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). $\endgroup$ Commented Feb 11, 2012 at 17:30
  • 1
    $\begingroup$ @MartinArgerami I think you're wrong. Put $xt = u$ and see how $t$ vanishes. $\endgroup$
    – Pedro
    Commented Feb 11, 2012 at 18:10
  • $\begingroup$ you are right about the first integral, my bad. $\endgroup$ Commented Feb 11, 2012 at 18:35

1 Answer 1


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. $\endgroup$ Commented Feb 11, 2012 at 21:40

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