Manipulation of the Integrals into Fresnel Integrals

Having satisfied ourselves that we may use integrals to approximate the FFT sums, we now apply algebraic manipulations to convert the integrals in equations 24 and 25 into Fresnel integrals.

First, we consider the argument of these sinusoids as a quadratic in $n$, for which we may complete the square to obtain:

$$\alpha n^2 -2\pi k n/ K = \alpha\left(n-\frac{\pi k}{K\alpha}\right)^2 - \frac{\pi^2 k^2}{K^2\alpha}.$$

This gives

$$\Re{Y_a(k)} = \int_{-N/2}^{N/2} \cos\left(\alpha\left(n-\frac{\pi k}{K\alpha}\right)^2 - \frac{\pi^2 k^2}{K^2\alpha}\right) dn$$ (50)

$$\Im{Y_a(k)} = \int_{-N/2}^{N/2} \sin\left(\alpha\left(n-\frac{\pi k}{K\alpha}\right)^2 - \frac{\pi^2 k^2}{K^2\alpha}\right) dn.$$ (51)

Now, applying the trigonometric identities

$$\cos(A-B) = \cos(A)\cos(B)+\sin(A)\sin(B)$$

$$\sin(A-B) = \sin(A)\cos(B)-\cos(A)\sin(B)$$

yields

$$\Re{Y_a(k)} = \cos\left(\frac{\pi^2 k^2}{K^2\alpha}\right) \int_{-N/2}^{N/2} \c... ...t_{-N/2}^{N/2} \sin\left(\alpha\left(n-\frac{\pi k}{K\alpha}\right)^2\right) dn$$ (52)

$$\Im{Y_a(k)} = \cos\left(\frac{\pi^2 k^2}{K^2\alpha}\right) \int_{-N/2}^{N/2} \s... ..._{-N/2}^{N/2} \cos\left(\alpha\left(n-\frac{\pi k}{K\alpha}\right)^2\right) dn.$$ (53)

Next, we make the change of variables
$\parbox{1.5in}{\begin{eqnarray*} m &=& n-\frac{\pi k}{K\alpha}\\ dn &=& dm \end{eqnarray*}}$$\parbox{5in}{\begin{eqnarray*} n &=& -N/2 \Rightarrow m = -N/2 - \... ...\ n &=& N/2 \Rightarrow m = N/2 - \frac{\pi k}{K\alpha}\\ \end{eqnarray*}}$
giving

$$\Re{Y_a(k)} = \cos\left(\frac{\pi^2 k^2}{K^2\alpha}\right) \int_{-N/2 - \frac{\... ...c{\pi k}{K\alpha}}^{N/2 - \frac{\pi k}{K\alpha}} \sin\left(\alpha m^2\right) dm$$ (54)

$$% \Im{Y_a(k)} = \cos\left(\frac{\pi^2 k^2}{K^2\alpha}\right) \int_{-N/2 - \frac{\... ...{\pi k}{K\alpha}}^{N/2 - \frac{\pi k}{K\alpha}} \cos\left(\alpha m^2\right) dm.$$ (55)

Our final step is the substitution
$\parbox{1.5in}{\begin{eqnarray*} \frac{\pi}{2}u^2 &=& \alpha m^2 \... ...{2\alpha}{\pi}}m\\ dm &=& \pm\sqrt{\frac{\pi}{2\alpha}} du \end{eqnarray*}}$$\parbox{5in}{\begin{eqnarray*} m &=& -N/2 - \frac{\pi k}{K\alpha} ... ...\alpha}{\pi}}\left(\frac{N}{2}-\frac{\pi k}{\alpha K}\right), \end{eqnarray*}}$
which yields

$$\fbox{$ \displaystyle \Re{Y_a(k)} = \pm\sqrt{\frac{\pi}{... ...}\right)} \sin\left(\frac{\pi}{2}u^2\right) du \Bigg) $}\\$$

$$\fbox{$ \displaystyle \Im{Y_a(k)} = \pm\sqrt{\frac{\pi}{... ... K}\right)} \cos\left(\frac{\pi}{2}u^2\right) du \Bigg). $}$$

We observe that we have obtained closed form expressions in terms of Fresnel integrals.