Can a derivative operation commute over an integral operation irrespective of the properties of the function under the integral ?


Not in general. I recommend Gelbaum and Olmsted's Counterexamples in Analysis, which is where I turned to find a counterexample to your question. Namely, example 15 on page 123 is titled

A function $f$ for which $d/dx\int_a^b f(x,y)dy\neq\int_a^b[\partial/\partial x f(x,y)]dy$, although each integral is proper.

The example is

$$f(x,y) = \left\{ \begin{array}{lr} \frac{x^3}{y^2}e^{-x^2/y} & : y>0, \\ 0 & : y=0, \end{array} \right. $$ integrated with respect to $y$ from $0$ to $1$. Actually, differentiating under the integral sign works here except where $x=0$.

The function and its partial derivative are not jointly continuous. When they are jointly continuous, differentiation and integration commute.

  • $\begingroup$ the reason is because int_a^b f(x,y)dy is not differentiable at x=0. This can be seen easily by sustituting y = 1/t. I guess the commuting is permitted if and only if the derivative exists. here by derivative i mean that of the integral and not the partial derivative. $\endgroup$ – Rajesh Dachiraju Nov 3 '10 at 4:25
  • $\begingroup$ @Rajesh D: No, the integral is differentiable at $x=0$. The integral is $xe^{-x^2}$ for all real $x$. $\endgroup$ – Jonas Meyer Nov 3 '10 at 5:33
  • $\begingroup$ @Jonas: yeah got it, thanks for pointing. $\endgroup$ – Rajesh Dachiraju Nov 3 '10 at 6:37
  • $\begingroup$ @Jonas: could you please explain what is 'jointly continous' ? $\endgroup$ – Rajesh Dachiraju Nov 3 '10 at 6:38
  • 2
    $\begingroup$ @Rajesh: It basically just means continuous on the relevant subset of $\mathbb{R}^2$, and the terminology is to distinguish from "separate" continuity. The function and its partial are continuous in $x$ for each fixed $y$ and in $y$ for each fixed $x$. So it's continuous "separately", or "in each variable", but not continuous at $(0,0)$. As mentioned in the book, one way to see it is not continuous is to approach the point $(0,0)$ along the curve $y=x^2$. $\endgroup$ – Jonas Meyer Nov 3 '10 at 7:32

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.