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How does $$\int\limits_{k_0-(\Delta k/2)}^{k_0+(\Delta k/2)}e^{ikx}dk$$ equal to $(e^{ik_0x}/x)2\sin (\Delta k \cdot x/2)$?

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Use the indefinite integrals given in your previous question, to get $$ \begin{eqnarray} \int\limits_{k_0-(\Delta k/2)}^{k_0+(\Delta k/2)}e^{ikx}dk &=& \frac{1}{i x} e^{ikx}\Biggr|_{k_0-(\Delta k/2)}^{k_0+(\Delta k/2)}\\ &=& \frac{1}{i x} e^{i(k_0+(\Delta k/2))x} -\frac{1}{i x} e^{i(k_0-(\Delta k/2))x}\\ &=& \frac{e^{ik_0x}}{i x} \left(e^{i((\Delta k/2))x} -e^{i(-(\Delta k/2))x}\right)\\ &=& \frac{e^{ik_0x}}{x} 2\sin(x\Delta k/2). \end{eqnarray} $$ And to come back to your first question: Let $k_0=0$, $\Delta k/2=\frac1a$, forget the $2$ and look at $$ \lim_{a\to 0 } \frac{1/a}{x/a} \sin(x/a)=\lim_{a\to 0 } \frac{1}{a} \text{sinc}(x/a)=\delta(x), $$ yet another representation of Dirac's $\delta$ function.

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