Came across the following integral today:$$\int_{\rho=0}^R \int_{\phi=0}^{2\pi}\int_{z=0}^H \frac{\rho d\rho d\phi dz}{z^2+R^2+\rho^2-2R\rho\cos(\phi)}$$
Here is my try: First we convert this to straightforward cartesian coordinates, which changes the integral to $$\int_{-R}^R \int_{-R}^R\int_0^H \frac{dx dy dz}{z^2+y^2+(R-x)^2}$$ Now if we change the variables $(x,y,z) \to (R-x,y,z)$, we can simply turn the integral into $$\int_{-R}^R \int_{-R}^R\int_0^H \frac{dx dy dz}{z^2+y^2+x^2}$$ Basically the Jacobian is $-1$ and the limits of $x$ simply get interchanged, thereby nullifying the effect of the $-1$. Now changing this back to cylindrical coordinates, we get $$\int_{\rho=0}^R \int_{\phi=0}^{2\pi}\int_{z=0}^H \frac{\rho d\rho d\phi dz}{z^2+\rho^2}$$ which can be evaluated without much difficulty to $$2\pi H\left[\frac RH\tan^{-1}\left(\frac HR\right)+\frac 12\ln\left(1+\frac{R^2}{H^2}\right)\right]$$ But this is not the answer given. THe naswer that is given is $$\pi H\left[1+\frac{2R}H-\sqrt{1+\frac{4R^2}{H^2}}+\ln\left(\frac12\left[1+\sqrt{1+\frac{4R^2}{H^2}}\right]\right)\right]$$ I know that i haven't made any mistake in the evaluation of the final integral in cylindrical coordinates. So, my mistake lies in the substitutions I am performing. Could someone point out where in the substitution I am going wrong, and suggest the correct substitution?
Here's a sketch of an evaluation in cylindrical coordinates:
After the substitutions $\rho = R r$ and $z = R u$ the integral takes the form $$ I \equiv R \int \limits_0^{H/R} \mathrm{d} u \int \limits_0^1 \mathrm{d}r \int \limits_0^{2 \pi} \mathrm{d} \phi \, \frac{r}{u^2 + 1 +r^2 - 2 r \cos(\phi)} \, . $$ The integral over $\phi$ can be evaluated using the residue theorem (the method is described here) or a tangent half-angle substitution. The result is $$ I = 2 \pi R \int \limits_0^{H/R} \mathrm{d} u \int \limits_0^1 \mathrm{d}r \, \frac{r}{\sqrt{(u^2 + 1 + r^2)^2 - 4 r^2}} \, .$$ Now let $t = r^2$, rewrite the denominator (by completing the square) and recognise the derivative of the inverse hyperbolic sine: \begin{align} I &= \pi R \int \limits_0^{H/R} \mathrm{d} u \int \limits_0^1 \mathrm{d}t \, \frac{1}{\sqrt{(t + u^2 - 1)^2 + 4 u^2}} \\ &= \pi R \int \limits_0^{H/R} \mathrm{d} u \left[\operatorname{arsinh} \left(\frac{u}{2}\right) - \operatorname{arsinh} \left(\frac{u^2-1}{2 u}\right)\right] \, . \end{align} Substitute $u = 2 v$, introduce $\alpha \equiv \frac{H}{2 R}$, use the definition of $\operatorname{arsinh}$ in terms of the logarithm and simplify the resulting expression to arrive at $$ I = 2 \pi R \int \limits_0^\alpha \mathrm{d} v \ln \left(\frac{1+\sqrt{1+v^{-2}}}{2}\right) \, . $$ Finally, integration by parts yields \begin{align} I &= 2 \pi R \left\{ \left[v \ln \left(\frac{1+\sqrt{1+v^{-2}}}{2}\right) \right]_{v=0}^{v=\alpha} + \int \limits_0^\alpha \mathrm{d} v \frac{v}{(1+\sqrt{1+v^{-2}})\sqrt{1+v^{-2}} \, v^3} \right\} \\ &= 2 \pi R \left\{ \alpha \ln \left(\frac{1+\sqrt{1+\alpha^{-2}}}{2}\right) + \int \limits_0^\alpha \mathrm{d} v \frac{1}{(v+\sqrt{1+v^2})\sqrt{1+v^2}} \right\} \\ &= 2 \pi R \left\{ \alpha \ln \left(\frac{1+\sqrt{1+\alpha^{-2}}}{2}\right) + \left[ - \frac{1}{v+\sqrt{1+v^2}} \right]_{v=0}^{v=\alpha}\right\} \\ &= 2 \pi R \alpha \left[1 + \alpha^{-1} - \sqrt{1+\alpha^{-2}} + \ln \left(\frac{1+\sqrt{1+\alpha^{-2}}}{2}\right) \right] \\ &= \pi H \left[1 + \frac{2R}{H} - \sqrt{1+ \frac{4 R^2}{H^2}} + \ln \left(\frac{1}{2}\left\{1+\sqrt{1+\frac{4 R^2}{H^2}}\right\}\right) \right] \, , \end{align} which agrees with your result.