# Leibniz's Rule for differentiation under the integral.

If we have

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

Then

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

and

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

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:

If

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

then

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

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:

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

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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). – Martin Argerami Feb 11 '12 at 17:30
@MartinArgerami I think you're wrong. Put $xt = u$ and see how $t$ vanishes. – Pedro Tamaroff Feb 11 '12 at 18:10
you are right about the first integral, my bad. – Martin Argerami Feb 11 '12 at 18:35

For your integral $$\int\limits_0^1 {\frac{{{x^\alpha } - 1}}{{\log x}}dx},$$ 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 repace the continuity assumption with boundedness (since $(x,\alpha)$ will be staying within a rectangle).