help-1-x-ln-x-dx-

Question Number 207906 by nachosam last updated on 30/May/24

h e l p

\int_{1}^{\infty} x^{- l n (x)} d x

Answered by Berbere last updated on 30/May/24

\int_{0}^{\infty} {(e^{t})}^{(- t)} e^{t} d t

= \int_{0}^{\infty} e^{- t^{2} + t} d t = \int_{0}^{\infty} e^{- {(t - \frac{1}{2})}^{2} + \frac{1}{4}} d t

= e^{\frac{1}{4}} \int_{- \frac{1}{2}}^{\infty} e^{- x^{2}} d x = e^{\frac{1}{4}} \int_{0}^{\infty} e^{- x^{2}} d x + e^{\frac{1}{4}} \int_{- \frac{1}{2}}^{0} e^{- x^{2}} d x

A = e^{\frac{1}{4}} . \frac{\sqrt{π}}{2} - e^{\frac{1}{4}} \frac{\sqrt{π}}{2} e r f (- \frac{1}{2})

e r f (x) = \frac{2}{\sqrt{π}} \int_{0}^{x} e^{- t^{2}} d t \Rightarrow e r f (- x) = - e r f (x)

A = \frac{\sqrt{π}}{2} e^{\frac{1}{4}} (1 + e r f (\frac{1}{2}))

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