Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Here is a consequence of Lebesgue dominated convergence theorem on differentiation under integral sign.

Function $f(x, t)$ is differentiable at $x_0$ for almost all $t \in A$, and $t \to f(x, t)$ is integrable. Moreover, there exist an integrable function $g(t)$ such that $|(\partial{f}/\partial{x})(x,t)| \le g(t)$. Then we have $$\frac{\mathrm{d}}{\mathrm{d}x} \left( \int_A f(x,t) ~ \mathrm{d}t \right)\bigg|_{x=x_0} = \int \frac{\partial{f}}{\partial{x}}(x_0,t) ~ \mathrm{d}t. $$


Of course, here the word "integrable" means "Lebesgue integrable". But, if we read the word "integrable" as "improper Riemann integrable" and add an assumption that the partial derivative is continuous, then I think the statement is still true. Weierstrass M-test for integrals guarantees uniform convergence of the improper integral and we can interchange the differentiaion and integration. If $f$ is integrable in both of (improper) Riemann and Lebesgue sense, the only gain of interpreting the integral as Lebesgue one is gettig rid of the continuity condition of the partial derivative. Is it right?

share|improve this question
    
In your right hand side is it supposed to be df/dx(x,t)dt evaluated at x = x0 and not df/dx(x0,t)? –  Michael Smith Dec 5 '12 at 5:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.