I must show that the integral $$\frac{1}{(2\pi)^{3}}\int_{\vec{k}}d^{3}k\frac{\cos(\vec{k}\cdot\vec{x})}{\left({\sqrt{|\vec{k}|^2+m^{2}}}\right)^{s}}=\delta^{3}(\vec{x})$$ when $s=0$ by using spherical coordinates. This should be true since $\delta^{3}(\vec{x})=\frac{1}{(2\pi)^{3}}\int_{-\infty}^{\infty}e^{i\vec{k}\cdot\vec{x}}d^{3}k$

My problem is just the lack of a factor $\frac{1}{2}$ at the final result.

Here is my procedure: (in spherical coordinates) $$\begin{aligned} \frac{1}{(2\pi)^{3}}\int_{-\infty}^{\infty}d^{3}k\frac{\cos(\vec{k}\cdot\vec{x})}{\left({\sqrt{|\vec{k}|^2+m^{2}}}\right)^{s}} &=\frac{1}{(2\pi)^{3}}\int_{0}^{\infty}dk\int_{0}^{2\pi}d\phi\int_{0}^{\pi}d\theta \sin(\theta) \frac{k^{2}}{(k^2+m^2)^{s/2}}\cos(kx \cos(\theta))\\ &=\frac{1}{(2\pi)^{2}}\int_{0}^{\infty}\frac{k^{2}}{(k^2+m^2)^{s/2}}\frac{2\sin(kx)}{kx}dk \\ &=\frac{1}{2\pi^{2}x}\int_{0}^{\infty}\frac{k\sin(kx)}{(k^{2}+m^{2})^{s/2}}dk \end{aligned}$$ Now, if $s=0$, then $$\begin{aligned} &=\frac{1}{2\pi^{2} x} \int_{0}^{\infty} k\sin(kx)dk \\ &=\frac{1}{4 \pi^{2} x} \frac{1}{i}\int_{-\infty}^{\infty} ke^{ikx}dk \\ &=\frac{1}{2 \pi x} \left(\frac{1}{2 \pi i}\int_{-\infty}^{\infty} ke^{ikx}dk\right) \\ &=\frac{1}{2\pi x}\left(-\frac{\partial \delta(x)}{\partial x}\right) \\ &=\frac{1}{2\pi x^{2}} \delta(x) \end{aligned}$$ where I used the identity $x\delta'(x)=-\delta(x)$.

This final result is almost the desired one since the Dirac delta in spherical coordinates in this case should be $$\delta^{3}(\vec{x})=\frac{1}{4 \pi x^{2}}\delta(x)$$ Where is the factor $\frac{1}{2}$ that I'm missing? Greetings and thank you a lot.

  • $\begingroup$ Please note that check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. $\endgroup$ Jan 6, 2021 at 5:37
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    $\begingroup$ Even when the delta function is on the boundary of the range of integration...? $\endgroup$
    – kaylimekay
    Jan 6, 2021 at 6:31
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    $\begingroup$ Indeed. Maybe it's helpful to think of the gamma as one of its resolved versions, like a narrow gaussian. Try the calculation and then take the limit. $\endgroup$
    – kaylimekay
    Jan 6, 2021 at 6:48
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    $\begingroup$ I meant delta, not gamma. This is what happens when you comment on here when you're in the middle of another calculation. $\endgroup$
    – kaylimekay
    Jan 6, 2021 at 7:06
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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Jan 7, 2021 at 5:38

1 Answer 1


The general delta function in 3D curved space is $$ \frac{\delta^{3}({\bf x}-{\bf x}_c)}{\sqrt{ g }} $$ where $g$ is determinant of metric. For spherical coordinates, you should have $\sqrt{g}=4\pi r^2$, if you have spherical symmetry. Thus you should get $$ \frac{\delta( r- r_c)}{4\pi r^2} $$

Edit: Delt function in spherical coordinates. Since $\sqrt{g}=r^2\sin\theta$, the full delta function without any additional symmetries should be $$ \frac{1}{r^2\sin\theta} \delta( r- r_c) \delta( \theta- \theta_c) \delta( \phi- \phi_c) $$ if you have axial symmetry, the delta function becomes $$ \frac{1}{2\pi r^2\sin\theta} \delta( r- r_c) \delta( \theta- \theta_c) $$

  • $\begingroup$ How does this answer the question? $\endgroup$ Jan 6, 2021 at 6:54
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    $\begingroup$ Doesn't answer the question and doesn't address the case where there is a coordinate singularity at $\mathbf x_c$ which is precisely the case here. $\endgroup$
    – kaylimekay
    Jan 6, 2021 at 6:56
  • $\begingroup$ @NiharKarve, kaylimekay, I just want to show that delta function in spherical coordinates is not $\delta(r)/(4\pi r^2)$. $\delta(r)/(4\pi r^2)$ is a delta function in spherical coordinates with an additional symmetry. $\endgroup$
    – user142288
    Jan 6, 2021 at 7:02
  • $\begingroup$ Hello, what is this additional symmetry? and how the function would be if this additional symmetry is not taken into account? $\endgroup$
    – Amadeus
    Jan 7, 2021 at 3:26
  • $\begingroup$ @Amadeus, sorry for not answering your question directly. See the update. $\endgroup$
    – user142288
    Jan 7, 2021 at 5:21

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