Can the Leibniz integral rule be used on integrals with singular endpoints? I'd like to know if the Leibniz integral rule has an extension or generalization that can handle convergent integrals whose endpoints are singular. This post attempts to ask a similar question but doesn't give a good example. Here's my example:
$$\int_a^\infty \frac{1}{x\sqrt{x^2-a^2}}\,dx=\frac{\pi}{2a}.$$
Suppose we take $d/da$ of both sides. Clearly we have:
$$\frac{d}{da}\left(\int_a^\infty \frac{1}{x\sqrt{x^2-a^2}}\,dx\right)=\frac{-\pi}{2a^2}.$$
But the Liebniz integral rule seems to give:
$$\frac{d}{da}\left(\int_a^\infty \frac{1}{x\sqrt{x^2-a^2}}\,dx\right)=\int_a^\infty \frac{a}{x(x^2-a^2)^{3/2}}\,dx-\frac{1}{a\sqrt{a^2-a^2}}.$$
Of course, the rightmost term is a problem, which results from the fact that the endpoint of the original integral was singular. Suppose the original integral weren't so easily solvable and we actually needed the Leibniz integral rule to make progress—is there a valid way to apply it?

One idea I have is to change the original integral to 
$$\lim_{\epsilon\to 0^+}\int_{a+\epsilon}^\infty \frac{1}{x\sqrt{x^2-a^2}}\,dx$$
and then take the derivative using the Leibniz rule, which should result in a difference of two expressions that individually diverge as $\epsilon\to 0$, but whose difference converges. But this would require justifying bringing the derivative $d/da$ inside the $\lim_{\epsilon\to 0}$ operation, which would require a uniform convergence proof. Is there a shortcut that gets around this difficulty, or a guarantee that this maneuver is always allowed?
 A: In [Fich] are stated the following versions of Leibniz integral rule for improper integrals.
Functions with possible singular values are considered in [Fich, 511].
Theorem 3*. Let both a function $f(x,y)$ and its derivative $f’_y(x,y)$ are continuous in a rectangle $[a,b]\times [c,d]$. Let $g(x)$ be an absolutely (but not necessarily properly) integrable on $[a,b]$. Then 
$$\left(\int_a^b f(x,y)\cdot g(x) dx\right)’_y=\int_a^b f’_y(x,y)\cdot g(x) dx.$$ 
The proof is similar to the proof of the Leibnitz integral rule (when $g\equiv 1$) and also uses the following 
Theorem 1*. If a function $f(x,y)$ with fixed $y$ is integrable with respect to $x$ in $[a,b)$ and when $y\to y_0$ tend to a (bounded) function $\phi(x)$ uniformly with respect to $x$ then $$\lim_{y\to y_0}\int_a^b f(x,y)\cdot g(x) dx= \int_a^b \varphi(x)\cdot g(x) dx.$$
Functions with unbounded domains are considered in [Fich, 520].  
Theorem 3. Let both a function $f(x,y)$ and its derivative $f’_y(x,y)$ are continuous for $x\ge a$ and $y\in [c,d]$. If an integral $$I(y)=\int_a^\infty f(x,y) dx$$  converges for each $y\in [c,d]$ and an integral 
$$J(y)=\int_a^\infty f’_y(x,y) dx$$ converges uniformly with respect to $y$ in $[c,d]$ then for each $y\in [c,d]$ we have  $I’(y)=J(y)$. 
The proof (in Russian) is here and here.
Also there is stated that essentially the same formulations and arguments provide a generalization of Theorem 3*. For this it suffices to replace in them a point $x=\infty$ by a point $x=b$.
References
[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, v. II, 7-th edition, M.: Nauka, 1970. (in Russian).
A: When the integrand becomes singular at one of the integration limits, the best policy is to reexpress it to remove this singularity
and then apply the Leibniz rule. Thus, for your example:
\begin{align*}
\frac{d}{da}\int_a^{\infty} \frac{dx}{x\sqrt{x^2-a^2}}&=
\frac{d}{da}\int_a^{\infty}\left[\frac{d}{dx}\left(\frac{\sqrt{x^2-a^2}}{x^2}\right)+\frac{2\sqrt{x^2-a^2}}{x^3}\right]\,dx \\
&=\frac{d}{da}\left[0+\int_a^{\infty}\frac{2\sqrt{x^2-a^2}\,dx}{x^3}\right] \\
&=-2a\int_a^{\infty}\frac{dx}{x^3\sqrt{x^2-a^2}}-0 \\
&=-\frac{\pi}{2a^2}.
\end{align*}
