Expert answers to "Evaluate the improper integral ∫0∞dxex+e−x"

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Answered question

2021-09-29

Evaluate the improper integral

\int_{0}^{\infty} \frac{d x}{e^{x} + e^{- x}}

Leonard Stokes

Skilled2021-09-30Added 98 answers

Step 1
Consider the integral,

\int_{0}^{\infty} \frac{d x}{e^{x} + e^{- x}}

Step 2
To solve the indefinite integrals,

\int \frac{d x}{e^{x} + e^{- x}} = \int \frac{d x}{e^{x} + \frac{1}{e^{x}}}

= \int \frac{e^{x}}{e^{2 x} + 1} d x

Let,

e^{x} = t

e^{x} d x = d t

so,

= \int \frac{d t}{t^{2} + 1}

= \tan^{- 1} (t) + C

Back substitution,

e^{x} = t

\int \frac{d x}{e^{x} + e^{- x}} = \tan^{- 1} (e^{x}) + C

Step 3
To solve the given integral,
Compute the boundaries,

\int_{0}^{\infty} \frac{d x}{e^{x} + e^{- x}} = {[\tan^{- 1} (e^{x} + C)]}_{0}^{\infty}

= (\tan^{- 1} (e^{\infty}) + C) - (\tan^{- 1} (e^{0}) + C)

= \tan^{- 1} (\infty) - \tan^{- 1} (1)

= \frac{π}{2} - \frac{π}{4}

= \frac{π}{4}

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