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I am trying to derive the intensity variation function for a single slit diffraction. 
Sorry for the poor diagram...
So I decided to take the amplitudes of the waves originating from the slit on the left (wherein the variable that denotes distance within the slit is $l$) and integrate the amplitudes over the entire slit width, taking some point at a distance $x$ on the screen to achieve the resultant amplitude of the waves that strike the screen. With this function, I decided I would use the standard expression for intensity (i.e. $I=\kappa A^2)$
The amplitude for a wave originating from a point on the slit should be: $$ y=a\sin{kr}$$ where $r$ is the distance between the point of origin on the slit and point of contact on the screen (and $k$ is the angular wave-number). So: $$ r^2=D^2+(x+l)^2$$ and on approximating: $$ r\approx D+\frac{1}{2D}(x+l)^2$$ So I took the amplitude function (for the screen) as $A(x)$ and: $$ A(x)=a\int_{-l/2}^{l/2}\sin{kD+\frac{k}{2D}(x+l)^2} dl$$ substituting $k(x+l)/2D=u$ (ignoring limits for now): $$ A(x)=a\sqrt{\frac{2D}{k}}(\sin{kD}\int_{l_1}^{l_2}\cos{u^2}du+\cos{kD}\int_{l_1}^{l_2}\sin{u^2}du)$$ I looked these integrals up so I know that they are Fresnel Integrals, but more importantly that they are transcendental functions.
So my questions are:
- Are my assumptions flawed?
- Is there a flaw somewhere in the procedure?
- If what I've done is correct, how shall I proceed?
To answer your questions:
1) What is your basis for assuming what the amplitude of a wave originating from the slit is? Your assumption is not clear to me and does not appear to be correct.
2) See 1).
3) What is it that you want to do?
I will derive for you an expression for the field amplitude at a point on the screen. Let us start from first principles: the Huygens-Fresnel principle. If you are unfamiliar, I suggest reading up at the link or, better yet, Born & Wolf.
By Huygens-Fresnel, we have
$$E(x) = -\frac{i}{\lambda} \int_{-\ell/2}^{\ell/2} dx' \frac{e^{i 2 \pi r/\lambda}}{r} \frac{1+\cos{\chi}}{2} $$
where $r = \sqrt{(x-x')^2+D^2}$ and $\chi$ is the so-called angle of inclination. Here we will assume that, due to small angles, we may set $\chi=0$. Further, we will expand $r$ out to second order in $(x-x')/D$ in the phase term, but to zeroth order in the amplitude term. Thus, we may write, as the Fresnel approximation,
$$E(x) = -\frac{i}{\lambda D} e^{i 2 \pi D/\lambda} \int_{-\ell/2}^{\ell/2} dx' \, e^{i \pi (x-x')^2/(\lambda D)} $$
which we may now express in terms of Fresnel integrals as follows:
$$E(x) = -\frac{i}{\sqrt{2 \lambda D}} e^{i 2 \pi D/\lambda} \left [\mathcal{F}\left [\sqrt{\frac{2}{\lambda D}} \left (\frac{\ell}{2}-x \right ) \right ]-\mathcal{F}\left [-\sqrt{\frac{2}{\lambda D}} \left (\frac{\ell}{2}+x \right ) \right ] \right ] $$
where
$$\mathcal{F}(u) = \int_0^u du' \, e^{i \pi u'^2/2} $$
is the Fresnel integral.