= Solution
Let $\delta(\mathbf x)$ be a mean-zero <density contrast>. For a statistically homogeneous and isotropic field, its <two-point correlation function> is $\xi(r)=\langle\delta(\mathbf x)\delta(\mathbf x+\mathbf r)\rangle$, depending only on $r=|\mathbf r|$. For a galaxy point process the corresponding distinct-pair definition is $dP_{12}=\bar n^2[1+\xi(r)]dV_1dV_2$, the excess pair probability relative to independent uniform positions. Self-pair shot noise is separated when using the density-field version.
Specify the Fourier convention by
$$
\delta(\mathbf x)=\int\frac{d^3k}{(2\pi)^3}\delta(\mathbf k)e^{i\mathbf k\cdot\mathbf x},\qquad
\langle\delta(\mathbf k)\delta^*(\mathbf k')\rangle=(2\pi)^3\delta_D^{(3)}(\mathbf k-\mathbf k')P(k).
$$
The diagonal covariance follows from <statistical homogeneity>, and isotropy makes the <matter power spectrum> a function of $k$ alone. Substitute the inverse transforms into the correlation and integrate the delta function:
$$
\xi(r)=\int\frac{d^3k}{(2\pi)^3}P(k)e^{i\mathbf k\cdot\mathbf r}.
$$
Choose the polar axis parallel to $\mathbf r$. Angular integration gives $\int d\Omega_k e^{ikr\cos\theta}=4\pi\sin(kr)/(kr)$. Hence
$$
\boxed{\xi(r)=\frac1{2\pi^2}\int_0^\infty k^2P(k)\frac{\sin(kr)}{kr}\,dk.}
$$
Conversely, when the transform is defined, $P(k)=4\pi\int_0^\infty r^2\xi(r)\sin(kr)/(kr)\,dr$. The stated convention fixes every $2\pi$ factor.
Back to article page
