In this paper is calculated the square modulus of the radial fourier transform of the function $\rho(r)$
$$\left|F(q)\right|^2=\left| \int_{\mathbb{R}^3} e^{i\mathbf{q}\cdot\mathbf{r}}\rho(\mathbf{r}) d^3r\right|^2=\left| \frac{4\pi}{q}\int_0^{+\infty}\sin(qr)r\rho(r) dr\right|^2$$
where $\rho(r)$ is defined as the sum of three gaussians
$$ \rho(r) = \sum_{k=1}^{3} \rho_k \exp[-(r-\delta_k)^2/(2\sigma_k^2)]$$
They claim that
$$ F(q)=\frac{2}{q}\sum_{k=1}^{3}\rho_k\sigma_k\exp(-q^2\sigma_k^2/2)[\delta_k\sin(q\delta_k)+\sigma_k^2q\cos(q\delta_k)] $$
but if I do the math I obtain
$$ F(q)=\frac{4\pi}{q}\int_0^{+\infty}\sin(qr)r\rho(r) dr =\\= \frac{2\pi}{iq}\sum_{k=1}^3\left[\int_0^{+\infty}r \exp(iqr)\exp[-(r-\delta_k)^2/(2\sigma_k^2)] dr - c.c. \right]$$ where $c.c.$ is the complex conjugate of the same integral.
$$ \int_0^{+\infty}r \exp(iqr)\exp[-(r-\delta_k)^2/(2\sigma_k^2)] dr=\\=\int_0^{+\infty}r \exp(-(r+\beta)^2/)\exp(-q^2\sigma^2/2)\exp(iq\delta_k) dr $$ where $\beta=-\delta_k-iq\sigma_k^2$. Then $$ \int_0^{+\infty}r \exp(-(r+\beta)^2/2\sigma_k^2)dr=-\beta\int_{-\beta}^{-\beta+\infty} \exp(-y^2/2\sigma_k^2)dy+\int_{-\beta}^{-\beta+\infty} y\exp(-y^2/2\sigma_k^2)dy=(\delta_k+iq\sigma_k^2)\sigma_k\sqrt\frac{\pi}{2} +\sigma_k^2\exp(q^2\sigma_k^2/2)$$
Putting all the terms together, I obtain:
$$ F(q)=\frac{4\pi}{q}\sum_{k=1}^{3}\rho_k\left\{\sqrt{\frac{\pi}{2}}\sigma_k\exp(-q^2\sigma_k^2/2)[\delta_k\sin(q\delta_k)+\sigma_k^2q\cos(q\delta_k)]+\sigma_k^2\sin(q\delta_k)\right\}$$
So, I have an additional term: $\sigma_k^2\sin(q\delta_k)$. Is the paper wrong or am I missing something?