# Detail from Integral Evaluation in Szabo (Appendix)

This is a step from Appendix A in Szabo and Ostlund's Quantum Chemistry. In short, the question is how to get from \ref{A.20} to \ref{A.21}. We can factor out some constants $K$ in the expressions to get:

\begin{align} (A| -Z_C/r_{1C} |B) &= K \left(\frac{1}{2\pi^2}\right) \int \mathrm{d}\mathbf{k} \; e^{-k^2/4p} k^{-2} \exp\big(-i\,\mathbf{k}\cdot(\mathbf{R}_P - \mathbf{R}_C)\big) \tag{A.20}\label{A.20}\\ &= K \left(\frac{2}{\pi|\mathbf{R}_P-\mathbf{R}_C|}\right) \int\limits_0^\infty \mathrm{d}k \; e^{-k^2/4p} k^{-1} \sin(k|\mathbf{R}_P-\mathbf{R}_C|) \tag{A.21}\label{A.21} \end{align}

Part of the note between these steps is that (paraphrasing) we can let $\mathbf{R}_P-\mathbf{R}_C$ lie on the $z$-axis and use that $k\cdot (\mathbf{R}_P-\mathbf{R}_C)= k|\mathbf{R}_P-\mathbf{R}_C|\cos\theta$, "then we can easily perform the angular part of the integration over $\boldsymbol{k}$ to obtain [\ref{A.21}]."

What I have done: It seems that Euler's formula must be applied to the second exponent in \ref{A.20} but using the hint we'd have something like $$\cos(k|\mathbf{R}_P-\mathbf{R}_C|\cos\theta) + i\sin \dots \text{etc}. \tag{???}$$ I guess that the sine in \ref{A.21} follows integration of the (real) cosine in Euler's expansion. In line with this, \ref{A.21} is multiplied by a factor of $2\pi$.

I will keep at this but I think if it puzzles me, then others may have stumbled over it so I am posting.
Hopefully removal of the common factors ($K$) makes this easier to read but the problem is on pages 413-414 of Szabo and Ostlunds book.

• Don't have my Szabo at hand (heresy, I know), will look into it today/tomorrow...
– user37142
Aug 21 '18 at 14:05
• There is an integration 'missing' effectively $\int \exp(-a\cos(\theta) ) d\theta$. Substituting $z=\sin(\theta)$ and some fiddling about makes the integral easier. Aug 21 '18 at 19:12
• Using porphyrin's substitution and Euler's relation I get an answer which is off by a factor of $2\pi ~k^2.$ If a spherical volume was intended by the original vector notation then this seems to be OK. Would still like an authoritative answer... Aug 24 '18 at 18:33

As commented, do note that the integral (A.20) is taken over all $\mathbf{k}$-space, whereas the integral (A.21) is one-dimensional. The explicit calculation of this integral follows as
\begin{align} \int\text{d}\mathbf{k}\,\Psi(k)e^{-i\mathbf{k\cdot r}} &=\int_0^\infty\text{d}k\,k^2\int_0^\pi\text{d}\theta\int_0^{2\pi}\text{d}\varphi\,\sin\theta\Psi(k)e^{-ikr\cos\theta}\\ &= 2\pi\int_0^\infty\text{d}k\,k^2\Psi(k)\int_0^\pi\text{d}\theta\,\sin\theta e^{-ikr\cos\theta}\\ &= 2\pi\int_0^\infty\text{d}k\,k^2\Psi(k)\left[\frac{e^{ikru}}{ikr}\right]_{u=-1}^1\\ &= \frac{4\pi}{r}\int_0^\infty\text{d}k\,k\Psi(k)\sin(kr).\end{align}
In the first equality we convert this integral from Cartesian to spherical coordinates, in the second equality we evaluate the integral over $\varphi$, in the third equality we substitute $u = -\cos\theta$, and in the fourth equality we make use of Euler's formula to get the exponential representation of $\sin(kr)$.
• This looks right to me...$\Psi(k)$ has absorbed the $k^{-2}$ in A.20, right? Aug 28 '18 at 18:42
• @daniel, yes.$\$ Aug 28 '18 at 18:46