# Conversion a three-dimensional integral into a one-dimensional one

I want to convert this integral $$\int_{\mathbb{R}^{3}}d^{3}k\frac{1}{|\mathbf{k}|^{2}}e^{i\mathbf{k}\cdot(\mathbf{x}-\mathbf{x}')}$$ to a one-dimensional one (here $d^{3}k$ means an element of volume of $\mathbb{R}^{3}$). I found a formula, but I really don't know how to prove it. The formula says $$\int_{\mathbb{R}^{3}}d^{3}kf(K)e^{i\mathbf{k}\cdot\mathbf{x}}=\frac{4\pi}{R}\int_{0}^{\infty}dK\, K\, f(K)\,\sin(KR)$$ where $K:=|\mathbf{k}|$ and $R:=|\mathbf{x}|$.

Can you help me to understand why this is true? Maybe proving the formula or telling me a book where I can find the proof. Thanks.

Note. Also it would be helpful if you tell me how to calculate the first integral (over all $k$-space) by another method.

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So I assume your initial domain of integration is $\mathbb{R}^3$ and the notation $d^3k$ stands for $dxdydz$, right? – 1015 Feb 15 '13 at 13:28
@julien Yes, the integral is over $\mathbb{R}^{3}$. And $d^{3}k$ means an element of volume of $\mathbb{R}^{3}$. – Ana Feb 15 '13 at 13:31
I think there shouldn't be an $x'$ in the lhs of your formula. Or maybe $R=|x-x'|$. – 1015 Feb 15 '13 at 13:36
@julien You are right, the correction has been done. – Ana Feb 15 '13 at 13:39
Strange, the edit record doesn't show $\mathbf x'$ ever appearing on the left-hand side of the second equation -- where did you correct it? – joriki Feb 15 '13 at 13:48

$$\mathrm e^{\mathrm i\mathbf k\cdot\mathbf r}=\mathrm e^{\mathrm iKR\cos\theta}=\sum_{l=0}^\infty\mathrm i^l(2l+1)j_l(KR)P_l(\cos\theta)\;.$$
Since you're integrating the plane wave with a spherically symmetrical function, the integral over all terms with $l\gt0$ vanishes by symmetry, so you get
\begin{align} \int\mathrm d^3kf(K)\mathrm e^{\mathrm i\mathbf k\cdot\mathbf x} &= \int\mathrm d^3kf(K)j_0(KR) \\ &= \int\mathrm d\Omega\int\mathrm dKK^2f(K)j_0(KR) \\ &= \int\mathrm d\Omega\int\mathrm dKK^2f(K)\frac{\sin(KR)}{KR} \\ &= \frac{4\pi}R\int\mathrm dKKf(K)\sin(KR)\;. \end{align}