How to show that for $C, d>0$, the integral $ \int_1^{\infty} \exp(-\frac{x^d}{2C})dx<\infty $? How to show that for $C, d>0$, the integral $$ 
\int_1^{\infty} \exp(-\frac{x^d}{2C})dx<\infty
$$
As $d=1$,
$$ 
\int_1^{\infty} \exp(-\frac{x}{2C})dx=2C\int_1^{\infty} \exp(-\frac{x}{2C})d(\frac{x}{2C})=2Ce^{-1}<\infty
$$
But how to show this integral is finite for $d>1$?

Note that by Taylor expansion of $\exp$ function, for any $k \ge 1$ and $x \ge 0$ we have $\exp(x) \ge \frac{x^k}{k!}$.
Hence for any $d > 0$, $C>0$, $k \ge 1$ and $x > 0$ we get $$ \exp(\frac{x^d}{2C}) \ge \frac{(x^d)^k}{(2C)^kk!} $$
Choose $k$ such that $dk \ge 2$. Let $M = \frac{1}{(2C)^k k!}$ (it's independent of $x$ !). Then for $x \ge 1$ we get $\exp(\frac{x^d}{2C}) \ge Mx^2$. Hence $$ \int_1^{\infty}\exp(-\frac{x^d}{2C})dx \le \int_{1}^\infty\frac{1}{Mx^2}dx < \infty$$
