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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?

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  • $\begingroup$ One must be very careful, especially with infinite bounds. For example, with $u=xt$, it is clear that we have, for $x>0$,$$\int_0^\infty\frac{\sin(xt)}t~\mathrm dt=\int_0^\infty\frac{\sin(u)}u~\mathrm du$$but naively differentiating both sides gives you$$\int_0^\infty\cos(xt)~\mathrm dt=0$$ $\endgroup$ – Simply Beautiful Art Aug 13 at 15:24
  • $\begingroup$ One may also note that the last limit does not converge uniformly in $a$. Consider $a\to0^+$. $\endgroup$ – Simply Beautiful Art Aug 13 at 15:27
  • $\begingroup$ Oh god. What rule is broken in the above example when you take the $d/dx$ under the integral sign? $\endgroup$ – WillG Aug 13 at 15:30
  • $\begingroup$ It's simply not possible to pass the derivative under the integral sign. $\endgroup$ – Simply Beautiful Art Aug 13 at 15:31
  • $\begingroup$ But surely sometimes we can pass the derivative under the integral sign—usually there are precise conditions that must be met, according to rigorous theorems that say when you can and can't. I think the issue with your example is that the integrand is not Lebesgue integrable. My example is Lebesgue integrable, so I'm not convinced your argument applies to it. $\endgroup$ – WillG Aug 13 at 22:16

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