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If $a$ is a (WLOG) positive real number, ignoring all constants of integration we have $$ \int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) $$ $$ \int \frac{1}{x^2-a^2}\,dx = \frac{1}{2a}\log \left(\frac{a-x}{a+x}\right) $$ $$ \int \frac{1}{x^2}\,dx = \frac{-1}{x} $$Is it the case that the first two functions should approach $-x^{-1}$ in the limit as $a\to0$, and if so why? I can't imagine why or how we'd be able to pass the limit inside the integral first.

Integrand
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    Remember that antiderivatives are only defined up to constant. One can select a sequence of constants $C_a$ so that $F_a(x)+C_a\to-1/x$ when $a\to0$ and $x\neq0$ for both those functions. It is easier to see with definite integrals like $\int_1^x \frac{1}{x^2+a^2},dx$. – Conifold Apr 15 '20 at 02:10

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Remember that antiderivatives are only defined up to constant. One can select a sequence of constants $C_a$ so that $F_a(x)+C_a\to-1/x$ when $a\to0$ and $x\neq0$ for both those functions. It is easier to see with definite integrals like $\int_1^x \frac{1}{t^2+a^2}\,dt$, where it is justified to move the limit inside the integral because the convergence on $[1,x]$ is uniform. This gives us for $a\to0$:

$$ \int_1^x \frac{1}{t^2+a^2}\,dt=\frac{1}{a}\arctan\left(\frac{x}{a}\right)-\frac{1}{a}\arctan\left(\frac{1}{a}\right)\\ \to\int_1^x \frac{1}{t^2}\,dt=-\frac{1}{x}+1. $$

Therefore,

$$ \frac{1}{a}\arctan\left(\frac{x}{a}\right)-\left(\frac{1}{a}\arctan\left(\frac{1}{a}\right)+1\right)\to -\frac{1}{x}. $$ The other case is analogous. There is a deeper sense to reconciling those two cases that comes from passing to the complex domain, namely:

$$ \arctan(x)=\frac{1}{2i}\ln \left(\frac{x-i}{x+i} \right)+C. $$

Conifold
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    Ironic: I was trying to ignore the constants of integration for simplicity, but they're actually what make the problem work. Since I didn't think to use definite integrals, would you be amenable to using $t$ for your dummy variable of integration instead of $x$, to avoid confusion? – Integrand Apr 15 '20 at 02:30
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    @Integrand Done. I thought the indefinite/definite parallel would be more recognizable with $x$ even though it is an abuse of notation. – Conifold Apr 15 '20 at 02:33