# find a certain $L^1(R)$ function

Hi I am just calculating an integral and I want to put the differential under the integral sign . I know that to get $D_x \int_E f(x,y)dy = \int_E D_xf(x,y)dy$ , need a $g(y)\in L^1(R)$ s.t $|D_xf(x,y)|\leq g(y)$ .

now my $f(x,y)$ is $e^{-ixy}$ and E is a bounded interval on the real line , and I cant find an integrable fucntion which bounds $|D_xf(x,y)|$. Could anyone here help me ? Thanks in advance.

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Since $$|\partial_x f(x,y)| = |y|,$$ you can take $g(y)=y$ on $E$ and zero otherwise.
@studenthp $|y| 1_{[a,b]}$ will do it then, where $[a,b]$ is any interval containing the projection of your compact set. –  N. S. Aug 28 '12 at 13:24
Sorry, I did not notice the boundedness of $E$. I changed my answer. –  Siminore Aug 28 '12 at 13:28
@Siminore OP forgot to add it in the original question, it was only added after your answer. You need to change a little the answer, keep in mind that $E$ is two dimensional, so your definition of $g$ is not perfect... –  N. S. Aug 28 '12 at 13:34
@N.S. It seems to me that $E$ is an interval. –  Siminore Aug 28 '12 at 13:38