What is the integral from 0 to infinity of e^{-bx^2} ?

The integral from 0 to infinity of e^{-bx^2} is (sqrt(pi)-sqrt(pi)\erf(0*sqrt(b)))/(2sqrt(b))

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integral from 0 to infinity of e^{-bx^2}

\int_{0}^{\infty} e^{- b x^{2}} d x

\frac{π - π erf ( 0 \cdot b )}{2 b}

Solution steps

\int_{0}^{\infty} e^{- b x^{2}} d x

Compute the indefinite integral:

\int e^{- b x^{2}} d x = \frac{π}{2 b} erf (b x) + C

Compute the boundaries:

\int_{0}^{\infty} e^{- b x^{2}} d x = \frac{π}{2 b} - \frac{π}{2 b} erf (b \cdot 0)

= \frac{π}{2 b} - \frac{π}{2 b} erf (b \cdot 0)

Simplify

= \frac{π - π erf ( 0 \cdot b )}{2 b}

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What is the integral from 0 to infinity of e^{-bx^2} ?
The integral from 0 to infinity of e^{-bx^2} is (sqrt(pi)-sqrt(pi)\erf(0*sqrt(b)))/(2sqrt(b))