find-0-ln-1-t-2-t-2-dt-

Question Number 32478 by prof Abdo imad last updated on 25/Mar/18

f i n d \int_{0}^{\infty} l n (\frac{1 + t^{2}}{t^{2}}) d t

Commented by prof Abdo imad last updated on 30/Mar/18

I = \int_{0}^{+ \infty} l n (1 + \frac{1}{t^{2}}) d t b y p a r t s

I =] {[t l n (1 + \frac{1}{t^{2}})]}_{0}^{+ \infty} - \int_{0}^{+ \infty} t \frac{- 2}{t^{3}} {(1 + \frac{1}{t^{2}})}^{- 1} d t

= 2 \int_{0}^{\infty} \frac{1}{t^{2}} . \frac{1}{1 + \frac{1}{t^{2}}} d t = 2 \int_{0}^{\infty} \frac{d t}{t^{2} + 1} = 2 \frac{π}{2} = π

l e t p r o v e t h a t {l i m}_{t \to + \infty} t l n (1 + \frac{1}{t^{2}}) = 0

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

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

= 0 f i n a l l y I = π