If we have

$$F\left( \alpha \right) = \int\limits_a^b {f\left( {\alpha ,x} \right)dx} $$


$$\frac{{F\left( {\alpha + \Delta \alpha } \right) - F\left( \alpha \right)}}{{\Delta \alpha }} = \frac{{\Delta F}}{{\Delta \alpha }} = \int\limits_a^b {\frac{{f\left( {\alpha + \Delta \alpha ,x} \right) - f\left( {\alpha ,x} \right)}}{{\Delta \alpha }}dx} $$


$$\mathop {\lim }\limits_{\Delta \alpha \to 0} \frac{{\Delta F}}{{\Delta \alpha }} = \frac{{dF}}{{d\alpha }} = \mathop {\lim }\limits_{\Delta \alpha \to 0} \int\limits_a^b {\frac{{f\left( {\alpha + \Delta \alpha ,x} \right) - f\left( {\alpha ,x} \right)}}{{\Delta \alpha }}dx} $$

However, this doesn't always mean

$$\mathop {\lim }\limits_{\Delta \alpha \to 0} \frac{{\Delta F}}{{\Delta \alpha }} = \frac{{dF}}{{d\alpha }} = \int\limits_a^b {\mathop {\lim }\limits_{\Delta \alpha \to 0} \frac{{f\left( {\alpha + \Delta \alpha ,x} \right) - f\left( {\alpha ,x} \right)}}{{\Delta \alpha }}dx} $$

$$\mathop {\lim }\limits_{\Delta \alpha \to 0} \frac{{\Delta F}}{{\Delta \alpha }} = \frac{{dF}}{{d\alpha }} = \int\limits_a^b {\frac{{\partial f\left( {\alpha ,x} \right)}}{{\partial \alpha }}dx} $$

I know that in other cases, for example in the integration of a series of functions or in sequences of functions, if $s(x)_n \to s(x)$ or $f_n(x) \to f(x) $ uniformly then we can integrate term by term (in the series) or change the order of integration and of taking the limit (in the sequence), i.e:


$${s_n}\left( x \right) = \sum\limits_{k = 0}^n {{f_k}\left( x \right)} $$


$$\mathop {\lim }\limits_{n \to \infty } \int\limits_a^b {{s_n}\left( x \right)dx} = \int\limits_a^b {s\left( x \right)dx} $$

and for the other case:

$$\mathop {\lim }\limits_{n \to \infty } \int\limits_a^b {{f_n}\left( x \right)dx} = \int\limits_a^b {\mathop {\lim }\limits_{n \to \infty } {f_n}\left( x \right)dx} $$

However Leibniz's rule is used in cases such as:

$$\int\limits_0^1 {\frac{{{x^\alpha } - 1}}{{\log x}}dx} $$

Which isn't even continuous in $[0,1]$. How can we then justify this procedure?


One particular example is

$$f(t) = \int\limits_0^\infty {\frac{{\sin \left( {xt} \right)}}{x}} dx =\frac{\pi}{2}$$

Which wrongly yields:

$$f'\left( t \right) = \int\limits_0^\infty {\cos \left( {xt} \right)dx} = 0$$

  • $\begingroup$ There are a couple problems with the last two integrals. In the one for $f(t)$, the integral does depend on $t$, i.e. its value is not always $\pi/2$ (it's $t\pi/2$, if I'm not wrong). The last integral doesn't exist (which still proves your point). $\endgroup$ – Martin Argerami Feb 11 '12 at 17:30
  • $\begingroup$ @MartinArgerami I think you're wrong. Put $xt = u$ and see how $t$ vanishes. $\endgroup$ – Pedro Tamaroff Feb 11 '12 at 18:10
  • $\begingroup$ you are right about the first integral, my bad. $\endgroup$ – Martin Argerami Feb 11 '12 at 18:35

Take a look at http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign

For your integral $$ \int_0^1 {\frac{{{x^\alpha } - 1}}{{\log x}}dx}, $$ I guess you need $\alpha>1$ (at least to apply the theorem the way it appears in the Wikipedia article). Be careful that $x$ in the article is your $\alpha$.

A more general result is Lebesgue's Dominated Convergence Theorem, where you can replace the continuity assumption with boundedness (since $(x,\alpha)$ will be staying within a rectangle).

  • 1
    $\begingroup$ You'll find it in any book on measure and integration. Some of the most common are Rudin's Real and Complex Analysis, Halmos' Measure Theory, Wheeden-Zygmund's Measure and Integral, Royden's Real Analysis. But there are many others, you will find some on your library. $\endgroup$ – Martin Argerami Feb 11 '12 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.