Relation between $\int_0^{\infty} \frac{e^{-ax}}{1+x^2}\,\,dx$ and $\int_0^{\infty} \frac{e^{-2ax}}{1+x^2}\,\,dx$ Suppose $a > 1$. I want to compare
$$\int_0^{\infty} \frac{e^{-ax}}{1+x^2}\,\,dx$$ and $$\int_0^{\infty} \frac{e^{-2ax}}{1+x^2}\,\,dx$$
My instinct suggests that after a certain value of $a$, $$\int_0^{\infty} \frac{e^{-2ax}}{1+x^2}\,\,dx < e^{-a}\int_0^{\infty} \frac{e^{-ax}}{1+x^2}\,\,dx$$
but I cannot prove it. Is this correct intuition, and if it is so, what would be the method to prove it?
 A: One can show that your integral behaves like $1/a$ as $a\to\infty$. In particular the integral does not decay exponentially, and your claim does not hold.
We can use the substitution $y=ax $ to rewrite the integral:
$$\int_0^\infty \frac {e^{-ax}}{1+x^2}dx= \int_0^\infty \frac {a e^{-y}}{a^2+y^2}dy.$$
Then we can use Lebesgue's Dominated Convergence Theorem to compute
$$\lim_{a\to\infty}a\int_0^\infty \frac {e^{-ax}}{1+x^2}dx= \lim_{a\to\infty} \int_0^\infty \frac {a^2 e^{-y}}{a^2+y^2}dy= \int_0^\infty e^{-y}dy=1. $$
This shows
$$ \int_0^\infty \frac {e^{-ax}}{1+x^2}dx\sim a^{-1}$$
as $a\to\infty $.
A: Note that for $a>0$ and $x \geq 1$, we have
$$
e^{-2ax} = e^{-ax}e^{-ax} \leq e^{-a}e^{-ax}
$$
Thus, we have
$$
\int_1^\infty \frac{e^{-2ax}}{1+x^2} \leq e^{-a}\int_1^\infty \frac{e^{-ax}}{1+x^2}
$$
Perhaps this will suffice for your purposes.
A: I think this is not true. For example, let $a=2$. Then
$$ \int_0^\infty\frac{e^{-4x}}{1+x^2}dx\approx 0.229193> e^{-2}\int_0^\infty\frac{e^{-2x}}{1+x^2}dx\approx 0.0540016. $$
