What is the integral from 1 to infinity of e^{-st} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to infinity of e^{-st}

\int_{1}^{\infty} e^{- s t} d t

\frac{e ^{- s}}{s}

Solution steps

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

= 0 - (- \frac{e ^{- s}}{s})

Simplify

= \frac{e ^{- s}}{s}

\frac{1}{( s ^{2} + 1 ) ( s ^{2} + 4 )}

x (\frac{d y}{d x}) = \frac{( 1 - 4 y ^{2} )}{3 y}

x x + x^{2} x

\frac{\partial}{\partial x} (\frac{5}{x})

x \to 0 lim ((\frac{1}{2})^{x})

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