Does $\int_{0}^{\infty} \frac{e^{-2x}}{ x^2 + 1} \,dx $ converge?

Question

Does the following integral converge? I will post my solution, but I am unsure if it is true.

$$\int_{0}^{\infty} \frac{e^{-2x}}{ x^2 + 1} \,dx $$

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My solution: I will use limit comparison test.

Let $$ f(x) = \frac{e^{-2x}}{ x^2 + 1}$$ and $$ g(x) = \frac {e^{-2x}}{x^2} $$

Then $$ \lim_{ x \to +\infty } \frac{f(x)}{g(x)} = \lim_{ x \to +\infty } \frac{\frac{e^{-2x}}{x^2 + 1}}{\frac {e^{-2x}}{x^2}} = \lim_{ x \to \infty } \frac{x^2}{x^2 + 1} = 1 - \lim_{ x \to \infty } \frac{1}{x^2 + 1} = 1 $$

Whatever whatever one does, so does the other

$$ \int_{0}^{\infty} g(x) dx = \int_{0}^{\infty} \frac{e^{-2x}}{ x^2 } dx = e^{-2}\int_{0}^{\infty} x^{3-1} *e^{-x}dx = e^{-2}*Γ(3) = e^{-2}*3! = e^{-2}*6$$

Therefore g(x) converges, thus f(x) converges.

Is this right? I'm totally not confident about it.

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Apparently on my solution silly mistakes were made. $$ e^{-2x} \neq e^{-x} * e^{-2} $$ and $$\frac {1}{x^2} \neq x^{3-1} $$

Thanks for the great answers. Sometimes the solution is east and in front of you but you fail to see it

Answer 1

It is not O.K., since $e^{-2}e^{-x}=e^{-(x+2)}$ and not $= e^{-2x}$and $ \frac{1}{x^2} \ne x^{3-1}.$

My solution: let $g(x)=e^{-2x}$ then we have $0 \le f \le g$ on $[0, \infty)$ and $\int_{0}^{\infty} g(x) dx$ is convergent.

$\int_{0}^{\infty} g(x) dx=1/2$.

Answer 2

Just easier, $$\left|\frac{e^{-2x}}{1+x^2}\right|\leq \frac{1}{1+x^2},$$ which is integrable.

Answer 3

$$ \int_{0}^{\infty} \frac{e^{-2x}}{ x^2 + 1} \,dx \;\leqslant\; \int_{0}^{\infty} \!e^{-2x} \,dx =-\frac12e^{-2x}\Big|_{x=0}^{x=\infty} =\frac12 $$ and $$ \frac{e^{-2x}}{ x^2 + 1} \geqslant0 $$ Thus the integral is well-defined and finite.