Question

Evaluate ${\int }_{-\infty }^{\infty }\frac{dx}{e^x+e^{-x}}$

• A. $\pi /2$
• B. $-\pi /2$
• C. $\pi /3$
• D. $-\pi /3$

Answer 1

Answer: (A)

$\pi /2$

Given : ${\int }_{-\infty }^{\infty }\frac{dx}{e^x+e^{-x}}$

${\int }_{-\infty }^{\infty }\frac{dx}{e^x+e^{-x}}={lim}_{c\to -\infty }{\int }_c^0\frac{dx}{e^x+e^{-x}}+{lim}_{b\to \infty }{\int }_0^b\frac{dx}{e^x+e^{-x}}={lim}_{c\to -\infty }{\int }_c^0\frac{e^{x}dx}{e^{2x}+1}+{lim}_{b\to \infty }{\int }_0^b\frac{e^{x}dx}{e^{2x}+1}$

Let $z=e^{x}x\to cz\to e^x$

$dz=e^{x}dxx\to 0z\to 1$

${\int }_{-\infty }^{\infty }\frac{dx}{e^x+e^{-x}}={lim}_{c\to -\infty }{\int }_{e^c}^0\frac{dz}{z^2+1}+{lim}_{b\to \infty }{\int }_0^{e^b}\frac{dz}{z^2+1}={lim}_{c\to -\infty }{\left[{\tan }^{-1}z\right]}_{e^c}^0+{lim}_{b\to \infty }{\left[{\tan }^{-1}z\right]}_0^{e^b}={lim}_{c\to -\infty }(0-{\tan }^{-1}{e}^c)+{lim}_{b\to \infty }({\tan }^{-1}{e}^b-0)=\frac{\pi }{2}$

Hence the correct answer is $\frac{\pi }{2}$