Comparison Test Question: How to Determine Convergence of a Complex Integral?

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Homework Statement

I'm just curious as to how to think about the following form of equation.

Homework Equations

\int_{3}^{\infty } \frac{1}{x + e^x} \,dx

The Attempt at a Solution

What you're trying to do is to test it;

\frac{1}{x \ + \ e^x} \ < \ \frac{1}{x}

\frac{1}{x} diverges

\frac{1}{x \ + \ e^x} \ < \ \frac{1}{e^x}

\lim_{t \to \infty} \int_{3}^{t} e^{-x}\,dx \ = \ \lim_{t \to \infty} - e^{-x} | \ _3 ^t \ = \ \lim_{t \to \infty} - {\frac{1}{e^t} \ + \ \frac{1}{e^3}

so this converges to \frac{1}{e^3}.

I don't get how this means the original eq. will also converge?

Both \frac{1}{x} and \frac{1}{e^x} are bigger than the original eq. with one converging and the other diverging.

\frac{1}{x} is bigger than \frac{1}{e^x} but the test of \frac{1}{x^p} is ringing in my ears as a kind of explanation, but I am confused.

Is there an easy way to link all of this together?

 

[Tedjn]

Indeed, because for [strike]positive x[/strike] x > 1 when p > 1, 1/x^p is always less than 1/x. In effect, its graph has less area underneath it for the integration. So there is no contradiction that 1/x leaves just too much area, i.e. it diverges, while 1/x^p for p > 1 does not, i.e. it converges. Meanwhile, 1/e^x just hugs the x-axis massively more than 1/x^p for any constant p, hence it of course converges. Finally, 1/(x+e^x) lies even closer to the x-axis and leaves even less area; it converges.

Knowing 1/x diverges does not mean we know 1/(x+e^x) diverges or converges. The graph of the latter has less area, but by how much? Enough to stop the integral from diverging? However, knowing that 1/e^x converges does tell us that 1/(x+e^x) converges, since it has less area than a finite amount.

EDIT: It occurred to me that I also need x > 1, not just x positive. Sorry for the mistake.

 

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Ahh Brilliant!

I think that idea of hugging the x-axis makes everything even clearer, thanks a lot for that