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I'm trying to integrate this:

$$\int^\infty_0 \frac{8}{\sqrt{e^{x}-x}} \,dx$$

And use the Direct Comparison Test to find out whether it diverges or converges.

I looked at a similar problem:

another improper integral and I can see how the integrals on the lefthand side are less than the integrals on the righthand side, since the rightmost right-side integral is squared from the rightmost left-side integral, but:

Why is the 5 dropped? Is it because a small added constant ultimately wouldn't affect the behavior of x on its way to infinity?

Why is the integral from 1 to infinity squared, out of all the possible operations we could perform on it?

And is this the correct next step in my own integration?

$$\int^\infty_0 \frac{8}{\sqrt{e^{x}-x}} \,dx = \int^1_0 \frac{8}{\sqrt{e^{x}-x}} \,dx + \int^\infty_1 \frac{8}{\sqrt{e^{x}-x}} \,dx < \int^1_0 \frac{8}{\sqrt{e^{x}-x}} \,dx + \int^\infty_1 \frac{1}{\sqrt{e^{x}}} \,dx $$

Thank you in advance if you're able to help clarify this.

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    $\begingroup$ Removing the +5 from the denominator makes the term larger. So the inequality holds, because it says the left side is smaller than the right. Yours does not work, because removing the square root makes the denominator bigger and thus the whole term becomes smaller. $\endgroup$ – Chantry Cargill Sep 21 '14 at 0:31
  • $\begingroup$ That makes sense. Am I able to remove the -x, then, and have it be 8/\sqrt{e^x}? Edited my post. $\endgroup$ – Calculistening helplessly Sep 21 '14 at 0:36
  • $\begingroup$ Sorry, my previous comment was incorrect. $\endgroup$ – user84413 Sep 21 '14 at 0:45
  • $\begingroup$ How so? Are you not able to do this? $\endgroup$ – Calculistening helplessly Sep 21 '14 at 0:49
  • $\begingroup$ @Calculisteninghelplessly No, that would make the numerator larger and thus the whole term smaller. $\endgroup$ – Chantry Cargill Sep 21 '14 at 0:50
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Note that $e^x-x \geq x^4$ for all sufficiently large $x$. So there exists some $N > 0$ such that $e^x-x \geq x^4$ for all $x \geq N$. Since $$ \sqrt{e^x-x} \geq \sqrt{x^4} = x^2 \quad \Rightarrow \quad \frac{1}{\sqrt{e^x-x}} \leq \frac{1}{x^2} $$ for all $x \geq N$, we have \begin{align*} \int_0^\infty \frac{dx}{\sqrt{e^x-x}} & = \int_0^N \frac{dx}{\sqrt{e^x-x}} + \int_N^\infty \frac{dx}{\sqrt{e^x-x}} \\ & \leq \int_0^N \frac{dx}{\sqrt{e^x-x}} + \int_N^\infty \frac{dx}{x^2}. \end{align*} The two integrals are finite so the integral you consider is convergent.

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Hint

$$ \frac{1}{\sqrt{e^x - x}} < \frac{1}{\sqrt{e^x - \frac{1}{2}e^x}}$$

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