IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

Sushmi

_[n]

_[

_] ( n!/n^n) ^1/n

_[n]

_[

_]log ( n!/n^n) ^1/n

= _[n]

_[

_] 1/n log ( 1.2.3.....n)/(n.n.n...n)

=_[n]

_[

_] 1/n _{[r =1]}

^[n] log r/n

=_[0]

^[1] log x dx

= [ log x . x -

x.1/x dx ] within limits from 0 to 1

= [x log x - x] within limits from 0 to 1

= - 1 - _[x]

_[0] x log x

= -1 - _[x]

_[0] log x / 1/x

= -1 - _[x]

_[0] 1/x / -1/x² ( by L'Hospital's theorem)

= - 1 - _[x]

_[0] (-x) = -1

now we know that _[x]

_[a] f(x) = e ^ _[x]

_[a] log f(x) = e ^-1 = 1/e