relative denominator error is bounded by ${ϵ}^2/{\delta }^2$, henceIntegration by parts yieldsFor the last step split the remaining integral at one: ${\int }_{\delta }^{1}dx/x=-\log \delta$, and replacing $e^{-x}$ by one on that interval has a bounded error.

In the inner interval, $|e^{-x}-1|\le x$ gives an explicit error estimate:The $1/\delta$ terms cancel between regions. Every displayed remainder is $o({ϵ}^{-1})$ whenever $ϵ/\delta \to 0$ and $\delta \to 0$; for example $ϵ({ϵ}^2/{\delta }^3)=(ϵ/\delta {)}^3$. Therefore the leading answer is independent of the arbitrary intermediate cutoff:

The integral construction requires differentiation under the integral and vanishing boundary terms. With those assumptions, integration by parts givesThe standard construction therefore sets $g^{\prime }={\zeta }^{2}g$, giving $g=C{e}^{{\zeta }^3/3}$. Convergent contours run between rays where $\Re {\zeta }^3<0$: convenient central rays have arguments $-\pi /3$, $\pi /3$ and $\pi$. Two choices are the oriented pathsThe cubic exponential dominates the linear one at both ends. To verify independence directly, put $A={\int }_0^{\infty }{e}^{-r^3/3}dr>0$ and $B={\int }_0^{\infty }r{e}^{-r^3/3}dr>0$. At $z=0$ the value and derivative of the integral along rays ${\theta }_1,{\theta }_2$ are $A(e^{i{\theta }_2}-e^{i{\theta }_1})$ and $B(e^{2i{\theta }_2}-e^{2i{\theta }_1})$. For the two contours above their derivative-to-value ratios are respectively $B/A$ and $(B/A)(e^{i\pi /3}-1)$, which differ. Hence their Wronskian is nonzero.

For the second equation multiply by $z$ before integrating by parts:Setting the residual weight to zero gives $g=C(1+{\zeta }^2{)}^{-1/2}$. One contour is the segment from $-i$ to $i$, with the square root positive at $\zeta =0$. Its integral is a nonzero multiple of $J_0(z)$: substituting $\zeta =it$ gives $i{\int }_{-1}^1{e}^{izt}(1-t^2{)}^{-1/2}dt$, an entire function nonzero at zero.

A second contour starts at $i$ and follows $\zeta =i-t$, $0\le t<\infty$, choosing a continuous branch along the path. For $\Re z>0$ it converges at infinity; at the branch endpoint its integrable singularity is $t^{-1/2}$. In both cases the boundary term contains $\sqrt{1+{\zeta }^2}$ and vanishes at branch endpoints. The second integral has a nonzero logarithmic divergence as $z\to 0$ through positive real values, because its large-$t$ integrand is a nonzero multiple of $e^{-zt}/t$. It is therefore independent of the entire first solution; analytic continuation supplies solutions elsewhere away from the equation's singular point.

Strictly, nonzero $w$ and an arbitrary contour alone do not prove the printed necessity for $g$. Necessity also needs injectivity of the chosen Laplace representation, while the vanishing-integrand argument above is the usual sufficient construction. For a concrete counterexample to unconditional necessity in the Bessel case, take a positively oriented circle $|\zeta |=R>1$, choose the branch $(1+{\zeta }^2{)}^{-1/2}\sim 1/\zeta$ outside the cut $[-i,i]$, and add one to that weight. The added entire integrand integrates to zero. Expanding the original weight on this circle gives integral $2\pi i{J}_0(z)$, which is nonzero and solves the equation, although the modified weight is not a constant multiple of the printed one. The standard weights and contours above still give the requested independent solutions.

The only nondegenerate stationary point of a phase of the phase $t^2$ is the endpoint zero. Its usual leading stationary phase method term is proportional to the amplitude there, but ${\tan }^2(0)=0$. Thus the nominal $x^{-1/2}$ stationary contribution vanishes.

Expand one order further in the local amplitude: ${\tan }^{2}t=t^2+O(t^4)$. The rescaling $t=u/\sqrt{x}$ then gives a first nonzero stationary contribution of order $x^{-3/2}$, with coefficient $-\sqrt{2\pi }/8$ in the real integral, as in part (i). Meanwhile the upper endpoint is nonstationary and has nonzero amplitude; its integration by parts contribution is $2\sin ({\pi }^{2}x/16)/(\pi x)$, of order $x^{-1}$.

Consequently the nonstationary endpoint dominates the first nonzero saddle contribution. A stationary phase method formula retaining only the usual leading saddle term would miss the answer. The full stationary phase method, with the vanishing-amplitude correction and the endpoint term, agrees with method of steepest descent. This is quadratic stationary endpoint with quadratically vanishing amplitude.

For an admissible Laplace integral solution, differentiation under the integral sign and integration by parts giveChoose endpoints for which the boundary term vanishes. Setting the density in braces to zero givesTwo suitable contours are the rays ${\Gamma }_{+}:0\to +i\infty$ and ${\Gamma }_{-}:0\to -i\infty$, with a continuous square-root branch on each ray. The Gaussian decays at infinity, and $\zeta g=O({\zeta }^{1/2})$ tends to zero at the finite endpoint. The integrals converge for every finite $z$.

To check independence rather than merely exhibit different paths, defineWith unit normalization, the two solutions satisfy $w_{\pm }(0)=e^{\pm i\pi /4}{J}_0$ and $w_{\pm }^{\prime }(0)=\pm i{e}^{\pm i\pi /4}{J}_1$. Their Wronskian at zero is $-2i{J}_0{J}_1\ne 0$, so they form an independent pair.

Strictly, the printed necessity claim requires an injective contour-transform class as well as vanishing endpoint terms. An arbitrary contour representation may contain a null weight in a contour Laplace representation, so a nonzero solution alone does not force every representing $g$ to have the displayed form. The calculation constructs the canonical density and two valid independent solutions; it does not assume uniqueness for unrestricted contour densities.

Let $g_x(t)=(1+xt{)}^{-1}$. Repeated integration by parts using the factor $e^{-t}$ givesBoundary terms at infinity vanish, and $g_x^{(n)}(t)=(-1{)}^{n}n!x^n(1+xt{)}^{-n-1}$. ThereforeFor every fixed $N$, this is $o(x^N)$ as $x\downarrow 0$, provingFor each fixed $x>0$ the displayed infinite series diverges, since successive term magnitudes have ratio $(n+1)x\to \infty$. Its usefulness is precisely that of an asymptotic expansion, not a convergent power series.

Find the potential from $\mathbf{E}=-\nabla \varphi$ and $\nabla \cdot (ϵ\nabla \varphi )=-{\rho }_f$. One must specify boundary data: a potential on each Dirichlet boundary, a normal displacement on each Neumann boundary, or an appropriate mixture. Conductors are equipotentials; at dielectric interfaces $\varphi$ is continuous and the normal displacement jumps by the specified free surface charge. With no singular dipole sheet, tangential electric field is continuous.

To establish uniqueness of Poisson equation, subtract two solutions with the same charge and boundary data, writing their difference as $u$. Integration by parts givessince $\nabla \cdot (ϵ\nabla u)=0$. The right side vanishes for identical Dirichlet data, identical Neumann data, or their usual mixed combination. Interface terms cancel under the homogeneous difference conditions. With positive permittivity, $\nabla u=0$: the field is unique, and the potential is unique if any suitable reference potential is fixed. For a pure Neumann problem an additive constant remains, and the prescribed outward displacement flux must equal the enclosed free charge. This theorem justifies image-charge solutions once their boundary data have been checked.

Generalized coordinates incorporate holonomic constraints without solving separately for every constraint force. For conservative mechanics use $L=T-V$. The principle of stationary action makes the action $\int Ldt$ stationary for variations vanishing at the endpoints. Integration by parts givesThe canonical momentum is $p_i=\partial L/\partial {\dot{q}}_i$. A cyclic coordinate gives conserved $p_i$; absent explicit time dependence, ${\sum }_i{p}_i{\dot{q}}_i-L$ is conserved. These are symmetry–conservation relationships, not assumptions that all momenta are constant. Nonconservative generalized forces enter the right side.

For example, a pendulum has $L=m{\ell }^2{\dot{\theta }}^2/2+mg\ell \cos \theta$, giving $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0$. The method extends directly to coupled coordinates and velocity-dependent interactions when the correct Lagrangian is used; $T-V$ with a position-only potential is not the most general form.