# Wolfram Alpha wrong Integral

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?

• Just a sign error. As $x\to-\infty$, $F(x)\to-\frac1R$, so$$F(0)-\lim_{x\to-\infty}F(x)=0-\left(-\frac1R\right) = \frac1R$$ Feb 8 at 20:33
• 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. Feb 8 at 20:34
• You've assumed $\frac{x}{\sqrt{x^2}}=1.$ Feb 8 at 20:45

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 $$a\ne0$$.
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 $$x>0.$$