What is the integral from 0 to infinity of (e^{-st}) ?

The integral from 0 to infinity of (e^{-st}) is 1/s

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to infinity of (e^{-st})

\int_{0}^{\infty} (e^{- s t}) d t

\frac{1}{s}

Solution steps

\int_{0}^{\infty} e^{- s t} d t

Compute the indefinite integral:

\int e^{- s t} d t = - \frac{1}{s} e^{- s t} + C

Compute the boundaries:

\int_{0}^{\infty} (e^{- s t}) d t = 0 - (- \frac{1}{s})

= 0 - (- \frac{1}{s})

Simplify

= \frac{1}{s}

\int_{- 1}^{4} \frac{1}{x ^{2}} d x

\int_{1}^{2} \frac{x ^{3} + 1}{x ^{2}} d x

\int_{\frac{1}{2}}^{\frac{1}{3}} cos (π t) d t

\int_{- 1}^{1} 8 - 8 x^{2} d x

\int_{0}^{t} u e^{2 u} sin (3 u) d u

What is the integral from 0 to infinity of (e^{-st}) ?
The integral from 0 to infinity of (e^{-st}) is 1/s