I will give a small background and explain the variables and the system first. I have two images which are observed and are constant and we can treat them as continuous functions and I will call them $r$ and $f$. In my problem, I am trying to find a continuous transform (which is very non-linear) that makes $f$ looks like $r$ according to some similarity criteria or cost function. I will call this transformation function $t$ and I am trying to estimate its parameters $w$.
So, the integral I need to compute turns out to be
$$ Z = \int_{-\infty}^{\infty} \exp-{\frac{\left( r(i) - f\left(t(w)\right)\right)^2}{2\sigma^2}} \, dw $$
where $\sigma$ is a constant. Now, given a constant linear function $A$, $f(t(w))$ is computed as:
$$ f(t(i, w)) = (\lceil{Aw}\rceil - Aw) * f(\lfloor{Ax}\rfloor) + (Aw - \lfloor{Aw}\rfloor) * f(\lceil{Ax}\rceil) $$
where $\lceil \rceil$ gives the ceiling function and $\lfloor \rfloor$ is the floor function. This basically means that I am using linear interpolation to make the transformation function continuous. This is because the images and the transformation are defined in the digital domain and are computed only on a uniform grid (corresponding to the pixel locations) and the transformation $t$ is telling me what the location of a pixel $i$ in image $r$ is in image $f$ through $w$.
Can someone tell me if I can compute such an integral? My first instinct was to use Taylor series to linearise $t(w)$ but then I realised it is not a good idea as $t(w)$ is in the integral and we are integrating over $w$. So the higher order terms will not cancel out and I cannot justify that approximation.
Yes, it can be solved using double integration. For simplicity, we integrate $\int_{ - \infty }^{\infty} {{e^{ - {x^2}}}dx}$. Consider the circular disc ${D_b}:{x^2} + {y^2} \le {b^2}$ with polar coordinates $(r,\theta)$ in the set $\Gamma :0 \le \theta \le 2\pi ,0 \le r \le b $. Therefore, \begin{align} \int_{{D_b}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } &= \int_\Gamma {\int {{e^{ - {r^2}}}rdrd\theta } } \\ &= \int_0^{2\pi } {\int_0^b {{e^{ - {r^2}}}rdrd\theta } } = \int_0^{2\pi } {\frac{1}{2}\left( {1 - {e^{ - {b^2}}}} \right)d\theta } = \pi \left( {1 - {e^{ - {b^2}}}} \right) \end{align} Let $S_a$ be the square $-a\le x \le a$, $-a\le y\le a$. Since $D_a \subseteq S_a \subseteq S_{2a}$ and ${{e^{ - \left( {{x^2} + {y^2}} \right)}}}$ is positive, \begin{align} \int_{{D_b}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } \le \int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } \le \int_{{D_{2a}}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } \end{align} It follows that \begin{align} \pi \left( {1 - {e^{ - {a^2}}}} \right) \le \int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } \le \pi \left( {1 - {e^{ - 4{a^2}}}} \right) \end{align} As $a \to \infty$, $\pi \left( {1 - {e^{ - {a^2}}}} \right) \to \pi$ and $\pi \left( {1 - {e^{ - 4{a^2}}}} \right) \to \pi$. Therefore, \begin{align} \mathop {\lim }\limits_{a \to \infty } \int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } = \pi. \end{align} But \begin{align} \int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } = \int_{ - a}^a {\int_{ - a}^a {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } &= \left( {\int_{ - a}^a {{e^{ - {x^2}}}dx} } \right)\left( {\int_{ - a}^a {{e^{ - {y^2}}}dy} } \right) \\ &= {\left( {\int_{ - a}^a {{e^{ - {x^2}}}dx} } \right)^2}. \end{align} Thus, \begin{align} \mathop {\lim }\limits_{a \to \infty } \int_{ - a}^a {{e^{ - {x^2}}}dx} = \mathop {\lim }\limits_{a \to \infty } {\left( {\int_{{S_a}} {\int {{e^{ - \left( {{x^2} + {y^2}} \right)}}dxdy} } } \right)^{1/2}} = \sqrt \pi. \end{align}