Help with Seemingly Hopeless Double Integral
As a mathematician, I would divide by force in the numerator and denominator by $d^3$, substitute $t/d$ by something, thus reducing to the case $d=1$. But here, let it be, we conserve the homogeneous setting as a control of the computations.
We split the numerator, compute first $$ \begin{aligned} J_1 &= \int_0^\rho dt \int_0^\pi \frac {t^2\sin\phi\cdot t\cos\phi} {\left[t^2\sin^2\phi+\left(t\cos\phi-D\right)^2\right]^{3/2}}\; d\phi \\ &= \int_0^\rho dt \int_0^\pi \frac {t^2(-\cos\phi)'\cdot t\cos\phi} {\left[t^2-2Dt\cos\phi+D^2\right]^{3/2}}\; d\phi \\ &\qquad\text{ Substitution: }u=\cos \phi\ , \\ &= \int_0^\rho dt \int_{-1}^1 \frac {t^3\; u} {\left[t^2-2Dt\;u+D^2\right]^{3/2}}\; du \\ &\qquad\text{ Substitution (for $u$, fixed $t$) of the radical }v=\sqrt{t^2-2Dt\;u+D^2}\ , \\ &\qquad u=\frac 1{2Dt}(t^2+D^2-v^2)\ ,\ du=-\frac v{Dt}\; dv\\ , \\ &= - \int_0^\rho dt \int_{\sqrt{t^2+2Dt+D^2}}^{\sqrt{t^2-2Dt+D^2}} \frac {t^3\; \frac 1{2Dt}(t^2+D^2-v^2)} {v^3}\; \frac v{Dt}\; dv \\ &= \int_0^\rho t\;dt \int_{D-t}^{D+t} \frac 1{2D^2} \cdot \frac {t^2+D^2-v^2} {v^2}\; dv \\ &= \int_0^\rho t\;dt \;\frac 1{2D^2} \left[ -(t^2+D^2)\frac 1v -1 \right]_{v=D-t}^{v=D+t} \\ &= \int_0^\rho dt \;\frac t{2D^2} \left[ (t^2+D^2)\left(\frac 1{D-t}-\frac 1{D+t}\right) - 2t \right] \\ &= \int_0^\rho dt \left[ \frac D{D+t} +\frac D{D-t} -2\frac{D^2+t^2}{D^2} \right] \\ &= D\ln\frac {D+t}{D-t} - 2\rho\left(1+\frac {\rho^2}{3D^2}\right) \ . \end{aligned} $$ Computer check for $D=2$, $\rho=1$ (pari/gp code):
? D=2; r=1;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%19 = 0.030557910669552716123823807178384744388
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%20 = 0.030557910669552716123823807178384742634
?
? D=223; r=101;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%22 = 1.9969022076015148346071622544965636670
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%23 = 1.9969022076015148346071622544965636629
The other integral. I will integrate here first w.r.t. $t$.
$$ \begin{aligned} J_2 &= -D \int_0^\pi d\phi \int_0^\rho \frac {t^2} {\left[(t-D\cos\phi)^2+D\sin^2\phi\right]^{3/2}} \; dt \\ &\qquad\text{ and we consider separately (without the factor $-D$)} \\ J_2(\phi) &= \int_0^\rho \frac {t^2} {\left[(t-D\cos\phi)^2+D\sin^2\phi\right]^{3/2}} \; dt \\ &= \int_{0-D\cos\phi}^{\rho-D\cos\phi} \frac {(u+D\cos\phi)^2} {(u^2+a^2)^{3/2}} \; du\ ,\qquad a:= D\sin\phi \ . \\ &\qquad \text{ Now the integrals can be computed} \\ \int \frac{u^2} {(u^2+a^2)^{3/2}} \; dt &= -\frac t{(u^2+a^2)^{1/2}}+\operatorname{arcsinh} \frac ta+C\ , \\ \int \frac{u} {(u^2+a^2)^{3/2}} \; dt &= -\frac 1{(u^2+a^2)^{1/2}}+C\ , \\ \int \frac{1} {(u^2+a^2)^{3/2}} \; dt &= -\frac {a^2\;u}{(u^2+a^2)^{1/2}}+C\ , \end{aligned} $$ and the computation goes on. If my calculus is ok, then $$ \begin{aligned} J_2(\phi) &= \int_0^\pi d\phi\; \Bigg[ \operatorname{arcsinh} \frac{t-D\cos \phi}{D\sin\phi} \\&\qquad\qquad\qquad+ \frac{t-D\cos\phi}{(t^2-2Dt\cos\phi+D^2)^{1/2}\sin^2\phi} \\&\qquad\qquad\qquad\qquad\qquad\qquad +\frac2{(t^2-2Dt\cos\phi+D^2)^{1/2}} \Bigg]_0^\rho\ . \end{aligned} $$ I have to submit, hope this is helpful to check with the own computations. I'll be back, but typing kills a lot of time.
Hint:
With the change of variable $u=\cos\phi$, the integral on $\phi$ becomes
$$\int_{-1}^1\frac{t^2(tu-d)}{\sqrt{(u-dt)^2+d^2(1-t^2)}}du.$$
By decomposition of the numerator, you will get a term
$$c(t)\log((u-dt)^2+d^2(1-t^2))$$
and another
$$c'(t)\arctan\frac{u-dt}{d\sqrt{1-t^2}}.$$
These terms do not simplify at the bounds of the integration interval.
The integral on $t$ (cubic in $t$ at the denominator) is worse. I am not optimisitc about existence of a closed-form.
\begin{align*} &\iint \frac{t^2 \sin(\phi) (t \cos(\phi) - d)}{(t^2 \sin^2(\phi) + (t \cos(\phi) - d)^2)^{3/2}} \,\mathrm{d}t\,\mathrm{d}\phi \\ &= \frac{\sqrt{d^2 + t^2 - 2 d t \cos(\phi)}(d^2 - 2 t^2 - 2 d t \cos(\phi) - 3 d^2 \cos(2 \phi))}{6d^2} \\ &+ d \cos(\phi) \ln\left(t - d \cos(\phi) + \sqrt{d^2 + t^2 - 2 d t \cos(\phi)}\right) \sin^2(\phi) \text{,} \end{align*} as one can readily verify. Then $I = \frac{-2 \rho^3}{3 d^2}$.
I think for your case $1$, you mean $I \propto \frac{1}{d^2}$. The integral can't be positive because:
- $t^2 \geq 0$ and
- $\sin(\phi) \geq 0$ since $\phi \in [0,\pi]$, but
- $t \cos(\phi) - d < 0$ because $0 < t < \rho < d$, while
- the denominator is $\geq 0$, so
- the integrand is (zero or) negative everywhere.