What is the integral from 1 to infinity of x^{-1/2} ?

The integral from 1 to infinity of x^{-1/2} is diverges

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integral from 1 to infinity of x^{-1/2}

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

diverges

Solution steps

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

Compute the indefinite integral:

\int x^{- \frac{1}{2}} d x = 2 x^{\frac{1}{2}} + C

Compute the boundaries:

\int_{1}^{\infty} x^{- \frac{1}{2}} d x =

diverges

= diverges

\int \frac{( u - 1 )}{( u + 1 )} d u

n = 1 \sum \infty 6 \frac{1}{n}

f (x) = x^{2} + 3 x + 4 π, at x = - 3

\frac{d}{d x} (arccos (\frac{x}{b}))

\frac{d ^{4}}{d x ^{4}} (\frac{8 x}{1 - x})

What is the integral from 1 to infinity of x^{-1/2} ?
The integral from 1 to infinity of x^{-1/2} is diverges