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The Dirichlet hyperbola method gives $D(x)=2{\sum }_{a\le r}\lfloor x/a\rfloor -r^2$, where $r=\lfloor \sqrt{x}\rfloor$. The harmonic number estimate $H_r=\log r+\gamma +O(1/r)$, with $\gamma$ the Euler's constant, yields $D(x)=2x(\log r+\gamma )-r^2+O(x/r+r)$. Substitution of $r=\sqrt{x}+O(1)$ gives the displayed asymptotic expansion. Comparison of ${\sum }_{n\le x}\log n$ with its integral consequently gives ${\sum }_{n\le N}(\tau (n)-\log n-2\gamma )=O(\sqrt{N})$ for integer $N$.

The Gamma reflection formula gives $(\sin \pi z/\pi )\Gamma (z)=1/\Gamma (1-z)$, analytic at zero with value one. A finite-parameter Watson's lemma representation of Ramanujan's function therefore gives the displayed coefficients. In particular $c_0=1$, $c_1=-\gamma$ and $c_2={\gamma }^2-{\pi }^2/6$, with $\gamma$ Euler's constant. Obtaining this representation requires taking an imaginary part before sending its Mellin exponent to zero; the original complex integral is divergent at that boundary.

For fixed $a>0$ and $r>1$, the factor $\exp (-\lambda e^{-x})$ is a translated Gumbel distribution cumulative function, with transition near $x=L$. Integration by parts transforms the integral to an exponentially small boundary term plus $(r-1{)}^{-1}{\int }_0^{\lambda e^{-a}}{e}^{-u}(L-\log u{)}^{1-r}du$. Expanding the latter against its finite logarithmic moments gives the displayed two terms because ${\int }_0^{\infty }{e}^{-u}\log udu=-\gamma$, with $\gamma$ Euler's constant. The smooth cutoff shifts the first correction even though replacing it by a step gives the leading tail.

Put $L=\log \lambda$ and integrate by parts using $x^{-2}=-d(x^{-1})/dx$. The exponential cutoff has derivative $\lambda e^{-x}{e}^{-\lambda e^{-x}}$, so the endpoint at infinity vanishes and the lower endpoint givesWith $u=\lambda e^{-x}$ this becomesThe mass of the transformed integral lies at $u=O(1)$, whereTo control this asymptotic expansion, split at $u=e^{-L/2}$ and $u=e^{L/2}$. On the central interval the geometric-series remainder is bounded by $2(\log u{)}^2/L^3$, whose integral against $e^{-u}$ is finite. The lower tail has length $e^{-L/2}$ and denominator at least $L$; the upper tail is exponentially small in $e^{L/2}$, with the original denominator at least one. The omitted tails of the logarithmic moments are likewise negligible. Thus extending the moments to $(0,\infty )$ is justified.

The Gamma function gives ${\int }_0^{\infty }{e}^{-u}du=1$, and the first logarithmic moment is $-\gamma$, with $\gamma$ Euler's constant. ThereforeThis is the moving Gumbel cutoff in an inverse-power tail: replacing the original cutoff by a sharp step at $x=\log \lambda$ gives the first term, but its smooth transition supplies the Euler-constant correction.

The relevant boundary values are $a=\pi$ and $\sigma =0$, but the original complex integral defining $g(0)$ requires $\sigma >0$. Substituting zero into it is invalid: its real part has a nonintegrable $dx/(x\log x)$ singularity near zero. Its imaginary part is different, since for positive $\sigma$This has a finite limit as $\sigma \downarrow 0$. On $(0,1)$, $x^{\sigma -1}\le x^{-1}$, whose displayed logarithmic denominator makes the integral finite; at infinity, a fixed upper bound on $\sigma$ and exponential damping provide an integrable dominator. Thus dominated convergence gives $N(t)={\lim }_{\sigma \downarrow 0}F(t,\pi ,\sigma )$.

For the asymptotics, take the imaginary-part parameter identity before passing to that limit, with a fixed $0<s<1/2$. Its endpoint factor converges toby the Gamma reflection formula. Near zero, $\sin (\pi r)\Gamma (\sigma +r)$ is bounded uniformly for small $\sigma \ge 0$, because the sine zero compensates the gamma pole. At $s>0$, the remainder integral itself remains absolutely convergent at $\sigma =0$ and is $O(t^{-s})$. ConsequentlyThe endpoint function is analytic with value one, so Watson's lemma proves the logarithmic expansion of Ramanujan's functionIn particular, with $\gamma$ Euler's constant,This use of imaginary-part regularization at a zero Mellin exponent matters: the positive-$\sigma$ coefficient ${\beta }_0$ in part (b) is always zero, whereas the boundary endpoint function has $h_0(0)=1$. The fixed-positive-parameter expansion is not uniform as $\sigma \downarrow 0$, so its coefficients cannot simply be sent to that boundary term by term.

Let $\mathcal{H}=a^{\prime }/a$ and ${\theta }_N=i\mathbf{k}\cdot {\mathbf{v}}_N$. Choose the synchronous gauge rest frame of cold dark matter, so ${\theta }_C=0$. Its continuity equation then gives $h^{\prime }=-2{\delta }_C^{\prime }$. Substitution in the trace Einstein equation givesFor radiation in cosmology the continuity and Euler equations are ${\delta }_R^{\prime }+4{\theta }_R/3+2h^{\prime }/3=0$ and ${\theta }_R^{\prime }=k^2{\delta }_R/4$. Differentiate the first, eliminate ${\theta }_R^{\prime }$ and $h^{\prime \prime }$, and obtainradiation in cosmology pressure supplies the acoustic restoring term while its energy density contributes twice as strongly to the trace gravitational source through $1+3w_R=2$.

Deep in radiation domination, $a\propto \tau$, $\mathcal{H}=1/\tau$, ${\Omega }_R\simeq 1$ and ${\Omega }_C\simeq 0$. At long wavelengths $k\tau \ll 1$, the adiabatic initial conditions require ${\delta }_C=3{\delta }_R/4$. The cold-matter equation reduces toA power ${\tau }^p$ gives $p^2-4=0$. The regular growing solution therefore hasThe other adiabatic solution diverges as ${\tau }^{-2}$ and is discarded for the regular primordial growing mode. The relation between the density contrasts fixes their relative entropy to zero, not their subsequent equality inside the acoustic horizon.

One can also solve the leading radiation-era equations explicitly, which proves that the same regular mode develops logarithmic cold-matter growth after horizon entry. Put $x=k\tau /\sqrt{3}$. The first equation gives ${\delta }_R=(x^2{\delta }_{C,xx}+x{\delta }_{C,x})/3$. Substituting in the second givesDefineIts useful derivative is $F^{\prime }=1/x-2\sin x/x^2+2(1-\cos x)/x^3$. Direct differentiation verifies the fourth-order equation and gives the associated radiation in cosmology solution:At zero, $F=x^2/8+O(x^4)$ and ${\delta }_R=K{x}^2/6+O(x^4)$, so this is precisely the regular adiabatic radiation-era cold-dark-matter transfer solution with $K=24A/k^2$. The other indicial behaviors of the fourth-order equation are $1,x,x^{-2}$; they are either nonadiabatic at leading order or singular and do not add another regular growing adiabatic mode.

For $x\gg 1$, the integral is $\gamma +\ln x-\mathrm{Ci}(x)$, where $\gamma$ is the Euler's constant and $\mathrm{Ci}$ the cosine integral. Its oscillatory tail cancels the explicit leading $\sin x/x$ term, givingThus the requested late sound-horizon behavior isAn additive constant in ${\delta }_C$ is absorbed into the reference ${\tau }_{\ast }$; the regular adiabatic solution fixes the acoustic phase, with $D=0$ in the explicit time-origin convention above. A general acoustic mixture has both terms. The physical expansion parameter is $k\tau /\sqrt{3}$; comparisons with $2\pi /k$ describe the same long- and short-wavelength regimes up to fixed numerical factors.

The contrast grows much more slowly after entry during radiation domination: radiation in cosmology undergoes pressure-supported acoustic oscillations and cold matter only the logarithmic Mészáros effect. Modes entering well before matter-radiation equality consequently have suppressed late amplitudes relative to modes that stay outside the horizon until near equality. After matter domination begins, pressureless growing modes can grow as $a$. This scale-dependent history produces the turnover and small-scale suppression of the matter transfer function and affects the initial conditions for large-scale structure of the universe. The early subhorizon limit requires both $k\tau \gg 1$ and $\tau \ll {\tau }_{\mathrm{eq}}$; it is not applicable to a mode that enters only in the matter domination.

Write $1(n)=1$ and let $\epsilon$ be the Identity for Dirichlet convolution. The Möbius function uses the empty-product convention $\mu (1)=1$; the Von Mangoldt function has $\Lambda (1)=0$. We first establish all the arithmetic function identities needed below. Every integer divisor of $n$ contributes one to $\tau (n)$, so $\tau =1\ast 1$. If $n>1$ has $r$ distinct prime factors, the squarefree divisors correspond to their subsets andAt $n=1$ the sum is one. Thus $\mu \ast 1=\epsilon$. Writing $n={\prod }_p{p}^{v_p(n)}$, the only divisors with nonzero Von Mangoldt function are $p,p^2,\dots ,p^{v_p(n)}$, and henceThe Dirichlet convolution is commutative by replacing $d$ with $n/d$, and associative because either bracketing sums $f(a)g(b)h(c)$ over $abc=n$. ConsequentlyThese are consequences of the definitions, rather than unproved invocations of Möbius inversion.

To identify the constant, integrate on each interval $[j,j+1)$:The tail is $O(1/J)$, since the integrand is at most $2/x^2$ for $x\ge 2$. Summation therefore gives $\gamma ={\lim }_{J\to \infty }(H_J-\log (J+1))$, where $H_J$ is the harmonic number, and $H_J=\log J+\gamma +O(1/J)$. Thus this is Euler's constant.

Put $r=\lfloor \sqrt{N}\rfloor$. Counting pairs of positive integers whose product is at most $N$, and subtracting the square counted twice, gives the Dirichlet hyperbola method identitySince $r=\sqrt{N}+O(1)$, the harmonic number estimate yieldsThe increasing natural logarithm also satisfies ${\sum }_{n\le N}\log n=N\log N-N+O(\log (2N))$ by comparison with its integral. Subtracting proves

For the second implication, denote the Mertens function by $M(x)$ and the Second Chebyshev function by $\psi (x)$. Apply the proved Dirichlet convolution identities to $b(n)=\tau (n)-\log n-2\gamma 1(n)$:Summing this identity over integers up to $N$ gives the exact relationFix an integer $K\ge 2$. The contribution to $E(N)$ from $d\le N/K$ has absolute value at mostFor $d>N/K$, group by $j=\lfloor N/d\rfloor$. This gives exactlyunder the assumed cancellation $M(x)=o(x)$; extending that assumption from integers to real $x$ uses $M(x)=M(\lfloor x\rfloor )$. Therefore ${\mathrm{limsup}}_{N\to \infty }|E(N)|/N\ll K^{-1/2}$. Letting $K\to \infty$ proves the prime number asymptotic from Mertens cancellation:

Change independent variable from proper time to the scale factor. Then $\dot{\delta }=\dot{a}{\delta }_a$ and $\ddot{\delta }={\dot{a}}^2{\delta }_{aa}+\ddot{a}{\delta }_a$. Under radiation domination, $\ddot{a}=-{\dot{a}}^2/a$ and $H^2\simeq 8\pi G{\bar{\rho }}_r/3$. Since ${\bar{\rho }}_m/{\bar{\rho }}_r=a/a_{\mathrm{eq}}$, the linear matter perturbation growth equation becomesPutting $\delta =au$ givesThis keeps the matter self-gravity term in a radiation-dominated background. It is not an exact background equation through radiation-matter equality.

At $a/a_{\mathrm{eq}}\ll 1$, neglecting that small term gives ${\delta }_{aa}+{\delta }_a/a=0$. Integration yieldsThus matter has at most logarithmic growth during this leading radiation approximation, together with a constant independent mode. The constant is non-growing, not a mode that literally falls as $a^{-1}$; that power belongs to $u$ rather than ${\delta }_m$.

For an increasing/decreasing basis of the displayed equation with matter self-gravity retained, put $x=\sqrt{6a/a_{\mathrm{eq}}}$. Its equation becomes $x^2{\delta }_{xx}+x{\delta }_x-x^2\delta =0$. HenceThe Modified Bessel function of the first kind gives $I_0(x)=1+3a/(2a_{\mathrm{eq}})+O((a/a_{\mathrm{eq}}{)}^2)$, which increases slowly. The Modified Bessel function of the second kind gives $K_0(x)=-\frac{1}{2}\ln (a/a_{\mathrm{eq}})+\frac{1}{2}\ln (2/3)-{\gamma }_E+O((a/a_{\mathrm{eq}})|\ln (a/a_{\mathrm{eq}})|)$, where ${\gamma }_E$ is Euler's constant; this mode decreases as $a$ increases. Their leading span is precisely the constant/logarithmic pair above. Mode labels depend on the chosen basis and normalization; no rapid matter-era growth $\delta \propto a$ occurs here. Neglected background corrections can change subleading terms, so the Bessel basis should not be extrapolated through equality.

Use the principal complex logarithm in the upper half-plane. Deform the interval upward into a half-strip: the vertical side at zero is traversed upward, and the side at one downward. The top side vanishes exponentially; a small indentation at zero contributes $O(r|\log r|)$ and vanishes. Thus the contour deformation gives the exact representation, for $x>0$,Since $\log (is)=\log s+i\pi /2$, scaling $u=xs$ and using the integral defining Euler's constant givesFor the second endpoint, $\log (1+is)=is+s^2/2+O(s^3)$. Watson's lemma, or integration of these powers against $e^{-xs}$ with an exponentially small tail, yieldsIn particular the required first terms areThe two different endpoint scales come from the logarithmic singularity at zero and the simple zero of the amplitude at one.

The expansion is singular near the moving lower limit, so expanding $e^{-\epsilon /x}$ throughout the original integral would lose its constant term. Introduce the symmetric exponential integralUnder $x=\epsilon /t$, the discarded piece becomesThis is a uniform Taylor expansion because $t\ge 1$. To derive the logarithmic expansion of a symmetric exponential integral without needing another special function, first exploit the same substitution to writeThe estimate follows from $0\le 1-e^{-\epsilon /x}\le \epsilon /x$. Here $\gamma$ is Euler's constant and the small-argument expansion of the exponential integral fixes the constant as well as the logarithm.

Differentiate under the integral and integrate the derivative of $e^{-x-\epsilon /x}/x$ over $(0,\infty )$; the exponentially small endpoint terms vanish. This gives the ordinary differential equationFor the Frobenius method at its regular singular point, write $J=A\log \epsilon +B+\epsilon (C_1\log \epsilon +D)+\cdots$. Substitution gives $C_1=A$ and $D=B-2A$. The preceding leading calculation sets $A=-1$, $B=-2\gamma$, henceSubtracting the moving-cutoff correction to a symmetric exponential integral yields the requested answer, including the logarithmically enhanced first-order term:Thus the $\epsilon \log \epsilon$ switchback term must be retained when asking for all contributions through order $\epsilon$.