Question-118757

Question Number 118757 by mohammad17 last updated on 19/Oct/20

Commented by mohammad17 last updated on 19/Oct/20

w i t h o u t x = t a n θ

Answered by mnjuly1970 last updated on 19/Oct/20

\frac{π}{4} \overset{?}{=} a n s w e r

m e t h o d 1 :

e u l e r r e f l e c t i o n f o r m u l a .

x^{2} = t \Rightarrow Ω = \frac{1}{2} \int_{0}^{\infty} \frac{t^{(\frac{- 1}{2} + 1) - 1}}{{(t + 1)}^{2}} d t

\Rightarrow Ω = \frac{1}{2} β (\frac{1}{2}, \frac{3}{2}) = \frac{1}{2} Γ (\frac{1}{2}) Γ (\frac{3}{2})

= {(\frac{1}{2})}^{2} Γ^{2} (\frac{1}{2}) \overset{Γ (\frac{1}{2}) = \sqrt{π}}{=} \frac{π}{4} ✓ ✓

\dots m . n . j u l y . 1970 \dots

\dots \dots \dots .

m e t h o t 2 :

x = \frac{1}{t} \Rightarrow Ω = \int_{0}^{\infty} \frac{d t}{t^{2} {(1 + \frac{1}{t^{2}})}^{2}}

= \int_{0}^{\infty} \frac{(t^{2} + 1 - 1)}{{(1 + t^{2})}^{2}} d t = \frac{π}{2} - Ω

2 Ω = \frac{π}{2} \Rightarrow Ω = \frac{π}{4} ✓ ✓

. m . n . 1970 . .

Commented by mohammad17 last updated on 19/Oct/20

y e s s i r i t s r i g h t

Commented by PRITHWISH SEN 2 last updated on 19/Oct/20

excellent

Commented by mnjuly1970 last updated on 19/Oct/20

t h a n k y o u

Commented by mnjuly1970 last updated on 19/Oct/20

s i n c e r e l y y o u r s

Answered by 1549442205PVT last updated on 19/Oct/20

Put x = \tan φ \Rightarrow dx = (1 + \tan^{2} φ) d φ

\int_{0}^{\infty} \frac{dx}{{(x^{2} + 1)}^{2}} = \int_{0}^{\frac{π}{2}} \frac{d φ}{(1 + \tan^{2} φ)} = \int_{0}^{\frac{π}{2}} \cos^{2} φ d φ

= \int_{0}^{\frac{π}{2}} \frac{1 + \cos 2 φ}{2} d φ = {(\frac{φ}{2} + \frac{1}{4} \sin 2 φ)}_{0}^{\frac{π}{2}}

= \frac{π}{4}