# 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?

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 $a\to\infty$.

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.

I think this is not true. For example, let $a=2$. Then $$\int_0^\infty\frac{e^{-4x}}{1+x^2}dx\approx 0.229193> e^{-2}\int_0^\infty\frac{e^{-2x}}{1+x^2}dx\approx 0.0540016.$$

• But will there be critical value of $a$ such that after $a$, the inequality direction is < ? – calgama Apr 29 '15 at 21:30