sinqr .pdf

A Question
Universe Generator
November 16, 2018
Hi, when we were calculating the fourier inverse of the coulomb potential
(for particle scattering) we encountered following integral
Z ∞
sin(qr)dr
0

I immediately thought: this does not converge.
But we used following ’trick’: we add a damping function e−µr and in the end
we take the limit µ → 0. It goes like
Z ∞
I=
e−µr sin(qr)dr
0
Z ∞
eiqr − e−iqr
dr
=
e−µr
2i
0
Z ∞

Z ∞
1
(iq−µ)r
(−iq−µ)r
=
e
dr −
e
dr
2i
0
0
The exponential, though complex, can just be integrated as usual.
Proof by wolfram alpha1 :
https://www.wolframalpha.com/input/?i=int+exp((i*q-m)*r)+dr,+r%3D0.
.infinity
We continue. . .

∞
∞ 
1
1
1
(iq−µ)r 
(−iq−µ)r 
I=
e
e
 −

2i iq − µ
−iq − µ
0
0


1
−1
1
=
+
2i iq − µ −iq − µ
1 2iq
=
2i q 2 + µ2
Now we do the magic: µ → 0 and we finally get:
Z ∞
1
sin(qr)dr =
q
0
We physicists often tend to sweep infinities under the rug but this one seems
particularly weird. Is this result mathematically correct? Is this some analytic
continuation of a function outside it’s initial definition?
1 Did

I mention I’m a physicist?

1