What is the integral of (e^{-u})/(sqrt(piu)) ?

The integral of (e^{-u})/(sqrt(piu)) is \erf(sqrt(u))+C

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integral of (e^{-u})/(sqrt(piu))

\int \frac{e ^{- u}}{π u} d u

erf (u) + C

Solution steps

\int \frac{e ^{- u}}{π u} d u

\frac{e ^{- u}}{π u} = \frac{1}{π} \frac{e ^{- u}}{u}

= \int \frac{1}{π} \cdot \frac{e ^{- u}}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{π} \cdot \int \frac{e ^{- u}}{u} d u

Apply u-substitution

: \int 2 e^{- v^{2}} d v

= \frac{1}{π} \cdot \int 2 e^{- v^{2}} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{π} \cdot 2 \cdot \int e^{- v^{2}} d v

Use the nonelementary integral:

\int e^{- v^{2}} d v = \frac{π}{2} erf (v)

= \frac{1}{π} \cdot 2 \cdot \frac{π}{2} erf (v)

Substitute back

v = u

= \frac{1}{π} \cdot 2 \cdot \frac{π}{2} erf (u)

Simplify

\frac{1}{π} \cdot 2 \cdot \frac{π}{2} erf (u) : erf (u)

= erf (u)

Add a constant to the solution

= erf (u) + C

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What is the integral of (e^{-u})/(sqrt(piu)) ?
The integral of (e^{-u})/(sqrt(piu)) is \erf(sqrt(u))+C