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For $\omega \ne 0$, insert the Gaussian damping factor $e^{-\epsilon x^2/2}$ and use the Gaussian integral:Taking $\epsilon \downarrow 0$ in ${\mathcal{S}}^{\prime }$ with the continuous square-root branch gives the Fresnel integralFor $\omega =0$, the function is constant and its Fourier transform is $\boxed{2\pi {\delta }_0}$.

The condition $2m={\hslash }^2$ makesForthe squared norm is the Gaussian integralSince ${\psi }_a^{\prime }=-ax{\psi }_a$, integration by parts gives the kinetic quotientThe potential quotient isThe variational principle therefore yields the Gaussian variational bound for an attractive Gaussian well

Insert a complete set of momentum eigenstates into the quantum-mechanical propagator:Completing the square and evaluating the resulting Gaussian integral, with the usual i-epsilon prescription, gives the free-particle propagatorThe square-root branch is fixed by requiring $K(x_f,T;x_i,0)\to \delta (x_f-x_i)$ as $T\downarrow 0$.

Because $\tilde{\mathbf{n}}\cdot (\tilde{\mathbf{n}}\times {\partial }_t\tilde{\mathbf{n}})=0$, the component of $\mathbf{q}$ parallel to $\tilde{\mathbf{n}}$ is $-\mathbf{B}\cdot \tilde{\mathbf{n}}$. The Gaussian integral over $\lambda$ removes precisely this longitudinal component. ThusUsing the vector triple-product identity and $|\tilde{\mathbf{n}}|=1$, this becomes the $O(3)$ nonlinear sigma model in a background field:The magnetic field acts as the temporal component of an $O(3)$ background gauge connection.

Write the phase and amplitude asThe saddle point relevant to the contour is $z=0$, since ${\varphi }^{\prime }(z)=\sinh z$ andThe local steepest-descent direction is therefore vertical: with $z=iy$,

The given contour approaches the lines $\mathrm{Im}z=-\pi$ at its left end and $\mathrm{Im}z=\pi$ at its right end. The poles of $f$ are at $z=\pm 4i$, outside the closed strip $|\mathrm{Im}z|\le \pi$. The Cauchy integral theorem therefore permits deformation to the contour consisting of the lower horizontal ray from $-\infty -i\pi$ to $-i\pi$, the vertical segment from $-i\pi$ to $i\pi$, and the upper horizontal ray from $i\pi$ to $\infty +i\pi$. On either horizontal ray,so those two integrals are exponentially small, of order $e^{-x}{x}^{-1/2}$.

The vertical segment, oriented upward, contributesIts real phase has a unique maximum at $y=0$. Sincethe simple-saddle contribution in steepest descent, equivalently the local Gaussian integral, givesThe factor $i$ comes from the upward tangent $dz=idy$ of the deformed contour.

For $n=2$, the coupling is a mass squared, $m^2=g$. After integration by parts, the quadratic action has kernel $K=-(\square +g-i0)$. Completing the square in the Gaussian integral and choosing the source-independent normalization so that $Z_2[0]=1$ givesEquivalently, in momentum space,The $i0$ prescription selects the Feynman propagator.

Use the Fourier transform conventionThe zero mode vanishes because the compositional order parameter has zero spatial average. Parseval identity and the Fourier transform of a derivative turn the quadratic part of the dimensionless free energy intoHere ${\sum }_{\mathbf{q}}^{+}$ is the positive-wavevector sum for a real field. Each independent complex amplitude has density proportional to $e^{-G(q)|{\varphi }_{\mathbf{q}}{|}^2}$, so its elementary Gaussian integral gives the static structure factorwhenever $G(q)>0$.

The stationary points of the Brazovskii model kernel obeyBecause $\kappa <0<\gamma$, the nonzero minimum is the nonzero-wavevector soft-mode sphereAt this wavevector,The first Gaussian field theory divergence therefore occurs at

Fourier expansion in volume $V$ diagonalizes the quadratic Landau-Ginzburg theory:whereEach independent real mode contributes a Gaussian integral proportional to $(T/K{)}^{1/2}$. Taking $-T\log Z$, dividing by $V$, and replacing the sum by an integral gives, up to field-independent conventions,

The induced four-$\varphi$ interaction is the tree-level diagram with two $\lambda {\varphi }^2\chi$ vertices joined by one internal $\chi$ propagator. The zero-dimensional propagator is $M^{-2}$, and the second-order expansion of the exponential supplies the factor $1/2$. The resulting effective-action coefficient is therefore $-{\lambda }^2/(2M^2)$, exactly the value of $\tilde{\lambda }$ found by the Gaussian integral.

Complete the square in the heavy variable:The shifted Gaussian integral is independent of $\varphi$. For $M>0$, choosing $N=M/\sqrt{2\pi }$ removes it and gives

The functions $g_n$ are nonnegative. The change of variables $u=nx$ and the given Gaussian integral implyFor every fixed $\delta >0$,Thus the mass concentrates at the endpoint zero. This is a one-sided approximate identity.

Because $f$ is continuous on the compact interval $[0,1]$, it is a bounded function; put $B={\sup }_{[0,1]}|f|$. Given $\epsilon >0$, choose $\delta \in (0,1)$ with $|f(x)-f(0)|<\epsilon$ for $0\le x<\delta$. Splitting the integral near and away from zero yieldsTaking the limit superior gives a bound of $\epsilon$, and $\epsilon$ was arbitrary. ConsequentlyOnly continuity at zero and boundedness on $[0,1]$ were needed for this final estimate; the assumed continuity everywhere also guarantees the ordinary integrals exist.

In first-quadrant polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, and the Jacobian determinant contributes $r$. The integrand and area element together become $r^2{e}^{-r^2}\sin \theta {\cos }^2\theta drd\theta$. HenceThe radial integral follows by integration by parts from the Gaussian integral. The isolated undefined ratio at the origin does not affect this convergent double integral.

The integrand is even and decays at both infinity and zero, since $a,b>0$. On $x>0$ set $u=\sqrt{a}x-\sqrt{b}/x$, a strictly increasing bijection onto the whole real line. ThenUsing evenness for the negative half-axis givesThe second summand is an integrable odd function, so its integral vanishes. The remaining Gaussian integral givesFor the second integral use $x=1/t$ separately on each half-axis. The factor $x^{-2}dx$ becomes $-dt$, and reversing the transformed limits yields $J(a,b)=I(b,a)$. ThusThese identities are instances of the Gaussian integral with an inverse-square term. Splitting at zero before either change of variables avoids treating a discontinuous transformation as a single bijection.

The integrand in the preceding expression is nonnegative. Tonelli's theorem permits squaring as a quadrant integral; then polar coordinates giveThe defining integral is positive, fixing the sign of the square root. ThereforeThis is the positive-quadrant proof of the Gaussian integral.

Take $\sigma >0$ and substitute $u=(x-\mu )/\sigma$. The density becomes the standard Gaussian distribution density $\phi (u)=e^{-u^2/2}/\sqrt{2\pi }$. Its normalization follows from the supplied Gaussian integral after scaling by $\sqrt{2}$. Since $u\phi (u)$ is an integrable odd function, its integral is zero, and therefore $\mathbb{E}X=\mu$.

For the second moment, ${\phi }^{\prime }=-u\phi$. Integration by parts givesThe endpoint term vanishes by Gaussian decay. Consequently

Choose the plane normal to be $+\hat{y}$. Only the second field component contributes, and on $y=0$ it is $e^{-(x^2+z^2)}$. Thus the surface integral factors into two Gaussian integrals:The original question does not specify an orientation; choosing $-\hat{y}$ instead gives $-\pi$.

The printed first component $\sin y/y$ is undefined on the entire integration plane. Its removable continuous extension there is $1$, since ${\lim }_{y\to 0}\sin y/y=1$. The stated flux uses this natural extension; in fact the normal component alone is already well-defined and integrable on the plane, so the first component does not affect either oriented value.

The indicated integration by substitution givesPointwise, the argument of $\phi$ tends to $x$. Since the Gaussian integral is $\int e^{-{\eta }^2}d\eta =\sqrt{\pi }$, the expected limit is $\phi (x)$. To justify it for unbounded data, use the approximate identity property rather than assuming a uniform bound on $\phi$.

In the original variable, normalization givesFor $|\xi -x|<\delta$, continuity makes the difference smaller than any prescribed $\epsilon$ if $\delta$ is small enough, and the kernel mass is at most one. On the complement the Gaussian mass tends to zero; the data-weighted tail tends to zero by heat-kernel convergence for Gaussian-integrable initial data. This provesFor heat-kernel convergence from a conditionally convergent Gaussian integral, the same tail conclusion follows using the bounded primitive of $(\phi (\xi )-\phi (x))e^{-{\xi }^2/(4T)}$. The factor $q=K(x-\xi ,t)e^{{\xi }^2/(4T)}$ has its maximum at $\xi =xT/(T-t)$, approaching $x$. For sufficiently small $t$, it is monotone on each of the two tails outside $|\xi -x|<\delta$, and its values at the tail endpoints tend to zero exponentially. Integration by parts bounds each tail by a constant times its endpoint value, proving the same limit without an absolute-convergence assumption. The approach $t\downarrow 0$ is from positive time.

Use integration by substitution with $x=2u$ and the Gaussian being an even function:The factor two from the differential exactly cancels the half-line factor in the given full-line Gaussian integral.

The integrand for $I_1$ is an odd function with absolute convergence, so $I_1=0$. For $I_2$, the recurrence gives $I_2=I_0/2$, where the supplied Gaussian integral is $I_0=\sqrt{\pi }$. HenceThe zero first moment is specific to the symmetric full-line domain.

Use differentiation under the integral sign, justified by the integrable bound $|x\sin (tx)e^{-x^2/2}|\le x{e}^{-x^2/2}$. This givesIntegration by parts makes the boundary term vanish: sine vanishes at zero and the Gaussian tends to zero at infinity. ThereforeThe integrating factor $e^{t^2/2}$ makes $(e^{t^2/2}F{)}^{\prime }=0$, so $F(t)=C{e}^{-t^2/2}$. At $t=0$, the given half-line Gaussian integral fixes $C=\sqrt{\pi /2}$. ThusThis is the even cosine-transform form of the Fourier transform of a Gaussian.