Understanding the Solution to an Integral: A 3D Case

[jdstokes]

Homework Statement

I'm trying to understand the solution to this integral

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

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

The Attempt at a Solution

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

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

 

[nrqed]

Homework Statement

I'm trying to understand the solution to this integral

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

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

The Attempt at a Solution

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

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

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