- #1
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- 19
I'm confused about differentiating an improper integral. Consider the function
[tex]F(r)=\int_0^\infty J_0(rx)\,dx=\frac{1}{r}\int_0^\infty J_0(m)\,dm=\frac{1}{r}[/tex]
where I've solved the integral by making the substitution [itex]m=rx[/itex] (I think this is OK). Now I would like to find [itex]\frac{\partial F}{\partial r}[/itex]. From the solution I know that this is [itex]-\frac{1}{r^2}[/itex], but I would like to do it another way, by differentiating inside the integral. I thought it was allowable to write
[tex]\frac{\partial F(r)}{\partial r}=\int_0^\infty \frac{\partial}{\partial r}J_0(rx)\,dx=\int_0^\infty -x\,J_1(rx)\,dx[/tex]
but this integral doesn't converge. Where have I gone wrong? Thanks!
[tex]F(r)=\int_0^\infty J_0(rx)\,dx=\frac{1}{r}\int_0^\infty J_0(m)\,dm=\frac{1}{r}[/tex]
where I've solved the integral by making the substitution [itex]m=rx[/itex] (I think this is OK). Now I would like to find [itex]\frac{\partial F}{\partial r}[/itex]. From the solution I know that this is [itex]-\frac{1}{r^2}[/itex], but I would like to do it another way, by differentiating inside the integral. I thought it was allowable to write
[tex]\frac{\partial F(r)}{\partial r}=\int_0^\infty \frac{\partial}{\partial r}J_0(rx)\,dx=\int_0^\infty -x\,J_1(rx)\,dx[/tex]
but this integral doesn't converge. Where have I gone wrong? Thanks!