For which values of $\alpha$, $E(|X|^{\alpha})$ is finite, if the density function of $X$ is $$(a) \quad e^{-x} \ x \geq 0, (b) \quad C(1+x^2)^{-m}, \ x \in \mathbb{R}.$$
For (b). I have: $$E(|X^{\alpha}|) = C\int_{-\infty}^{\infty} |x^{\alpha}|(1+x^2)^{-m}dx,$$ which can be reduced to $$E(|X^{\alpha}|) = 2C\int_0^{\infty} x^{\alpha}(1+x^2)^{-m}dx, \text{for } x \geq 0,$$ if my calculations are correct. Now, $$0 \leq E(|X^{\alpha}|) \leq 2C\int_0^{\infty} x^{\alpha}x^{-2m}dx = 2C\int_0^{\infty} x^{\alpha-2m}dx.$$
If we integrate, we get: $$2C\int_0^{\infty} x^{\alpha-2m}dx = 2C\left(\frac{x^{\alpha - 2m + 1}}{\alpha - 2m + 1}\right) \Big|_0^{\infty}$$.
This means that we require $\alpha - 2m + 1 < 0$, which gives $\alpha < 2m-1$. Also, since $0 \leq E(|X^{\alpha}|)$ and $\alpha - 2m + 1 < \alpha + 1$, we require $\alpha + 1 > 0$ which gives $\alpha > -1$, so $-1 < \alpha < 2m-1$.
For part (a), we have: $$E(|X^{\alpha}|) = \int_0^{\infty}x^{\alpha}e^{-x}dx,$$ that is, $E(|X^{\alpha}|) = \Gamma(\alpha+1)$, which is know to be convergent for $\alpha +1 >0$, that is $\alpha > -1$.