Let $X_1, X_2, \ldots, X_n$ be a random sample of size $n$ from a normal distribution. a) Show that an unbiased estimator of $\sigma$ is $cS$ where
$$c = \frac{\sqrt{n-1} \,\Gamma\left(\frac{n-1} 2\right)}{\sqrt2\Gamma \left( \frac n 2 \right)}$$
Recall that the distribution of $(n-1)S^2/\sigma^2$ is $\chi_{n-1}^2$, i.e. $\dfrac{(n-1)S^2}{\sigma^2} \sim \chi_{n-1}^2$
$$ \operatorname{E}(S) = \int_0^\infty s f_S(s)\,ds. $$ Here of course I am careful about where capital $S$ appears and where lower-case $s$ appears. \begin{align} f_S(s) & = \frac d{ds} F_S(s) = \frac d{ds} \Pr(S\le s) = \frac d {ds} \Pr(S^2 \le s^2) = \frac d {ds} \int_0^{s^2} f_{S^2} (x) \, dx \\[10pt] & = \frac d {ds} \int_0^{s^2} \left. \left( \frac x 2 \right)^{n/2} e^{-x/2} \, \left(\frac{dx/2} {x/2} \right) \right/ \Gamma(n/2) \\[10pt] & = \frac 1 {\Gamma(n/2)} \, \frac d {ds} \int_0^{s^2/2} u^{n/2} e^{-u} \, \frac{du} u \\[10pt] & = \frac 1 {\Gamma(n/2)} \left( \frac{s^2} 2 \right)^{n/2} e^{-s^2/2} \frac 1 {s^2/2} \cdot \frac d {ds} \, \frac {s^2} 2 \\[10pt] & = \frac 1 {\Gamma(n/2)} \left( \frac{s^2} 2 \right)^{n/2} e^{-s^2/2} \frac 1 s. \end{align} Therefore \begin{align} \operatorname{E}(S) & = \int_0^\infty sf_S(s)\,ds = \frac 1 {\Gamma(n/2)} \int_0^\infty \left( \frac{s^2} 2 \right)^{n/2} e^{-s^2/2} \, ds \\[10pt] & = \frac 1 {\Gamma(n/2) 2^{n/2}} \int_0^\infty s^{n-1} e^{-s^2/2} (s\, ds) \\[10pt] & = \frac 1 {\Gamma(n/2) 2^{n/2}} \int_0^\infty (2v)^{(n-1)/2} e^{-v} \, dv \end{align} and then the powers of $2$ simplify and the value of the integral is the value of the Gamma function at a certain point.