Example of non-integrable partial derivative

[jostpuur]

Can you give an example of a function $f:X\times Y\to\mathbb{R}$, where $X,Y\subset\mathbb{R}$, such that the integral

$$\int\limits_Y f(x,y) dy$$

converges for all $x\in X$, the partial derivative

$$\partial_x f(x,y)$$

exists for all $(x,y)\in X\times Y$, and the integral

$$\int\limits_Y \partial_x f(x,y) dy$$

diverges at least for some $x\in X$?

 

[Hurkyl]

Have you tried working backwards? Rather than guess at what f has to be to give a $\partial_x f$ with the desired property, instead choose $\partial_x f$ first.

 

[jostpuur]

If I choose $\partial_x f$ so that it cannot be integrated with respect to $y$, then I easily get a function $f$ which cannot be integrated with respect to $y$ either. It looks like a difficult task, either way you try it.

 

[Ben Niehoff]

I have a few infinite families of solutions. Do you want me to just post them or let you figure them out?

Hint: try $X, Y = \left[ 1, \infty \right)$

 

[jostpuur]

Feel free to post your example functions. I don't mind if you take, the right to the feel of discovery, away from me

I might try to obtain the feel of proving somebody wrong, when I check what's wrong with your example functions

 

[Ben Niehoff]

OK, the first one I thought of was

$$f(x,y) = \frac{1}{y^2} x^{-y}$$

where X and Y are both the interval [1, infinity).

$$\int_1^\infty f(x,y) \; dy$$

converges for all x in X, but

$$\int_1^\infty \partial_x f(x,y) \; dy = \int_1^\infty \frac{-1}{y} x^{-y-1} \; dy$$

diverges for x = 1.

 

[jostpuur]

I see. Very nice.

 

[Ben Niehoff]

And here is one where the sets X and Y are the entire real line:

$$f(x,y) = \frac{\sin{xy}}{y^2 + a^2}$$

Then

$$\int_R \partial_x f(x,y) \; dy = \int_R \frac{y \cos{xy}}{y^2 + a^2} \; dy$$

fails to converge for x = 0.