$\newcommand{\E}{\operatorname{E}}$Given the tail property of sub-gaussian random variable $\exists c>0, \forall \lambda >0, P(|X|\ge\lambda)\le2e^{-c\lambda^2}$, prove that $\exists a>0, \E e^{aX^2}\le2$. Although I have the result but I am getting two different expression using different approach.
Approach 1. Using $\E[g(x)] = g(a) +\int_a^\infty g'(x)(1-F_X(x)) \,dx $ if $P(X\ge a)=1$ [source https://en.wikipedia.org/wiki/Expected_value#General_definition]
$$Ee^{aX^2} = 1 + \int_0^\infty 2ate^{at^2}P(|X|>t)\,dt \le 1+\int_0^\infty 2at\cdot e^{at^2}\cdot2e^{-ct^2}dt=1+\frac{2a}{c-a}$$
Approach2. Let $Y=e^{aX^2}$ $$P(Y\ge y)=P(e^{aX^2}\ge y)=P\left(X^2\ge\frac 1 a \ln y\right) = P \left(|X|\ge\sqrt{\frac 1 a \ln y} \,\, \right) \le 2e^{\frac{-c}a \ln y}$$
Now using the fact that for positive random variable $\E[Y]=\int_0^\infty P(Y\ge y) \,dy$, we have
$$Ee^{aX^2}\le \int_1^\infty 2e^{\frac{-c}a \ln y} = \frac{2a}{c-a}$$
Not able to figure out the error?
Note that $P(Y > y) = 1$ for $0 \le y \le 1$, so you can't just skip the integral from $0$ to $1$. Then you get the same result.