$\operatorname P\left[X\ge\lambda\right]\le e^{-a\lambda}$ for all $\lambda$ implies $\operatorname E\left[e^X\right]\le\frac a{a-1}$

Let $$X$$ be a real-valued random variable and $$a>1$$ with $$\operatorname P\left[X\ge\lambda\right]\le e^{-a\lambda}\;\;\;\text{for all }\lambda\ge0\tag1.$$ I want to show that $$\operatorname E\left[e^X\right]\le\frac a{a-1}\tag2.$$ My first thought was to write $$\operatorname E\left[e^X\right]=\int_0^\infty\operatorname P\left[X\ge\ln\lambda\right]\:{\rm d}\lambda\le\int_0^\infty\frac1{\lambda^a}\:{\rm d}\lambda=\left.\frac{\lambda^{1-a}}{1-a}\right|_{\lambda\:=\:0}^{\lambda\:=\:\infty}\tag3,$$ but since the right-hand side is $$\infty$$, this doesn't seem to be the right approach. So, how can we show the claim?

$$Ee^{X} =\int_0^{1} P(X\geq \log \lambda ) d\lambda +\int_1^{\infty} P(X\geq \log \lambda ) d\lambda \leq 1+\frac {\lambda^{a-1}} {1-a}|_1^{\infty}\leq 1+\frac 1 {a-1}=\frac a {a-1}$$.
• @DaviBarreira There was a typo. It is not equal to $1$ but it is $\leq 1$. Jun 20, 2020 at 12:35
I think the idea with showing $$\mathbb E[\sup |M_n|] \le \frac{e}{e-1} \sup \mathbb E[|M_n|\ln^+(|M_n|)$$ for a martingale can be applied here. The idea is to, "get away" from points that are trouble (in your case $$0$$) and "get enough away" so that you can actually calculate what's left. Your idea is good, with a little change:
$$\mathbb E[e^X] = \int_0^\infty \mathbb P(e^X > \lambda)d\lambda = \int_0^\infty \mathbb P(X > \ln(\lambda))d\lambda = \int_0^1 \mathbb P(X > \ln(\lambda))d\lambda + \int_1^\infty \mathbb P(X > \ln(\lambda))d\lambda$$
The first one, you bound it brutally by $$1$$, for the second, we use estimate getting: $$\mathbb E[e^X] \le 1 + \int_1^\infty \frac{1}{\lambda^a}d\lambda = 1 + \frac{\lambda^{1-a}}{1-a}|_{\lambda =1}^{\lambda = \infty} = 1 - \frac{1}{1-a} = 1+\frac{1}{a-1} = \frac{a}{a-1}$$