Computational Physics

4.8 Integrals involving Gaussian functions 75

Furthermore, the Fourier transform of exp(−γr^2 )is(π/γ )^3 /^2 exp(−k^2 / 4 γ),so
substituting these transforms into (4.107), we obtain

〈1s,α,A|−Z/rC|1s,β,B〉

=−Z( 2 π)−^6

(

π

γ

) 3 / 2 ∫

d^3 rd^3 k 1 d^3 k 2 Ke−k

12 /(^4 γ)

eik^1 ·(r−RP)

× 4 πk 2 −^2 eik^2 ·(r−RC). (4.109)
In this equation, the expression(4.106)for the delta-function fork 1 +k 2 is
recognised, and this transforms the integral into

〈1s,α,A|−Z/rC|1s,β,B〉

=−ZK( 2 π^2 )−^1

(

π

γ

) 3 / 2 ∫

d^3 ke−k

(^2) /( 4 γ)
k−^2 e−ik·(RP−RC). (4.110)
Integrating over the angular variables leads to
〈1s,α,A|−Z/rC|1s,β,B〉=N

∫∞

0

dke−k

(^2) /( 4 γ)
1 /ksin(k|RP−RC|);
N=− 2 ZK(π|RP−RC|)−^1 (π/γ )^3 /^2. (4.111)
The integral (withoutN) can be rewritten as
I(x)≡

1

2

∫x

0

dy

∫∞

−∞

dke−k

(^2) /( 4 γ)
cos(ky) (4.112)
withx=|RP−RC|. The integral overkis easy, and the result is
I(x)=

1

2

√

π/γ

∫x

0

dye−γy

2

. (4.113)

So, finally, we have

〈1s,α,A|−Z/rC|1s,β,B〉

=− 2 πKZγ−^1 (γ^1 /^2 |RP−RC|)−^1

∫γ 1 / (^2) |RP−RC|
0
dye−y
2
,
and, using the definition
F 0 (t)=t−^1 /^2
∫t 1 / 2
0
dye−y
2
, (4.114)
the final result can be rewritten as
〈1s,α,A|−Z/rC|1s,β,B〉
=− 2 πZ(α+β)−^1 exp[−αβ/(α+β)|RA−RB|^2 ]
×F 0 [(α+β)|RP−RC|^2 ]. (4.115)