let-I-n-0-e-nt-1-e-t-n-1-dt-n-from-N-prove-the-existence-of-I-n-2-find-lim-n-I-n-

Question Number 66351 by mathmax by abdo last updated on 12/Aug/19

l e t I_{n} = \int_{0}^{\infty} \frac{e^{n t}}{{(1 + e^{t})}^{n + 1}} d t (n f r o m N^{★})

) p r o v e t h e e x i s t e n c e o f I_{n}

2) f i n d {l i m}_{n \to + \infty} I_{n}

Commented by mathmax by abdo last updated on 15/Aug/19

\forall A > 0 t h e f u n c t i o n t \to \frac{e^{n t}}{{(1 + e^{t})}^{n + 1}} i s c o n t i n u e o n [0, A] s o

i n t e g r a b l e l e t s e e w h a t h a p p e n a t + \infty w e h a v e

{l i m}_{t \to + \infty} t^{2} \frac{e^{n t}}{{(1 + e^{t})}^{n + 1}} = {l i m}_{t \to + \infty} t^{2} \frac{e^{n t}}{e^{(n + 1) t}} = {l i m}_{n \to + \infty} t^{2} e^{- t} = 0

s o I_{n} i s c o n v e r g e n t

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

I_{n} = \int_{1}^{+ \infty} \frac{x}{{(1 + x^{\frac{1}{n}})}^{n + 1}} \frac{d x}{n x} = \int_{1}^{+ \infty} \frac{d x}{{(1 + x^{\frac{1}{n}})}^{n + 1}}

= \int_{1}^{+ \infty} e^{- (n + 1) l n (1 + x^{\frac{1}{n}})} d x = \int_{R} f_{n} (x) d x w i t h

f_{n} (x) = e^{- (n + 1) l n (1 + x^{\frac{1}{n}})} χ_{[1, + \infty [} (x) w e h a v e f_{n}) c o n t i n u e s

a n d f_{n} \to^{c s} 0 \Rightarrow {l i m}_{n \to + \infty} I_{n} = \int_{R} {l i m}_{n \to + \infty} f_{n} (x) d x = 0