Can I compute this integral analytically? 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} }  &amp;= \int_\Gamma  {\int {{e^{ - {r^2}}}rdrd\theta } } \\
 &amp;= \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} }  &amp;= \left( {\int_{ - a}^a {{e^{ - {x^2}}}dx} } \right)\left( {\int_{ - a}^a {{e^{ - {y^2}}}dy} } \right) 
\\
&amp;= {\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}
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} }  &amp;= \int_\Gamma  {\int {{e^{ - {r^2}}}rdrd\theta } } \\
 &amp;= \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} }  &amp;= \left( {\int_{ - a}^a {{e^{ - {x^2}}}dx} } \right)\left( {\int_{ - a}^a {{e^{ - {y^2}}}dy} } \right) 
\\
&amp;= {\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}