Jackson - p. 35 - integral of certain expression

[bjnartowt]

Homework Statement

The expression {\textstyle{{3{a^2}} \over {{{({a^2} + {r^2})}^{5/2}}}}} has a volume integral equal to 4\pi for arbitrary "a".

Homework Equations

The Attempt at a Solution

\int_0^R {\int_0^{\pi /2} {\int_0^\pi {{\textstyle{{3{a^2}} \over {{{({a^2} + {r^2})}^{5/2}}}}} \cdot dr \cdot r \cdot d\theta \cdot r\sin \theta \cdot d\phi } } } = 12\pi {a^2}\int_0^R {{\textstyle{{{r^2}} \over {{{({a^2} + {r^2})}^{5/2}}}}} \cdot dr}

Maple claims this integral, before evaluating at endpoints, is,

\left( {\frac{{{r^3}}}{{3{a^2}{{({r^2} + {a^2})}^{3/2}}}}} \right)_{r = 0}^{r = R} = \left( {\frac{{{R^3}}}{{3{a^2}{{({R^2} + {a^2})}^{3/2}}}} - \frac{{{r^3}}}{{3{a^2}{{({r^2} + {a^2})}^{3/2}}}}} \right)

I can't evaluate the second term, because the possibility of a --> 0 comes up later in the problem, and as a --> 0 and r --> 0, the second term clearly diverges, as would a Dirac delta function. How am I to argue that the integral does boil down to 4*pi? It just seems a farfetched claim...

 

[tiny-tim]

hi bjnartowt!

(have a pi: π*)

I can't evaluate the second term, because the possibility of a --> 0 comes up later in the problem, and as a --> 0 and r --> 0, the second term clearly diverges, as would a Dirac delta function. How am I to argue that the integral does boil down to 4*pi? It just seems a farfetched claim...

i don't see the difficulty …

that r=0 expression is zero for any non-zero value of a