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Translate the variables by $u=x-x_0$, $v=y-y_0$, $w=z-z_0$. The numerator becomes $u+x_0$. The part odd in $u$ integrates to zero, while the constant part factors into three Gaussian integrals. ThusThe absolute integrability of a polynomial times a Gaussian justifies this separation by Fubini's theorem. Neither $y_0$ nor $z_0$ affects the result; when $x_0=0$, odd symmetry makes the integral zero.

Normalize the Rayleigh distribution by substituting $s=x^2/(2{\sigma }^2)$:For its sketch, the density starts at zero, is positive for $x>0$, and tends to zero at infinity. Differentiation givesIt therefore has a single maximum at $x=\sigma$, with height $e^{-1/2}/\sigma$.

The expected value isIntegrating by parts with $xdx{e}^{-x^2/(2{\sigma }^2)}=-{\sigma }^{2}d(e^{-x^2/(2{\sigma }^2)})$ makes the boundary term vanish and leaves ${\int }_0^{\infty }{e}^{-x^2/(2{\sigma }^2)}dx$. The Gaussian integral gives

Use integration by parts in $I_{n+1}$, writing the integrand as $x^{2n+1}(x{e}^{-x^2})$. Since an antiderivative of $x{e}^{-x^2}$ is $-e^{-x^2}/2$,The boundary term vanishes because the Gaussian exponential function decays faster than any fixed power. These improper integrals are absolutely convergent, justifying the limiting integration by parts.

Starting from the permitted Gaussian integral $I_0=\sqrt{\pi }$, the Gaussian moment recurrence gives $I_1=\sqrt{\pi }/2$ and thenMore generally, iteration gives $I_n=(2n-1)!!\sqrt{\pi }/2^n$, using the double factorial with $(-1)!!=1$.

For fixed $x>0$, integrate first over $y$:The integrand is nonnegative, so this order of the double integral is legitimate even before its finiteness is known. Applying the scaled Gaussian integral ${\int }_0^{\infty }{e}^{-a{x}^2}dx=\sqrt{\pi }/(2\sqrt{a})$ yields

For the Gaussian ansatz, partial differentiation givesCancel the positive exponential and compare the constant and quadratic powers of $x$ in the diffusion equation. The two coupled ordinary differential equations areThe initial profile requires $F(0)=T_0$ and $H(0)=L^{-2}$. The second equation is a separable differential equation:Substitute this into the first equation and integrate its logarithmic derivative:Thusand the solution isThis is Gaussian diffusion of a smooth initial profile: its width grows while its peak decreases. The Gaussian integral gives ${\int }_{-\infty }^{\infty }T(x,t)dx=\sqrt{\pi }{T}_{0}L$, independent of time, which checks conservation of the total amount under the diffusion equation.

Let $I$ be the positive Gaussian integral. Its improper integral converges because $e^{-x^2}\le e^{-|x|}$ for $|x|\ge 1$. Multiplying two copies and using the polar coordinates area element givesNonnegativity justifies the product double integral and the change of integration order. Since $I>0$,

At height $z>0$, the elliptic paraboloid has a circular cross-section $0\le r\le \sqrt{z}$, of area $\pi z$. The cylindrical coordinates volume element therefore reduces the volume integral toFor the specified $f$, the derivative is $e^{-z}(z^{-1/2}+\frac{1}{2}{z}^{-3/2})$. Thus $z{f}^{\prime }(z)$ is integrable at zero and decays exponentially at infinity. Apply integration by parts first on a truncated interval and then take the endpoint limits:Both boundary terms vanish because $zf(z)=-\sqrt{z}{e}^{-z}$. With $z=u^2$, integration by substitution gives the Gaussian integral:The integrand is singular at the apex but the resulting improper integral is convergent.

For $n=0$, the change of variables $s=\sqrt{a}x$ reduces the integral to the given Gaussian integral:For $n=1$, the integrand is an integrable odd function, so its contributions from the two half-lines cancel:The condition $a>0$ ensures convergence in both cases.

Start with the supplied Gaussian integral $I(0)=\sqrt{\pi }/2$ and part (i)'s $I(1)=1/2$. Repeated use of the Gaussian moment recurrence givesThe even indices inherit the $\sqrt{\pi }$ factor from $I(0)$, while the odd indices inherit rational values from $I(1)$.

The fundamental theorem of calculus gives $y^{\prime }(x)=e^{-x^2}>0$, so the graph is strictly increasing. Its second derivative is $y^{\prime \prime }(x)=-2x{e}^{-x^2}$: it curves upward for $x<0$ and downward for $x>0$, with an inflection point at zero.

By the Gaussian integral and the evenness of its integrand,Thus the only axis intersection is $(0,\sqrt{\pi }/2)$; there is no finite $x$-axis intersection because the integral is strictly positive for every finite $x$. The error function gives $y(x)=\frac{\sqrt{\pi }}{2}(1+\mathrm{erf}x)$.

For a nonzero real function, write $\Lambda =F/G$ with $G={\int }_a^{b}w{y}^{2}dx>0$. Multiplying the Sturm-Liouville equation by $y$ and integrating givesThus the stated vanishing boundary product proves $\lambda =\Lambda [y]$.

For the variational statement, the first variation isWith admissible boundary conditions that also make $[p{y}^{\prime }\eta {]}_a^b=0$ for every allowed variation, the fundamental lemma of the calculus of variations identifies stationary functions with the eigenfunctions and stationary values with their eigenvalues. Zero Dirichlet boundary conditions, zero Neumann boundary conditions, and matched periodic data give familiar examples. The single printed condition $[py{y}^{\prime }{]}_a^b=0$ proves the quotient identity but, by itself, does not specify the admissible variations needed for stationarity; the boundary conditions for variational Rayleigh quotients make this distinction explicit.

For the half-line equation, multiplication by $x^{n-1}$ gives $p=w=x^{n-1}$ and $q=x^{n+1}$. Use the regular radial boundary condition at zero and square-integrable decay at infinity, under which the proposed Gaussian is admissible. With $y=e^{a{x}^2}$, $a<0$, $y^{\prime }=2axy$, soThe ratio of the two Gaussian integrals is $n/(-4a)$, from a change of variables and the Gamma function recurrence. Differentiating gives ${\Lambda }^{\prime }(a)=-n[1-1/(4a^2)]$, so the minimum on $a<0$ occurs at $a=-1/2$. The Gaussian variational bound for a radial harmonic oscillator therefore givesHere the trial actually satisfies the differential equation with $\lambda =n$. Moreover, on the same regular decaying domain, integration by parts gives $F[y]-nG[y]={\int }_0^{\infty }{x}^{n-1}(y^{\prime }+xy{)}^{2}dx\ge 0$. Hence the estimate is exact, with $\boxed{{\lambda }_0=n}$. For $n=1$, regular radial data mean the Neumann boundary condition $y^{\prime }(0)=0$; imposing a zero Dirichlet boundary condition there would exclude this trial and give a different problem.

The convolution theorem turns the product $e^{-k^{2}t}\tilde{\theta }(k,0)$ into a spatial convolution. Its kernel isComplete the square:After rescaling the variable, the supplied translated Gaussian integral givesThis is the heat-kernel solution for diffusivity one. Its kernel integrates to one, preserving the total temperature integral, and approaches a Dirac delta function as $t\downarrow 0$. Its variance is $2t$; the factor is appropriate to ${\theta }_t={\theta }_{xx}$, rather than the alternative convention with diffusivity $1/2$.

For the Gaussian initial data for the heat equation, substitute the initial profile into the heat-kernel solution. Completing the square in $y$ givesThe remaining Gaussian integral yieldsThe centre stays at $x_0$ and the squared width grows linearly with time.

For the interval pulse, the convolution reduces to $B{\int }_a^{b}G(x-y,t)dy$. Using $u=(y-x)/(2\sqrt{t})$ and the error function givesThis is the translated and scaled heat equation with interval-indicator initial data. At its jump points the $t\downarrow 0$ limits are $B/2$; away from them the initial data are recovered pointwise.

For fixed $a<x<b$ at late time, both error function arguments are small. Since $\mathrm{erf}u=2u/\sqrt{\pi }+O(u^3)$,The estimate is uniform for $x\in [a,b]$. Equivalently, on this fixed interval the Gaussian in the convolution is $1+O((b-a{)}^2/t)$.

The spreading distance is more apparent on the growing spatial scale. For the Gaussian pulse the exponent is order one when $|x-x_0|=O(\sqrt{t})$. For the interval pulse, set $x=c+\xi \sqrt{t}$ with $c=(a+b)/2$ and fixed $\xi$. Uniformly for bounded $\xi$, its convolution givesThus both nonzero pulses spread over a distance $O(\sqrt{t})$, while their heights are $O(t^{-1/2})$ and their total integrals remain constant. More quantitatively, the variances of their normalized profiles are $1/(2a_0)+2t$ and $(b-a{)}^2/12+2t$, respectively, confirming the same width law.

Complete the square in the Fourier transform exponent:The supplied shifted Gaussian integral therefore givesThis is the Fourier transform of a Gaussian, with angular-wavenumber convention $e^{-ikx}$.

Take the spatial Fourier transform of the heat equation. Two integrations by parts, with decaying fields and first derivatives at infinity, give $\mathcal{F}[{\psi }_{xx}]=-k^2\tilde{\psi }$. Thus each mode satisfies the ordinary differential equationInserting the transformed initial condition and integrating the time equation givesFinally set $A=t+1/(4\lambda )>0$. The inverse Fourier transform, evaluated by the same Gaussian identity, givesEquivalently the Gaussian initial data for the heat equation evolve asThe profile broadens and its peak decreases, while its total integral remains $\sqrt{\pi /\lambda }$, as required by diffusion without loss on the whole line. Direct differentiation verifies the equation and the initial limit.

For the Gaussian function, direct differentiation gives $f^{\prime }=-tf$. Applying the Fourier transform identities from part (a),Solving this first-order ordinary differential equation givesIn fact, the normalization can also be fixed: the Gaussian integral gives $F(0)={\int }_{\mathbb{R}}{e}^{-t^2/2}dt=\sqrt{2\pi }$, so $C=\sqrt{2\pi }$ with the convention used here. The transform has the same Gaussian shape.

Normalize the Gaussian trial first. The Gaussian integral givesIts second and fourth moments are $⟨x^2⟩=1/(2\alpha )$ and $⟨x^4⟩=3/(4{\alpha }^2)$. Since ${\psi }^{\prime }=-\alpha x\psi$, the kinetic term is $\int ({\psi }^{\prime }{)}^2={\alpha }^2⟨x^2⟩=\alpha /2$. At the specified coupling, the energy is thereforeThe stationary condition isThe quadratic factor is positive for real $\alpha$, so the unique positive stationary width is $\alpha =2$. Also $E^{\prime \prime }={\alpha }^{-3}+9{\alpha }^{-4}>0$ and $E$ diverges at both ends of $\alpha >0$, proving a global minimum within this trial family. Its value isThe associated Sturm-Liouville problem isHere $\rho =1$, $\sigma =x^2+2x^4$ and $w=1$. The Rayleigh-Ritz variational principle makes the exact lowest eigenvalue the minimum over all normalized admissible states. Restricting to Gaussians can only raise this minimum, hence $\boxed{E_0\le 13/8}$. The inequality is strict because $H{\psi }_{\ast }/{\psi }_{\ast }=2-3x^2+2x^4$ is not constant: the trial is not an exact eigenfunction. This is the Gaussian trial for a quadratic-plus-quartic oscillator.

Take $\lambda >0$ and $t>0$, the positive diffusivity requirement for the forward heat kernel. The PDF does not explicitly specify the sign of $\lambda$; the stated real Gaussian function formula requires this assumption. For a direct calculation, initially assume sufficiently smooth and rapidly decreasing data, so that the Fourier transform, its inverse and the following integrations are justified. The resulting heat-kernel convolution extends to wider classes of initial data.

Using the stated Fourier transform convention and integration by parts,The transformed ordinary differential equation has solution $\tilde{u}(k,t)=e^{-\lambda k^{2}t}{\tilde{u}}_0(k)$. Fourier inversion then givesComplete the square:Part (a), after scaling by $\sqrt{\lambda t}$, evaluates the remaining Gaussian integral even though its shift is imaginary. ThereforeThe heat kernel has integral one and is an approximate identity as $t\downarrow 0$, which recovers the initial condition. Mere decay at spatial infinity does not by itself imply every Fourier integral is absolutely convergent; the regularity assumption above supplies a rigorous setting for the derivation.

Put $s=\lambda t>0$. In the heat-kernel convolution, complete the real square asThe Gaussian integral therefore givesAt $t=0$ the same formula gives $\beta =\alpha$. Equivalently, ${\beta }^{-2}={\alpha }^{-2}+4\lambda t$: the variance of this normalized Gaussian function grows from $1/(2{\alpha }^2)$ to $1/(2{\alpha }^2)+2\lambda t$.

The initial profiles have unit integral and concentrate at zero as $\alpha \to \infty$. Precisely, for every smooth compactly supported test function $\phi$,Thus the limit is the Dirac delta distribution, rather than an ordinary function: pointwise the profiles tend to zero away from the origin and diverge at the origin. At every fixed $t>0$,The limiting evolved profile is the heat kernel for a unit point source.

Let $G(k)=\tilde{g}(k)$. Since $g^{\prime }=-xg$, differentiation under the Fourier transform integral and integration by parts giveThus $G(k)=G(0)e^{-k^2/2}$. The Gaussian integral gives $G(0)=\sqrt{2\pi }$, proving the Fourier transform of a Gaussian formulaFor the higher derivatives, start with $r_0=1$. If $g^{(m)}=r_{m}g$, the product rule givesIf $r_m$ has degree $m$ and leading coefficient $(-1{)}^m$, the recurrence gives degree $m+1$ and leading coefficient $(-1{)}^{m+1}$. Induction proves the assertion for every $m$. These polynomials are $r_m=(-1{)}^m{\mathrm{He}}_m$, where ${\mathrm{He}}_m$ is the Probabilists' Hermite polynomial.

For the trial function, $y_T^{\prime }=-ax{y}_T$. The required Gaussian integrals areThe second formula follows by differentiating the first with respect to $a$. Thereforeand division by $G[y_T]$ gives the Gaussian variational bound for an attractive Gaussian well:This is a good trial family because it consists of smooth, positive, square-integrable functions decaying at both infinities. Each is an even function, matching the symmetry of the lowest bound state in this even potential, and the parameter varies its width. It also permits exact Gaussian integrals. Its Gaussian tail need not equal the exact exponential tail of a negative-energy bound state for the Rayleigh-Ritz variational principle to give a valid bound.