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DavideGenoa
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Let ##\boldsymbol{l}:\mathbb{R}\to\mathbb{R}^3## be the piecewise smooth parametrization of an infinitely long curve ##\gamma\subset\mathbb{R}^3##. Let us define $$\boldsymbol{B}(\boldsymbol{x})=\frac{\mu_0 I}{4\pi}\int_\gamma\frac{d\boldsymbol{l}\times(\boldsymbol{x}-\boldsymbol{l})}{\|\boldsymbol{x}-\boldsymbol{l}\|^3}=\frac{\mu_0 I}{4\pi}\int_{-\infty}^{+\infty}\frac{\boldsymbol{l}'(t)\times(\boldsymbol{x}-\boldsymbol{l}(t))}{\|\boldsymbol{x}-\boldsymbol{l}(t)\|^3}dt.$$A physical interpretation of the integral is that ##\boldsymbol{B}## represents the magnetic field generated by an infinitely long wire ##\gamma## carrying a current ##I##, whith ##\mu_0## representing vacuum permeabilty.
Can we be sure that the integral converges in general and, if we can, how can it be proved?
I am posting here rather than in calculus because I suppose that the best way to approach the problem is by using the techniques of Lebesgue integration.
I notice that every component of the integral is the difference of two terms having the form ##l_i'(t)(x_j-l_i(t))\|\boldsymbol{x}-\boldsymbol{l}(t)\|^{-3}##, and I see that ##|l_i'(t)(x_j-l_i(t))|\|\boldsymbol{x}-\boldsymbol{l}(t)\|^{-3}\le |l_i'(t)||x_j-l_i(t)|^{-2}##, but the absolute value does not allow me to use the rule ##l_i'(t)dt=dl_i##...
Thank you so much for any answer!
Can we be sure that the integral converges in general and, if we can, how can it be proved?
I am posting here rather than in calculus because I suppose that the best way to approach the problem is by using the techniques of Lebesgue integration.
I notice that every component of the integral is the difference of two terms having the form ##l_i'(t)(x_j-l_i(t))\|\boldsymbol{x}-\boldsymbol{l}(t)\|^{-3}##, and I see that ##|l_i'(t)(x_j-l_i(t))|\|\boldsymbol{x}-\boldsymbol{l}(t)\|^{-3}\le |l_i'(t)||x_j-l_i(t)|^{-2}##, but the absolute value does not allow me to use the rule ##l_i'(t)dt=dl_i##...
Thank you so much for any answer!
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