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I'm having some real trouble with lebesgue integration this evening and help is very much appreciated.

I'm trying to show that $f(x) = \dfrac{e^x + e^{-x}}{e^{2x} + e^{-2x}}$ is integrable over $(0,\infty)$.

My first thought was to write the integral as $f(x) = \frac{\cosh(x)}{\cosh(2x)}$ and then note $f(x) = \frac{\cosh(x)}{\sinh(x)^2 + \cosh(x)^2}$ so that $|f(x)| \le \frac{\cosh(x)}{\cosh(x)^2}$. These all seemed like sensible steps to me at this point, and I know the integral on the right hand side exists (wolfram alpha), but I'm having trouble showing it and am wondering if I have made more of a mess by introducing trigonometric functions.

Thanks

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  • $\begingroup$ Why you cannot proceed then $\frac{1}{\cosh(x)} < 2e^{-x}$. $\endgroup$ Jun 11 '14 at 19:50
  • $\begingroup$ I'm not sure who to show that $\frac{cosh(x)}{cosh(x)^2}$ is integrable? $\endgroup$
    – Wooster
    Jun 11 '14 at 19:52
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    $\begingroup$ It's $\frac{1}{\cosh x}$. Now, $\cosh x > \frac{1}{2}e^{\lvert x\rvert}$ for real $x$. $\endgroup$ Jun 11 '14 at 19:54
  • $\begingroup$ @DanielFischer Thanks again for your help today, I don't know how I did not spot that, it's been a long day. $\endgroup$
    – Wooster
    Jun 11 '14 at 19:58
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Hint: You should be able to show that $e^{-x}$ and $e^{-3x}$ are integrable without too much trouble (this should follow from the "standard" trick of writing $\int_{0}^{\infty}g=\int_{0}^{\infty}\lim_{n}g\chi_{(0,n)}$ and then switching limits using the monotone convergence theorem). Then you can do a comparison based on $g(x)<e^{-x}+e^{-3x}$.

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  • $\begingroup$ Also note that $-3x<-x$ allows you to do $e^{-x}>e^{-3x}$ so you can get the inequality mentioned by Alexander on his comment using this method but without going through $\cosh$ etc. $\endgroup$
    – UserB1234
    Jun 11 '14 at 19:58
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The given function $f$ is continuous on $(0,\infty)$ and has a finite limit at $x=0$ and $$f(x)\sim_\infty e^{-x}\in L^1(0,\infty)$$ so $f$ is integrable on $(0,\infty)$.

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For the numerator, observe that $e^x$ dominates $e^{-x}$ as $x \rightarrow \infty$.

For the denominator, observe that $e^{2x}$ dominates $e^{-2x} $ as $x \rightarrow \infty$.

So the integrand will decay like ${e^x \over e^{2x}}$ = $e^{-x}$ as $x \rightarrow \infty$ and the integral will converge. To get a formal proof, use the limit comparison test with $e^{-x}$.

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