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I want to evaluate the following integral: $$\int_{-\infty}^0\frac{R}{\sqrt{x^2+R^2}^3}\mathrm{d}x$$ If I do it with the substitution $x=R\tan(\varphi)$ I get: $$\int_{-\infty}^0\frac{R}{\sqrt{x^2+R^2}^3}\mathrm{d}x=\frac{1}{R}$$ Which is the same like Wolfram Alpha. But if I use the antiderivative $F(x)=\frac{x}{R\sqrt{x^2+R^2}}$ I get: $$\int_{-\infty}^0\frac{R}{\sqrt{x^2+R^2}^3}\mathrm{d}x=\underbrace{F(0)}_{=0}-F(-\infty)=-\frac{1}{R\sqrt{1+\frac{R^2}{x^2}}}\bigg|_{x=-\infty}=-\frac{1}{R}$$ So where is the mistake?

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    $\begingroup$ Just a sign error. As $x\to-\infty$, $F(x)\to-\frac1R$, so$$F(0)-\lim_{x\to-\infty}F(x)=0-\left(-\frac1R\right) = \frac1R$$ $\endgroup$
    – user170231
    Feb 8 at 20:33
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    $\begingroup$ Your mistake is in the evaluation of F. You cannot bring the $x$ inside the square root as you have done because it is negative. You must negate the square root if you do that. $\endgroup$
    – superckl
    Feb 8 at 20:34
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    $\begingroup$ You've assumed $\frac{x}{\sqrt{x^2}}=1.$ $\endgroup$ Feb 8 at 20:45

2 Answers 2

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The first approach gets $\frac1R\int_{-\pi/2}^0\cos\varphi d\varphi=\frac1R$; the second gets$$\frac1R[x(x^2+R^2)^{-1/2}]_{-\infty}^0=-\frac1R\lim_{x\to-\infty}x(x^2+R^2)^{-1/2}=-\frac1R\cdot-1=\frac1R,$$contra your sign error. You mistakenly rewrote $x(x^2+R^2)^{-1/2}$ as $(1+R^2/x^2)^{-1/2}$, rather than $-(1+R^2/x^2)^{-1/2}$, for $x<0$. Bear in mind $\frac{\sqrt{a^2+b}}{a}=\frac{\sqrt{1+b/a^2}}{\operatorname{sgn}a}$ for $a\ne0$.

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You've assumed $$\frac{x}{R\sqrt{R^2+x^2}}=\frac{1}{R\sqrt{1+\frac{R^2}{x^2}}},$$ which is equivalent to $\frac{x}{\sqrt{x^2}}=1.$ But that is only true if $x>0.$

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