What is the integral from 1 to infinity of 1/(x^e) ?

The integral from 1 to infinity of 1/(x^e) is -1/(-e+1)

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 1/(x^e)

\int_{1}^{\infty} \frac{1}{x ^{e}} d x

- \frac{1}{- e + 1}

Decimal Notation

0.58197 \dots

Solution steps

\int_{1}^{\infty} \frac{1}{x ^{e}} d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{1}^{\infty} \frac{1}{x ^{e}} d x = 0 - \frac{1}{- e + 1}

= 0 - \frac{1}{- e + 1}

Simplify

= - \frac{1}{- e + 1}

\int_{0}^{\frac{π}{4}} sin (2 θ) d θ

\int_{3}^{7} 4 d x

\int_{1}^{5} (\frac{6}{x}) d x

\int_{0}^{c} 4 e^{- x} d x

\int_{0}^{5} x^{2} + 5 x d x

What is the integral from 1 to infinity of 1/(x^e) ?
The integral from 1 to infinity of 1/(x^e) is -1/(-e+1)