kaniello
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Dear "General Math" Community,
my goal is to calculate the following integral $$\mathcal{I} = \int_{-\infty }^{+\infty }\frac{f\left ( \mathbf{\vec{x}} \right )}{\left | \mathbf{\vec{c}}- \mathbf{\vec{x}} \right |}d^{3}x $$ in the particular case in which f\left ( \mathbf{\vec{x}} \right )=f\left ( x \right ) where x=\left | \mathbf{\vec{x}} \right |.
I switched to spherical coordinates and wrote \left | \mathbf{\vec{c}- \mathbf{\vec{x}}} \right |= \sqrt{c^{2}+x^{2}-2cx\cos \vartheta } and d^{3}x=x^{2}\cos \vartheta d\varphi d\vartheta dx. After the integration in \varphi and \vartheta it just remains $$\mathcal{I} = \frac{1}{c}\int_{0}^{c}fx^{2}dx+\int_{c}^{+\infty }fxdx.$$
The integral can also be seen as the convoultion of the function f with the function \frac{1}{\left | \mathbf{\vec{x}} \right |} so I expect to find the same result if I evaluate $$\mathscr{F}^{-1}\left \{ \hat{f} \cdot \frac{1}{k^{2}}\right \}$$ where \hat{f} and \frac{1}{k^{2}} are the Fourier transforms of the function f and the Coulomb potential up to some coefficients respectively.
So now I can write $$\int_{0}^{2\pi }\int_{-\frac{\pi }{2}}^{\frac{\pi}{2}}\int_{0}^{\infty }\frac{\hat{f}}{k^{2}}k^{2}e^{i\vec{\mathbf{k}}\cdot \vec{\mathbf{x}}}\cos \vartheta d\vartheta dkd\varphi = 2\pi\int_{-\frac{\pi }{2}}^{\frac{\pi}{2}}\int_{0}^{\infty }\frac{\hat{f}}{k^{2}}k^{2}e^{ikx\sin \vartheta } \cos \vartheta d\vartheta dk=$$$$=2\pi \int_{0}^{\infty }\hat{f}dk\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}e^{ikx\sin \vartheta } \cos \vartheta d\vartheta.$$
Performing the integration in \vartheta yelds \frac{1}{kx}\sin \left ( kx \right ) which finally brings to $$\frac{1}{x}\int_{0}^{\infty }\frac{f}{k}\sin \left ( kw \right )dk.$$
I am stuck at this point and I do not see how to recover the first solution.
Can anybody help me out?
Thank you very much in advance
my goal is to calculate the following integral $$\mathcal{I} = \int_{-\infty }^{+\infty }\frac{f\left ( \mathbf{\vec{x}} \right )}{\left | \mathbf{\vec{c}}- \mathbf{\vec{x}} \right |}d^{3}x $$ in the particular case in which f\left ( \mathbf{\vec{x}} \right )=f\left ( x \right ) where x=\left | \mathbf{\vec{x}} \right |.
I switched to spherical coordinates and wrote \left | \mathbf{\vec{c}- \mathbf{\vec{x}}} \right |= \sqrt{c^{2}+x^{2}-2cx\cos \vartheta } and d^{3}x=x^{2}\cos \vartheta d\varphi d\vartheta dx. After the integration in \varphi and \vartheta it just remains $$\mathcal{I} = \frac{1}{c}\int_{0}^{c}fx^{2}dx+\int_{c}^{+\infty }fxdx.$$
The integral can also be seen as the convoultion of the function f with the function \frac{1}{\left | \mathbf{\vec{x}} \right |} so I expect to find the same result if I evaluate $$\mathscr{F}^{-1}\left \{ \hat{f} \cdot \frac{1}{k^{2}}\right \}$$ where \hat{f} and \frac{1}{k^{2}} are the Fourier transforms of the function f and the Coulomb potential up to some coefficients respectively.
So now I can write $$\int_{0}^{2\pi }\int_{-\frac{\pi }{2}}^{\frac{\pi}{2}}\int_{0}^{\infty }\frac{\hat{f}}{k^{2}}k^{2}e^{i\vec{\mathbf{k}}\cdot \vec{\mathbf{x}}}\cos \vartheta d\vartheta dkd\varphi = 2\pi\int_{-\frac{\pi }{2}}^{\frac{\pi}{2}}\int_{0}^{\infty }\frac{\hat{f}}{k^{2}}k^{2}e^{ikx\sin \vartheta } \cos \vartheta d\vartheta dk=$$$$=2\pi \int_{0}^{\infty }\hat{f}dk\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}e^{ikx\sin \vartheta } \cos \vartheta d\vartheta.$$
Performing the integration in \vartheta yelds \frac{1}{kx}\sin \left ( kx \right ) which finally brings to $$\frac{1}{x}\int_{0}^{\infty }\frac{f}{k}\sin \left ( kw \right )dk.$$
I am stuck at this point and I do not see how to recover the first solution.
Can anybody help me out?
Thank you very much in advance