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Question: The integral \(\int_{- \infty}^{\infty}{\frac{1}{e^{- x} + e^{x}}dx}\) is...

The integral ∫−∞∞1e−x+exdx\int_{- \infty}^{\infty}{\frac{1}{e^{- x} + e^{x}}dx} is

A

Convergent and equal to π/6

B

Convergent and equal to π/4

C

Convergent and equal to π/3

D

Convergent and equal to π/2

Answer

Convergent and equal to π/2

Explanation

Solution

I=∫−∞∞1e−x+exdx=∫−∞∞ex1+e2xdxI = \int_{- \infty}^{\infty}{\frac{1}{e^{- x} + e^{x}}dx} = \int_{- \infty}^{\infty}{\frac{e^{x}}{1 + e^{2x}}dx}

Put ex=t⇒exdx=dte^{x} = t \Rightarrow e^{x}dx = dt

∴ I=∫0∞11+t2dtI = \int_{0}^{\infty}{\frac{1}{1 + t^{2}}dt} ⇒ I=[tan⁡−1t]0∞=[tan⁡−1∞−tan⁡−10]I = \lbrack\tan^{- 1}t\rbrack_{0}^{\infty} = \lbrack\tan^{- 1}\infty - \tan^{- 1}0\rbrack

⇒ I=π⥂/⥂2I = \pi ⥂ / ⥂ 2, which is finite so convergent.