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?

  • $\begingroup$ 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$$\int_0^\infty\cos(xt)~\mathrm dt=0$$ $\endgroup$ Aug 13 '19 at 15:24
  • $\begingroup$ One may also note that the last limit does not converge uniformly in $a$. Consider $a\to0^+$. $\endgroup$ Aug 13 '19 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 '19 at 15:30
  • $\begingroup$ It's simply not possible to pass the derivative under the integral sign. $\endgroup$ Aug 13 '19 at 15:31
  • 1
    $\begingroup$ 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 '19 at 22:16

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 $x=b$.


[Fich] Grigorii Fichtenholz, Differential and Integral Calculus, v. II, 7-th edition, M.: Nauka, 1970. (in Russian).

  • $\begingroup$ My question is about integrals whose limits of integration are themselves functions of the variable with respect to which we are taking a derivative. It doesn’t look like your $a$ or $b$ endpoints vary as functions of $y$. In particular, I’m interested in cases where $f(x,y)$ is singular at $y=a$. Does your answer address this case? $\endgroup$
    – WillG
    Oct 11 '19 at 17:43

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*}


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