# Tricky Integral: $\int_{r_{\text{min}}}^{\infty} \frac{1}{r^{7/2}}\frac{1}{\sqrt{r - r_{\text{min}}} }.$

I have the following integral:

$$\int_{r_{\text{min}}}^{\infty} dr\, \frac{1}{r^{7/2}}\frac{1}{\sqrt{r - r_{\text{min}}} },$$

where $r_{\text{min}}$ is positive.

$$\frac{2\sqrt{r - r_{\text{min}}}\left(8 r^2 + 4 r r_{\text{min}} + 3 r_{\text{min}}^2\right) }{15 r^{5/2} r_{\text{min}}^3}\bigg\vert_{r_{\text{min}}}^{\infty} = \frac{16}{15 r_{\text{min}}^3}.$$

I have no idea how to get this result. Integration by parts doesn't seem to work, since then the powers of $r$ or $(r-r_{\text{min}})$ in the denominators only increase. I don't see any substitution that would help either.

Let $a = r_{\mathrm{min}}$ \begin{align} \tag{a} \int\limits_{a}^{\infty} \frac{1}{x^{7/2}} \frac{1}{\sqrt{x-a}} dx &= \int\limits_{0}^{\infty} \frac{1}{\sqrt{z}} \frac{1}{(z+a)^{7/2}} dz \\ \tag{b} &= \frac{1}{a^{3}} \int\limits_{0}^{\infty} \frac{y^{-1/2}}{(y+1)^{7/2}} dy \\ &= \frac{1}{a^{3}} \mathrm{B}(1/2,3) \\ &= \frac{16}{15a^{3}} \end{align}
a. $z=x-a$
b. $z=ay$
The substitution $$v=\frac{1}{x}-r_\min$$ turns this into an easy integral, but you have to be carefull about signs and square roots.