Now employ the Least Squares Lemma: $$ w_1 (x-A)^2 + w_2 (x-B)^2 = (w_1 + w_2)\left[x - \frac{w_1 A + w_2 B}{w_1 + w_2}\right]^2 + \frac{w_1 w_2}{w_1 + w_2} (A - B)^2 $$ Substitute $w_1 = v_x^2$ , $w_2 = v_y^2$ , $A = (x-s_x)/v_x$ , $B = (y-s_y)/v_y$ : $$ v_x^2.\left[t-(x-s_x)/v_x\right]^2 + v_y^2\left[(t-(y-s_y)/v_y\right]^2 = $$ $$ (v_x^2+v_y^2) \left[t - \frac{v_x^2.(x-s_x)/v_x+v_y^2.(y-s_y)/v_y}{v_x^2+v_y^2}\right]^2 + $$ $$ \frac{v_x^2.v_y^2}{v_x^2+v_y^2}\left[(x-s_x)/v_x - (y-s_y)/v_y\right]^2 = $$ $$ (v_x^2+v_y^2) \left[t - \frac{v_x.(x-s_x)+v_y.(y-s_y)}{v_x^2+v_y^2}\right]^2 + \frac{\left[v_y.(x-s_x) - v_x.(y-s_y)\right]^2}{v_x^2+v_y^2} $$ First introduce a couple of abbreviations: $$ v^2 = v_x^2 + v_y^2 \EN \mu = \frac{v_x.(x-s_x)+v_y.(y-s_y)}{v_x^2+v_y^2} $$ Then the exponent becomes: $$ v^2 (t-\mu)^2 \: + \: \left[ \frac{- v_y.(x-s_x) + v_x.(y-s_y)}{v}\right]^2 $$ The quotient $v_x/v$ may be set to the cosine and the quotient $v_y/v$ may be set to the sine of a certain angle $\phi$, giving for the second term: $$ \left[- \sin(\phi).(x-s_x) + \cos(\phi).(y-s_y)\right]^2 $$ This is recognized as the square of the function that defines a straight line, having a direction $(\cos(\phi),\sin(\phi))$: $$ - \sin(\phi).(x-s_x) + \cos(\phi).(y-s_y) = 0 $$ The introduction of an angle $\phi$ also affects the expression for $\mu$: $$ \sqrt{v_x^2+v_y^2} \; \mu = \frac{v_x.(x-s_x) + v_y.(y-s_y)}{\sqrt{v_x^2+v_y^2}} $$ $$ \hieruit v.\mu = \cos(\phi).(x-s_x)+\sin(\phi).(y-s_y) $$ $$ \hieruit v^2 (t-\mu)^2 = \left[ v.t - \left\{ \cos(\phi).(x-s_x)+\sin(\phi).(y-s_y) \right\} \right]^2 $$ This is recognized again as the square of the function that defines a straight line which is perpendicular to the previous one, having a normal $(\cos(\phi),\sin(\phi))$: $$ \cos(\phi).(x-s_x) + \sin(\phi).(y-s_y) = 0 $$ Now substitute the result into the integral: $$ F(x,y) = \vorm \int_{t_1}^{t_2} e^{-\half v^2 (t - \mu)^2 / \sigma^2 + \left[- \sin(\phi).(x-s_x) + \cos(\phi).(y-s_y)\right]^2 / \sigma^2} \; dt = $$ $$ e^{-\half \left[- \sin(\phi).(x-s_x) + \cos(\phi).(y-s_y)\right]^2 / \sigma^2} \; \vorm \int_{t_1}^{t_2} e^{-\half v^2 (t - \mu)^2 / \sigma^2} \; dt $$ Thus, the exponential function splits up into a part which is quite independent on the running parameter $t$ and another part which is still dependent on it. Only the latter has to be integrated further, of course. Due to our findings with respect to $v.\mu$, it is advantageous to introduce the length $s$ as the running parameter, instead of the (time) $t$, by: $s = v.t$ . Herewith the factor $v$ in the norm will cancel against the factor $v$ in $ds = dt/v$. Giving for the remaining integral: $$ I(x,y) = \norm \int_{s_1}^{s_2} e^{-\half \left[ \left(s - \left\{ \cos(\phi).(x-s_x) + \sin(\phi).(y-s_y) \right\} \right)/\sigma \right]^2} \; ds $$ The integral over a Gaussian can be expressed as the sum of two error functions, where the ERror Function ($\erf$) is defined as: $$ \erf(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{ - \half u^2} \; du $$ Now define as the new integration parameter: $$ u(s) = \frac{s-\left[\cos(\phi).(x-s_x)+\sin(\phi).(y-s_y)\right]}{\sigma} $$ Then $du = ds/\sigma$ and with $u_1 = u(s_1)$ , $u_2 = u(s_2)$ , we find : $$ I(x,y) = \erf(u_2) - \erf(u_1) \qquad \mbox{where} \qquad u_1 = u_1(x,y) \EN u_2 = u_2(x,y) $$ Giving as the end-result: $$ F(x,y) = \left[ \erf(u_2(x,y)) - \erf(u_1(x,y)) \right] e^{-\half \left[- \sin(\phi).(x-s_x) + \cos(\phi).(y-s_y)\right]^2 / \sigma^2} $$ For $(s_1,s_2) = (-\infty,+\infty)$ , hence $(u_1,u_2) = (-\infty,+\infty)$ , it is known that $\erf(u_1) = 0$ and $\erf(u_2) = 1$ . Hence: $$ F(x,y) = e^{-\half \left[- \sin(\phi).(x-s_x) + \cos(\phi).(y-s_y)\right]^2 / \sigma^2} $$ Herewith we find the fuzzyfication of a straight line, with infinite length.