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For $X\sim \mathrm{Beta}(\alpha ,\beta )$ with positive parameters, and real $s>-\alpha$, $t>-\beta$, multiply its density by $x^s(1-x{)}^t$. The resulting integral is a beta function with shifted parameters, henceThe restrictions on the powers are exactly the endpoint-integrability conditions. This includes ordinary moments, reciprocal moments when they exist, and quantities involving both the occupied fraction $X$ and remaining fraction $1-X$. In particular, $s=1,t=0$ gives $\mathbb{E}[X]=\alpha /(\alpha +\beta )$.^[ref]

The same integral works with both powers shifted:The assumptions $s>-\alpha$ and $t>-\beta$ say precisely that the exponents at each endpoint are greater than $-1$, ensuring integrability there. Values of a negative power at the endpoints themselves can be assigned arbitrarily, because those endpoints have probability zero. This establishes the mixed power moments of a beta distribution without requiring $s$ and $t$ to be integers.^[ref]

Multiply the probability density function by $x$ and integrate over its support:The middle equality is the beta function integral with parameters $\alpha +1$ and $\beta$; the last uses the recurrence for the gamma function. Both parameters are positive, so the integral converges. Equivalently, multiplying the beta distribution probability density function by $x$ converts its kernel to that of a beta distribution probability density function with its first parameter increased by one.^[ref]