# Fourier transform in crystallography: why do the bounds of the fourier integral have to be symmetric about the origin?

When analyzing the diffraction patterns of x-rays on crystals, we utilize the formula for the scattering intensity ($$I(\vec{K})$$):

$$I(\vec{K})\propto \left|\sum_G \rho_G\int_V e^{i(\vec{G}-\vec{K})\cdot \vec{r}}dV\right|^2 \tag{1}$$ where $$\rho(\vec{r})=\sum_G \rho_Ge^{i\vec{G}\cdot \vec{r}}$$ is the scatterer density (i.e electron density), and $$\vec{K}\equiv \vec{k'}-\vec{k}$$ is the difference between the scattered wave vector $$\vec{k'}$$ incident wave vector $$\vec{k}$$. Note that the integral above is performed over the volume of the whole crystal. My textbook then says that because the integral is over a plane wave with wave vector $$\vec{G} −\vec{K}$$ , if the crystal is very large, the integral will average over the crests and troughs of the wave and end up being very small (or even zero) unless $$\vec{K}=\vec{k'}-\vec{k}=\vec{G}$$ in which case the exponential becomes one and the integral will equal the crystal volume. It is this last paragraph that I take issue with for suppose that the crystal at hand is a perfect cube of side length $$L$$. Also suppose that $$\vec{K}\neq \vec{G}$$ so that $$\vec{G} −\vec{K}\equiv \vec{q}=(q_1,q_2,q_3)$$ where atleast one of the $$q_i$$ is non zero. In this case we have $$\int_V e^{i\vec{q}\cdot \vec{r}}dV=\int_{-L/2}^{L/2} \int_{-L/2}^{L/2} \int_{-L/2}^{L/2} e^{i\vec{q}\cdot \vec{r}} dxdydz=\frac{8 \sin \left(\frac{q_x L}{2}\right) \sin \left(\frac{q_y L}{2}\right) \sin \left(\frac{q_z L}{2}\right) }{q_x q_y q_z} \tag{2}$$ This result approaches $$L^3$$ as $$|\vec{q}|\rightarrow 0$$ and approaches zero as $$|\vec{q}|\rightarrow \infty$$ which are expected results. My primary issue is that the result of the above integral becomes totally different if we change the bounds of the integral from $$(\frac{-L}{2},\frac{L}{2})$$ to $$(0,L)$$. This should not be the case because we are still effectively integrating over the same volume. We are merely labeling its "corners" differently. Using these new bounds, we get that

$$\int_{0}^{L} \int_{0}^{L} \int_{0}^{L} e^{i\vec{q}\cdot \vec{r}} dxdydz=\frac{i \left(-1+e^{i L q_1}\right) \left(-1+e^{i L q_2}\right) \left(-1+e^{i L q_3}\right)}{q_1 q_2 q_3}$$ which is not only no longer similar to the sinc function but now it is imaginary. Why do we get this discrepancy when we simply choose to change the starting and ending points of the bounds for the integral?

• While changing the bounds of the integral when the integrand contains harmonic terms, one has to be careful since you aren't just labeling the corners differently rather you are shifting the volume. Think of it in this sense, how the changing of bounds of an integral which contains different symmetric properties(cosine and sine) affect the value of the integral itself. Aug 17, 2021 at 4:03
• @NormTrotsky Thanks for the response. Okay I see what your saying. Computing the limit of the integral with bounds $(0,L)$ as $(q_1,q_2,q_3) \rightarrow (0,0,0)$ I do in fact get the same result of $L^3$ which is the volume as required. But Mathematica doesn't seem to work when trying to compute the limit $(q_1,q_2,q_3) \rightarrow (\infty ,\infty ,\infty)$. But looking at the integrals result, it makes sense that it will tend to zero when $|q| \rightarrow \infty$. So basically despite the integral results looking different, we still get the same qualitative behavior which is all we need? Aug 17, 2021 at 10:47

The components of the allowed wave vectors defined on the cubic volume $$L^3$$ must fulfill the following Born- von Karman boundary conditions: $$q_i = \frac{2\pi}{L}\, m_i \quad \tag{1}$$
for $$i=,1,2,3$$ and $$m_i \in \mathbb Z$$. Then you see that both of your integrals yield the volume $$L^3$$ if $$q=0$$ and are vanishing whenever $$q\neq 0$$ - just plug equation $$(1)$$ with $$m_i \neq 0$$ into your results. After all, your integral is simply the orthonormality condition of plane waves in a cubic box (obeying the above periodic boundary conditions):
$$\delta_{kk^\prime} = \frac{1}{L^3} \int_{L^3}\mathrm dx\, e^{i(k^\prime-k)x} \quad .\tag{2}$$
As it was mentioned in the comments already, the problem is that if you want to make a 'transformation' of the form $$\int\limits_0^L \longrightarrow \int\limits_{-L/2}^{L/2} \quad ,$$ then you also have to make a substitution. For example, for a 1D integral (here all the integrals factor anyway) one could try $$a=L/2-x$$. In this case it will lead to a factor which assures that both integrals indeed yield the same result. However, it is equal to one when $$q=0$$ and constant whenever $$q\neq 0$$, which explains why both of your integrals still give the correct Kronecker-delta behaviour.