 Solution

Let $⟨{\Phi }_{\mathbf{k}}{\Phi }_{{\mathbf{k}}^{\prime }}^{\ast }⟩=(2\pi {)}^3{\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })P_{\Phi }(k)$. Statistical isotropy and $\int |Y_{\ell m}{|}^{2}d\Omega =1$ applied to the preceding expression give the CMB angular power spectrumFor $n=1$, define the constant dimensionless potential power ${\mathcal{A}}_{\Phi }=k^3{P}_{\Phi }(k)/(2{\pi }^2)$. ThenThe needed logarithmic spherical Bessel square integral can be proved directly. Put $I_{\ell }={\int }_0^{\infty }{j}_{\ell }^{2}dx/x$ and $J_{\ell }={\int }_0^{\infty }{j}_{\ell }{j}_{\ell -1}dx$. From the Spherical Bessel function recurrences,integration of the squares' derivatives gives $J_{\ell }=(\ell +1)I_{\ell }$, and for $\ell >1$ also $J_{\ell }=(\ell -1)I_{\ell -1}$. At $\ell =1$, $j_0(0)=1$ and $j_0(\infty )=0$ give $J_1=1/2$, hence $I_1=1/4$. Induction now proves $I_{\ell }=1/[2\ell (\ell +1)]$. ConsequentlyThis is the flat large-angle Sachs-Wolfe plateau. The useful integral printed in the PDF is missing the square on $j_{\ell }$: without that square the formula is false, since at $\ell =1$ its left-hand side is $\pi /4$, whereas its asserted value would be $1/4$. The squared identity above is the one required by the variance calculation. The plateau describes the matter-era large-angle approximation, not acoustic peaks or late integrated-potential corrections.