(a) Compute the moment generating function of $\chi^2_\nu$ random variable where $\chi^2_\nu$ is defined as $X\sim \chi^2_\nu$ if and only if $$f_X(x) = \begin{cases} \frac{1}{2^{\frac{\nu}{2}}\Gamma(\frac{\nu}{2}) }x^{\frac{\nu}{2}-1}e^{-\frac{x}{2}} & x\geq 0\\ 0 & \text{otherwise.} \end{cases}$$
This is my solution so far.
$M_X(s)=E(e^{sx})$
Thus $M_X(s)=\int_0^\infty \frac{1}{2^{\frac {\nu}{2}}\Gamma(\frac{\nu}{2}) }x^{\frac{\nu}{2}-1}e^{-\frac{x}{2}}e^{sx}$
Taking out the constants we have $\frac{1}{2^{\frac {\nu}{2}}\Gamma(\frac{\nu}{2}) } \int_0^\infty x^{\frac{\nu}{2}-1}e^{-\frac{x}{2}}e^{sx} $
Simplifying yields $\frac{1}{2^{\frac {\nu}{2}}\Gamma(\frac{\nu}{2}) }e^{s-\frac{1}{2}} \int_0^\infty x^{\frac{\nu}{2}-1}e^x $
(I am not sure whether or not I combined my $e$ terms correctly)
Now, if all is so far correct the simplified integral, without the constants, to me looks like
$\frac{\nu}{2}-1$ $\Gamma(\frac{\nu}{2}-1)$ however not fully since in the integral I have an $x$ not a $-x$.
Please help me figure out my mistakes.
(b) Use these moment generating functions give a simpler demonstration that if $X$ ∼ $\chi^2_\nu$ and $Y$ ∼ $\chi^2_k$ are independent, then $X +Y$ ∼ $\chi^2_{\nu+k}$.
Using the answer from below
Since $M_X(s)=(1-2s)^{-\frac{\nu}{2}}$ for $X$ ~ $\chi^2_\nu$ it follows that $M_Y(s)=(1-2s)^{-\frac{k}{2}}$ for $Y$ ~ $\chi^2_k$
Using the property $M_{X+Y}(s)=M_X(s)M_Y(s)$
We can conclude that $X+Y$ ~ $\chi^2_{\nu+k} = (1-2s)^{-\frac{\nu}{2}}(1-2s)^{-\frac{k}{2}}=(1-2s)^{-\frac{1}{2}(\nu+k)}$
No, $e^{sx}e^{-\frac{x}{2}} = e^{(s-\frac{1}{2})x}\neq e^{s-\frac{1}{2}}e^x$. Instead \begin{align*} M_X(s) &= \int_0^\infty e^{sx} \cdot\frac{1}{2^{\frac{\nu}{2}}\Gamma\left(\frac{\nu}{2}\right)} x^{\frac{\nu}{2}-1}e^{-\frac{x}{2}}\,dx\\ &=\frac{1}{2^{\frac{\nu}{2}}}\int_0^\infty \frac{1}{\Gamma\left(\frac{\nu}{2}\right)}x^{\frac{\nu}{2}-1} e^{\left(s-\frac{1}{2}\right)x}\,dx\\ &=\frac{\left(\frac{1}{2}-s\right)^{-\frac{\nu}{2}}}{2^{\frac{\nu}{2}}}\int_0^\infty \frac{1}{\Gamma\left(\frac{\nu}{2}\right)}\left(\frac{1}{2}-s\right)^{\frac{\nu}{2}}x^{\frac{\nu}{2}-1} e^{-\left(\frac{1}{2}-s\right)x}\,dx\\ &=\frac{[(1/2)(1-2s)]^{-\frac{\nu}{2}}}{2^{\frac{\nu}{2}}}\cdot 1\\ &= (1-2s)^{-\frac{\nu}{2}} \end{align*} where the integrand in the third line is the pdf of a $\operatorname{Gamma}\left(\frac{\nu}{2}, \frac{1}{2}-s\right)$.