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 $d>1$?

Using the change of variable $u = \frac{x^d}{2C} \iff x= \left(2Cu\right)^{\frac{1}{d}}$, you have
$$I =  
\int_1^{\infty} \exp(-\frac{x^d}{2C})\ dx=A\int_{\frac{1}{2C}}^{\infty} u^{\frac{1}{d}-1}e^{-u}\ dx$$ where $A$ is a constant. And the integral of the RHS is convergent.
