Problem related to the integration

Question

Recently I read that $$\int_{-p^{2}}^{+p^{2}} \frac{1}{\sqrt{x^{2}-p^{2}}}dx$$ tends to a finite real number as $p \to 0$. Can anyone explain me why this is true?

Answer 1

For $-1 < p < 1$, the integral

$$I(p) := \int_{-p^2}^{p^2} \frac{1}{\sqrt{x^2-p^2}}\,dx$$

has a purely imaginary value, since the radicand $x^2 - p^2$ is negative. Nevertheless, $\lim\limits_{p\to 0} I(p) = 0 \in \mathbb{R}$. Broadly, we can argue that the integrand has roughly the value $\pm \frac{i}{p}$, the sign depending on the chosen branch of the square root, and since the length of the interval of integration is $2p^2$, we have $I(p) \approx \pm 2ip \to 0$.

More precisely, we can substitute $x = py$, and obtain

$$I(p) = \int_{-\lvert p\rvert}^{\lvert p\rvert} \frac{1}{\sqrt{y^2-1}}\,dy.$$

In this form, it should be evident that $\lim\limits_{p\to 0} I(p) = 0$, since the integrand doesn't depend on $p$, and the interval of integration still shrinks to a single point.