Understanding the Solution to an Integral: A 3D Case

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SUMMARY

The integral under discussion is defined as \(\int_{\mathbb{R}^3} \frac{e^{i \mathbf{x} \cdot \mathbf{a}}}{\sqrt{r^2+1}}d\mathbf{x}\), where \(d\mathbf{x} = dxdydz\) and \(r = \sqrt{x^2+y^2+z^2}\). The initial transformation to spherical coordinates leads to the expression \(2\pi \int_0^{\infty}\frac{1}{\sqrt{r^2 +1}}\frac{e^{iua}-e^{-iua}}{iua}u^2du\). This transformation utilizes the relationship \(\vec{x} \cdot \vec{a} = x a \cos \theta\) and the volume element \(d^3 x = x^2 dx d\phi d \cos \theta\), integrating over \(\cos \theta\) from -1 to 1.

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Homework Statement



I'm trying to understand the solution to this integral

[itex]\int_{\mathbb{R}^3} \frac{e^{i \mathbf{x} \cdot \mathbf{a}}}{\sqrt{r^2+1}}d\mathbf{x}[/itex]

where [itex]d\mathbf{x} =dxdydz, r = \sqrt{x^2+y^2+z^2},\mathbf{a}\in \mathbb{R}^3[/itex]

The Attempt at a Solution



[itex]\int_{\mathbb{R}^3} \frac{e^{i \mathbf{x} \cdot \mathbf{a}}}{\sqrt{r^2+1}}d\mathbf{x} = 2\pi \int_0^{\infty}\frac{1}{\sqrt{r^2 +1}}\frac{e^{iua}-e^{-iua}}{iua}u^2du[/itex]

Could anyone please explain to me how this first step was obtained?
 
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jdstokes said:

Homework Statement



I'm trying to understand the solution to this integral

[itex]\int_{\mathbb{R}^3} \frac{e^{i \mathbf{x} \cdot \mathbf{a}}}{\sqrt{r^2+1}}d\mathbf{x}[/itex]

where [itex]d\mathbf{x} =dxdydz, r = \sqrt{x^2+y^2+z^2},\mathbf{a}\in \mathbb{R}^3[/itex]

The Attempt at a Solution



[itex]\int_{\mathbb{R}^3} \frac{e^{i \mathbf{x} \cdot \mathbf{a}}}{\sqrt{r^2+1}}d\mathbf{x} = 2\pi \int_0^{\infty}\frac{1}{\sqrt{r^2 +1}}\frac{e^{iua}-e^{-iua}}{iua}u^2du[/itex]

Could anyone please explain to me how this first step was obtained?

Just write [itex]\vec{x} \cdot \vec{a} = x a cos \theta[/itex] and the volume element as [itex]d^3 x =x^2 dx d\phi d \cos \theta[/itex] and integrate over cos theta from -1 to 1.
 

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