Question-160873

Question Number 160873 by SANOGO last updated on 08/Dec/21

Answered by TheSupreme last updated on 08/Dec/21

1) \int_{0}^{\infty} \frac{1}{a^{2}} \frac{d x}{1 + {(\frac{x}{a})}^{2}} = \frac{1}{a} a r c t a n \frac{x}{a} = \frac{π}{2 a}

Commented by SANOGO last updated on 08/Dec/21

m e r c i v u s t p l e r e s t e

Answered by puissant last updated on 13/Dec/21

2 - c)

\int_{- \infty}^{+ \infty} \frac{d t}{1 + t^{2}} = 2 \int_{0}^{\infty} \frac{d t}{1 + t^{2}} = 2 {[a r c t a n (t)]}_{0}^{\infty} = π .

3 .

\int_{0}^{\infty} \frac{e^{- u}}{\sqrt{u}} d u; p o s o n s u = t^{2} \Rightarrow d u = 2 t d t

\Rightarrow \int_{0}^{\infty} \frac{e^{- t^{2}}}{t} 2 t d t = 2 \int_{0}^{\infty} e^{- t^{2}} d t . .

Facebook Tweet Pin