2nalfq8

2022-07-01

I want to evaluate the following integral:
${\int }_{-\mathrm{\infty }}^{0}\frac{R}{{\sqrt{{x}^{2}+{R}^{2}}}^{3}}\mathrm{d}x$

Rafael Dillon

Expert

The first approach gets $\frac{1}{R}{\int }_{-\pi /2}^{0}\mathrm{cos}\phi d\phi =\frac{1}{R}$; the second gets
$\frac{1}{R}\left[x\left({x}^{2}+{R}^{2}{\right)}^{-1/2}{\right]}_{-\mathrm{\infty }}^{0}=-\frac{1}{R}\underset{x\to -\mathrm{\infty }}{lim}x\left({x}^{2}+{R}^{2}{\right)}^{-1/2}=-\frac{1}{R}\cdot -1=\frac{1}{R},$
contra your sign error. You mistakenly rewrote $x\left({x}^{2}+{R}^{2}{\right)}^{-1/2}$, rather than $-\left(1+{R}^{2}/{x}^{2}{\right)}^{-1/2}$, for $x<0$. Bear in mind $\frac{\sqrt{{a}^{2}+b}}{a}=\frac{\sqrt{1+b/{a}^{2}}}{\mathrm{sgn}a}$ for $a\ne 0$

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