0-x-ln-x-dx-

Question Number 8691 by FilupSmith last updated on 22/Oct/16

\int_{0}^{\infty} x^{- \ln (x)} d x

Commented by FilupSmith last updated on 22/Oct/16

x^{- \ln (x)} = e^{- \ln (x) \ln (x)} = e^{- \ln^{2} (x)}

u = \ln (x) \Rightarrow x = e^{u}

= e^{- u^{2}}

\frac{d u}{d x} = \frac{1}{x} \Rightarrow d x = x d u = e^{u} d u

x = 0 \Rightarrow u = - \infty

x = \infty \Rightarrow u = \infty

= \int_{- \infty}^{\infty} e^{- u^{2}} e^{u} d u

= \int_{- \infty}^{\infty} e^{u - u^{2}} d u

c o n t i n u e

Answered by prakash jain last updated on 23/Oct/16

= \int_{- \infty}^{\infty} e^{u - u^{2}} d u

= \int_{- \infty}^{\infty} e^{u - u^{2} - \frac{1}{4} + \frac{1}{4}} d u

= e^{1 / 4} \int_{- \infty}^{\infty} e^{- {(u - \frac{1}{2})}^{2}} d u

(u - \frac{1}{2}) = v \Rightarrow u = v + \frac{1}{2}

u = \infty, v = \infty, u = - \infty, v = - \infty

= e^{1 / 4} \int_{- \infty}^{\infty} e^{- v^{2} d v}

= e^{1 / 4} \sqrt{π}