# Improper Integral: Comparison Test

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.

• So we have $2016\int_{-\infty}^{\infty}\sech x\text dx$. What might you compare to? – abiessu Apr 4 '16 at 13:30
• Just for fun, evaluating the integral yields $1008\pi$. – Henricus V. Apr 4 '16 at 14:06

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.