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

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

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

Undefined - \frac{1}{- a + 1}

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

\int_{1}^{\infty} \frac{1}{x ^{a}} d x =

Undefined

- \frac{1}{- a + 1}

= Undefined - \frac{1}{- a + 1}

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

\int_{0}^{2} 2^{x} - 1 - \frac{3 x}{2} d x

\int_{0}^{2} \frac{1}{x ^{2} + 4 x + 3} d x

\int_{1}^{2} \frac{1}{( 3 x - 1 ) ^{2}} d x

\int_{π}^{2 π} sin (x) cos (n x) d x

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