Seems like you want to compute the CDF/pdf of the sample range (assume the underlying sample \( X \) is a continuous random variable) - the difference between the sample maximum and sample minimum, \( X_{(n)} - X_{(1)} \)

http://en.wikipedia.org/wiki/Range_(statistics)

By elementary multinomial-type arguments, the joint pdf of \( (X_{(1)}, X_{(n)})\) is

\( f_{X_{(1)}, X_{(n)}}(x_1, x_n) \)

\( = \frac {n!} {0!1!(n-2)!1!0!} F_X(x_1)^0f_X(x_1)^1[F_X(x_n) - F_X(x_1)]^{n-2}f_X(x_n)^1[1 - F_X(x_n)]^0 \)

\( = n(n-1)f_X(x_1)f_X(x_n)[F_X(x_n) - F_X(x_1)]^{n-2} \)

where \( f_X, F_X \) are the pdf and CDF of the orginal unordered random sample.

Using elementary transformation tricks, we know that the pdf of sample range

\( f_R(r) = \int_{-\infty}^{+\infty} n(n-1)f_X(x)f_X(x+r)[F_X(x+r) - F_X(x)]^{n-2}dx,

r > 0 \)

By integration, the CDF is

\( F_R(r) = \int_{-\infty}^{+\infty} nf_X(x)[F_X(x+r) - F_X(x)]^{n-1}dx, r > 0 \)

as listed in the above wiki link.

Therefore one just need to apply the above integral to find the answer. However many of them are quite ugly and have no closed-form solution. For your case, lets try to calculate the CDF:

\( n\int_0^{+\infty} \frac {x} {\sigma^2} \exp\left\{-\frac {x^2} {2\sigma^2}\right\}

\left[\exp\left\{-\frac {x^2} {2\sigma^2}\right\}

- \exp\left\{-\frac {(x+r)^2} {2\sigma^2}\right\}\right]^{n-1} dx \)

\( = \frac {n} {\sigma^2} \int_0^{+\infty} x \exp\left\{-\frac {nx^2} {2\sigma^2}\right\} \left[1 - \exp\left\{-\frac {2rx + r^2} {2\sigma^2}\right\}\right]^{n-1}dx\)

\( = \frac {n} {\sigma^2} \int_0^{+\infty} x \exp\left\{-\frac {nx^2} {2\sigma^2}\right\} \sum_{k=0}^{n-1} \binom {n-1} {k} (-1)^k \exp\left\{-\frac {2krx + kr^2} {2\sigma^2}\right\} dx\)

\( = \frac {n} {\sigma^2} \sum_{k=0}^{n-1} \binom {n-1} {k} (-1)^k \int_0^{+\infty} x \exp\left\{-\frac {nx^2 + 2krx + kr^2} {2\sigma^2}\right\} dx\)

\( = \frac {n} {\sigma^2} \sum_{k=0}^{n-1} \binom {n-1} {k} (-1)^k \) \( \int_0^{+\infty} x

\exp\left\{- \frac {n} {2\sigma^2} \left(x^2 + 2 \frac {kr} {n} x + \frac {k^2r^2} {n^2}\right)

+ \frac {1} {2\sigma^2} \left(\frac {k^2r^2} {n} - kr^2 \right)\right\} dx\)

\( = \frac {n} {\sigma^2} \sum_{k=0}^{n-1} \binom {n-1} {k} (-1)^k \) \( \int_0^{+\infty} x

\exp\left\{- \frac {n} {2\sigma^2} \left(x^2 + 2 \frac {kr} {n} x + \frac {k^2r^2} {n^2}\right)

- \frac {1} {2\sigma^2} \left(kr^2 - \frac {k^2r^2} {n}\right)\right\} dx\)

\( = \frac {n} {\sigma^2} \sum_{k=0}^{n-1} \binom {n-1} {k} (-1)^k

\exp\left\{-\frac {kr^2} {2\sigma^2} \left(1 - \frac {k} {n}\right) \right\} \) \( \int_0^{+\infty} x \exp\left\{- \frac {n} {2\sigma^2} \left(x + \frac {kr} {n}\right)^2 \right\}dx \)

Finally we consider:

\( \int_0^{+\infty} \left(x + \frac {kr} {n} \right) \exp\left\{- \frac {n} {2\sigma^2} \left(x + \frac {kr} {n}\right)^2 \right\}dx = \frac {\sigma^2} {n} \)

and

\( \int_0^{+\infty} \exp\left\{- \frac {n} {2\sigma^2} \left(x + \frac {kr} {n}\right)^2 \right\}dx \)

\( = \sqrt{2\pi\frac {\sigma^2} {n}} \int_0^{+\infty} \frac {1} {\sqrt{2\pi\frac {\sigma^2} {n}}} \exp\left\{- \frac {n} {2\sigma^2} \left(x + \frac {kr} {n}\right)^2 \right\}dx\)

\( = \sqrt{2\pi\frac {\sigma^2} {n}} \Pr\left\{\frac {\sigma} {\sqrt{n}} Z - \frac {kr} {n} > 0 \right\} \)

\( = \sqrt{2\pi\frac {\sigma^2} {n}}\left[1 - \Phi\left(\frac {kr} {\sigma\sqrt{n}}\right)\right] \)

where \( Z \) is the standard normal random variable and \( \Phi \) is its CDF.

As a result, the CDF is

\( = \frac {n} {\sigma^2} \sum_{k=0}^{n-1} \binom {n-1} {k} (-1)^k

\exp\left\{-\frac {kr^2} {2\sigma^2} \left(1 - \frac {k} {n}\right) \right\} \) \( \int_0^{+\infty} \left(x + \frac {kr} {n} - \frac {kr} {n} \right) \exp\left\{- \frac {n} {2\sigma^2} \left(x + \frac {kr} {n}\right)^2 \right\}dx \)

\( = \frac {n} {\sigma^2} \sum_{k=0}^{n-1} \binom {n-1} {k} (-1)^k

\exp\left\{-\frac {kr^2} {2\sigma^2} \left(1 - \frac {k} {n}\right) \right\} \) \( \left\{\frac {\sigma^2} {n} - \frac {kr} {n} \sqrt{2\pi\frac {\sigma^2} {n}}\left[1 - \Phi\left(\frac {kr} {\sigma\sqrt{n}}\right)\right] \right\} \)

But as you see it is quite a mess.