prove-0-cot-1-1-x-2-1-2-1-2-pi-

Question Number 163400 by mnjuly1970 last updated on 06/Jan/22

p r o v e

Ω = \int_{0}^{\infty} {c o t}^{- 1} (1 + x^{2}) = (\sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}}) π

Answered by Ar Brandon last updated on 06/Jan/22

Ω = \int \cot^{- 1} (1 + x^{2}) = x \cot^{- 1} (1 + x^{2}) + \int \frac{2 x^{2}}{x^{4} + 2 x^{2} + 1} d x

= x \cot^{- 1} (1 + x^{2}) + \int \frac{(x^{2} + 1) + (x^{2} - 1)}{x^{4} + 2 x^{2} + 1} d x

= x \cot^{- 1} (1 + x^{2}) + \int \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 4} d x + \int \frac{1 - \frac{1}{x^{2}}}{{(x + \frac{1}{x})}^{2}} d x

= x \cot^{- 1} (1 + x^{2}) + \frac{1}{2} \tan^{- 1} (\frac{x^{2} - 1}{2 x}) - \frac{x}{x^{2} + 1} + C

Answered by Kamel last updated on 07/Jan/22

p r o v e

Ω = \int_{0}^{\infty} {c o t}^{- 1} (1 + x^{2}) = (\sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}}) π

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

= \frac{\sqrt{π}}{2} \int_{0}^{+ \infty} \frac{s i n (t) e^{- t}}{t^{\frac{3}{2}}} d t = \frac{\sqrt{π}}{2} \int_{1}^{+ \infty} \int_{0}^{+ \infty} \frac{s i n (t)}{\sqrt{t}} e^{- a t} d t d a

= \sqrt{π} \int_{1}^{+ \infty} \int_{0}^{+ \infty} s i n (u^{2}) e^{- {a u}^{2}} d u d a

= \frac{π}{4 i} \int_{1}^{+ \infty} (\frac{1}{\sqrt{a - i}} - \frac{1}{\sqrt{a + i}}) d a = \frac{π}{2 i} (\sqrt{1 + i} - \sqrt{1 - i}) = π \sqrt{\sqrt{2}} s i n (\frac{π}{8})

s i n (\frac{π}{8}) = \sqrt{\frac{1 - c o s (\frac{π}{4})}{2}} = \sqrt{\frac{\sqrt{2} - 1}{2 \sqrt{2}}} = \frac{1}{\sqrt{\sqrt{2}}} \sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}}

∴ Ω = π \sqrt{\frac{1}{\sqrt{2}} - \frac{1}{2}}

Commented by mnjuly1970 last updated on 07/Jan/22

b r a v o \dots