How to solve this gamma integral

Question

Let we have the (p.d.f) of x which is:

$$f(x)=\frac {\Gamma{(n-1)/2)}}{\Gamma{(1/2)} \Gamma{(n-2)/2)}}x(1-x^2)^{(n/2)-2}$$

then to find the $E(x) = $$\int_{-\infty}^{\infty} x *(f(x) dx$ ; (Expectation of Mean)

Thus, we construct that

$$\int_{-\infty}^{\infty} (x) \frac{\Gamma{(n-1)/2)}}{\Gamma{(1/2)} \Gamma{(n-2)/2)}}(1-x^2)^{(n/2)-2}=??? $$

In my text book, it shows the result:

$$E(x) = 0 $$

Thanks

Answer 1

*HINT*Whatever is your problem, did you notice that the derivative of (1 - r^2) looks like r ? On the other hand, what does happen is you split your integral (from -1 to 0 and from 0 to 1) ? Can you continue with this ?