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I have the following improper integral: $$\int ^\infty _{-\infty}\frac{2016}{e^x+e^{-x}} \, dx$$ My question is how to prove that it is convergent or divergent by using the Comparison Test.

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  • $\begingroup$ So we have $2016\int_{-\infty}^{\infty}\sech x\text dx$. What might you compare to? $\endgroup$ – abiessu Apr 4 '16 at 13:30
  • $\begingroup$ Just for fun, evaluating the integral yields $1008\pi$. $\endgroup$ – Henricus V. Apr 4 '16 at 14:06
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Hint : Use $\frac{2016}{e^x}$ for the integral from $0$ to $\infty$ and $\frac{2016}{e^{-x}}$ for the integral from $-\infty$ to $0$ to show the convergence due to the majorant-criterion.

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