So the integral was $$kR^2\int^{+\infty}_{-\infty}(R^2+x^2)^{\frac{-3}{2}}dx$$
$$\implies kR^2\int^{+\infty}_{-\infty}R^{-3}\left(1+\frac{x^2}{R^2}\right)^{\frac{-3}{2}}dx$$
$$\implies kR^2\int^{+\infty}_{-\infty}R^{-3}\left(1+\tan^2\theta\right)^{\frac{-3}{2}}dx$$
$$\implies k\int^{+\infty}_{-\infty}\left(sec^2\theta\right)^{\frac{-1}{2}}d\theta$$ Because $\cfrac{d \tan \theta} {dx}=\cfrac{1}{R}$ and $\cfrac{d\tan\theta}{d\theta}=1+\cfrac{x^2}{R^2}$ Therefore $dx=R\left(1+\frac{x^2}{R^2}\right)d\theta$
Integral changes to $$k\left[\sin\theta\right]^{\pi/2}_{-\pi/2}=2$$
But if I change the integral to $$ k\left[\left(1+\frac{R^2}{x^2}\right)^{\frac{-1}{2}}\right]^{+\infty}_{-\infty}=0$$
The given answer is 2 in book.
- Why is there a difference in the values of integral?
- If above contradiction is not wrong,why and when to prefer one formula over other?
Such integrals are used a lot of times in physics (e.g. field due to infinite solenoid, field due to infinite wire.) Does it show a property of calculus often ignored in physics?( you know like we cancel $dx$ in physics though technically wrong)