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For integer $n\ge 2$, integration by parts gives $J_n=(n-1)J_{n-2}/2$. The boundary term $-x^{n-1}{e}^{-x^2}/2$ vanishes at both infinities because Gaussian decay dominates every power. The full-line initial values are $J_0=\sqrt{\pi }$ and $J_1=0$, the latter by odd symmetry. HenceUnlike the half-line Gaussian moment recurrence, full-line odd moments vanish and even moments are twice their half-line values. The domain of integration must be retained when transferring a recurrence between the two settings.

For $H=p^2/(2m)+k{x}^4$ with $k>0$, the normalized even trial wavefunction $f_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$ has Rayleigh quotientThe Gaussian moment recurrence gives the potential term, and integration by parts gives the kinetic term. Its minimum is at ${\alpha }^3=6mk/{\hslash }^2$, proving the displayed upper bound by the Rayleigh-Ritz variational principle.

An odd trial family $g_{\beta }=(4{\beta }^3/\pi {)}^{1/4}x{e}^{-\beta x^2/2}$ is orthogonal to the exact even ground state. The Sturm oscillation theorem identifies the lowest odd state as the first excited state, so minimizing within this family bounds $E_1$:Orthogonality to an approximate ground trial alone would not justify this excited-state bound; odd parity ensures orthogonality to the exact ground state. Both bounds exhibit the characteristic quartic-oscillator energy scale $({\hslash }^{4}k/m^2{)}^{1/3}$.

A confined ground state requires $k>0$; take $\alpha >0$ so the trial wavefunction is normalizable. Its normalized form isThe Gaussian moment recurrence gives $⟨x^2⟩=1/(2\alpha )$ and $⟨x^4⟩=3/(4{\alpha }^2)$. Also, integration by parts gives $⟨p^2⟩={\hslash }^2\int |f_{\alpha }^{\prime }{|}^{2}dx={\hslash }^2\alpha /2$. The Rayleigh quotient for the quartic oscillator is consequentlyDifferentiation yields ${\mathcal{E}}^{\prime }(\alpha )={\hslash }^2/(4m)-3k/(2{\alpha }^3)$, so the unique minimum is at ${\alpha }_{\ast }=(6mk/{\hslash }^2{)}^{1/3}$. Since the quotient diverges at both endpoints $0$ and $\infty$, this is the global minimum in the trial family. The Rayleigh-Ritz variational principle givesFor the first excited energy eigenstate, choose the odd trial familyThe even confining potential has an even, nodeless ground state, by the nodeless theorem for a one-dimensional ground state. By the Sturm oscillation theorem, its first excited state is odd with one node. The proposed family has the same parity operator eigenvalue and one node, and is orthogonal to the exact ground state, not merely to its Gaussian trial approximation. Minimization in the odd linear subspace therefore bounds the first excited energy from above. For completeness, its Rayleigh quotient is $3{\hslash }^2\beta /(4m)+15k/(4{\beta }^2)$, minimized at ${\beta }^3=10mk/{\hslash }^2$, giving $E_1\le (9{\hslash }^2/8m)(10mk/{\hslash }^2{)}^{1/3}$. These are Gaussian variational bounds for a quartic oscillator.

If $k=0$, there is no normalizable free particle ground state on the whole line, though the spectral infimum is zero. If $k<0$, the Hamiltonian operator is not bounded below, so a ground-energy bound of the requested kind does not exist.

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 $(-1)!!=1$.

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 $I(1)$.

For $n\ge 2$, write $x^n{e}^{-x^2}=x^{n-1}(x{e}^{-x^2})$ and apply integration by parts, using the antiderivative $-e^{-x^2}/2$ for the second factor:The boundary term is zero: at zero, $n-1>0$, and at infinity Gaussian decay dominates every power. Consequently the Gaussian moment recurrence is