Choosing which function to compare to for the Direct Comparison test

Question

$$\int_1^\infty (e^{-x^{2}})dx$$

why use $$e^{-x}$$

for the direct comparison test to determine convergence or divergence?

Answer 1

When choosing a function for direct comparison, you want it to have certain qualities:

• Its integral should be known to converge on some interval $[a,\infty)$.
• It should be greater than your function of interest, if it converges, or less than your function of interest, if it diverges.

The function $e^{-x}$ satisfies both of these, so it's very useful in this context. Another option would be $xe^{-x^2}$, because it converges, and we have $e^{-x^2}<xe^{-x^2}$ for $x>1$.

Answer 2

HINT. $$\int_1^\infty e^{-x^{2}}dx\leq \int_1^\infty e^{-x}dx$$ where second integral converges. In fact, if $x\geq 1$:

$$e^{-x^2}\leq e^{-x}$$

ans so $x\leq x^2$ true for $x\geq 1$.

Then, from comparison test, if $$\int_1^\infty e^{-x}dx$$ converges, even $$\int_1^\infty e^{-x^{2}}dx$$ converges.

Answer 3

For $x \ge 1$, we have $e^{-x^2} \le e^{-x}$, so if $\int_{1}^{\infty} e^{-x} dx$ converges, then $\int_{1}^{\infty} e^{-x^2} dx$ converges. Let me know if you need help proving the first part.