How would you derive the mean of this distribution with a degree of freedom p?
My course writes the distribution as follows:
T distribution
Edit: I was trying to proceed with calculus but nothing struck me at all. I didn't have much experience with t-distribution.
I don't know how you define the Student's T-distribution, but for me it's by definition the distribution of the random variable
$$T=\sqrt n \frac {\overline{X_n} - \mu} {\sqrt{\frac 1 {n-1}\sum(X_i-\overline{X_n})^2}}$$
where the $X_i$ are $n$ independent $\mathcal N(\mu, \sigma^2)$ variables. Note that at first glance this seems like a bad definition, since you would think the distribution of $T$ would depend on $\mu$ and $\sigma$. So in order to show that the T-distribution is well-defined in the first place, we need to justify that the distribution of the above expression does not depend on $\mu$ and $\sigma$ (see here). Once we know that, the fact that the mean is $0$ is actually a corollary.
Since $\mu$ and $\sigma$ don't affect the distribution of $T$, we can take $\mu=0$. Now consider the variables $Y_i=-X_i$ and form from them the T-distributed random variable:
$$T'=\sqrt n \frac {\overline{Y_n} - \mu} {\sqrt{\frac 1 {n-1}\sum(Y_i-\overline{Y_n})^2}}$$
But if you replace each $Y_i$ by its definition and work out the algebra, you'll see that in fact $T'=-T$. Thus $T$ and $-T$ have the same distribution, and $\mathbb E(T)=0$.