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

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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