What is the integral from 1 to infinity of x^{-1/9} ?

The integral from 1 to infinity of x^{-1/9} is diverges

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integral from 1 to infinity of x^{-1/9}

\int_{1}^{\infty} x^{- \frac{1}{9}} d x

diverges

Solution steps

\int_{1}^{\infty} x^{- \frac{1}{9}} d x

Compute the indefinite integral:

\int x^{- \frac{1}{9}} d x = \frac{9}{8} x^{\frac{8}{9}} + C

Compute the boundaries:

\int_{1}^{\infty} x^{- \frac{1}{9}} d x =

diverges

= diverges

\frac{2}{s ( s + 2 ) + 2}

y = \frac{23}{5} x - 12

\frac{d}{d x} (- 5 e^{- 5 x})

f (x) = \frac{1}{x ^{6}}, (1, 1)

What is the integral from 1 to infinity of x^{-1/9} ?
The integral from 1 to infinity of x^{-1/9} is diverges