show-that-0-x-log-x-x-log-x-dx-e-pi-

Question Number 82446 by M±th+et£s last updated on 21/Feb/20

s h o w t h a t

\int_{0}^{\infty} x^{- l o g (x)} x l o g (x) d x = e \sqrt{π}

Commented by abdomathmax last updated on 21/Feb/20

c h a n g e m e n t l o g x = t g i v e x = e^{t} \Rightarrow

I = \int_{- \infty}^{+ \infty} {(e^{t})}^{- t} e^{t} t e^{t} d t

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

= \int_{- \infty}^{+ \infty} t e^{- {(t - 1)}^{2} + 1} d t =_{t - 1 = u} e \int_{- \infty}^{+ \infty} (u + 1) e^{- u^{2}} d u

b u t \int_{- \infty}^{+ \infty} (u + 1) e^{- u^{2}} d u

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

= {[- \frac{1}{2} e^{- u^{2}}]}_{- \infty}^{+ \infty} + \sqrt{π} = 0 + \sqrt{π} \Rightarrow

I = e \sqrt{π}

Commented by M±th+et£s last updated on 21/Feb/20

t h a n k y o u s i r

Commented by mathmax by abdo last updated on 21/Feb/20

y o u a r e w e l c o m e .