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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

A phase-space path integral represents a quantum transition kernel by integrating both coordinates and conjugate momenta with action $\int (p\dot{q}-H)dt$. It follows from inserting position and momentum completeness relations between short-time evolution operators. For a quadratic momentum dependence the momentum variables can be integrated by a Gaussian integral.

For $\alpha >0$, polar coordinates reduce the integral to $2\pi {\int }_0^{\infty }r{e}^{-(\alpha -i\lambda )r^2}dr$. The result is $\pi /(\alpha -i\lambda )$. Its damped limit for real $\lambda >0$ is $i\pi /\lambda$, which fixes the sign of the oscillatory complex Gaussian integral. The Fresnel integral convention can be compared with NIST definitions.

For $\mathrm{Re}A>0$ and $\mathrm{Re}v>0$, rescaling the defining gamma function integral proves this representation, with compatible complex-power branches. It turns denominator powers into exponential factors, making Gaussian integrals possible. Combining two parameters by their sum and ratio gives the usual Feynman parameter identity. Outside its convergence domain a further continuation or pole prescription is needed.

For an isothermal thin atmosphere with number density $n(z)=n_0{e}^{-z/H}$ and extinction cross-section ${\sigma }_{\lambda }$, a tangent ray near radius $R_p+z$ has height $z+{\ell }^2/(2R_p)$ at distance $\ell$ along the ray. The resulting Gaussian integral gives slant optical depthThe effective transit height is near ${\tau }_{\lambda }\sim 1$, so $z(\lambda )=H\log {\sigma }_{\lambda }+\mathrm{constant}$ if composition is constant. This explains the scattering slope of a transmission spectrum.

In Laplace method for the Gamma function, scaling $t=x(1+s/\sqrt{x})$ gives a standard Gaussian integral. The first even correction polynomial is $s^2-7s^4/12+s^6/18$. Its expectation under a unit-variance centered Gaussian is $1-7\cdot 3/12+15/18=1/12$. The amplitude expansion contributes to this coefficient as well as the phase expansion; expanding only the latter gives the wrong Stirling formula correction.