prove-that-0-arctan-x-2-x-2-dx-pi-2-

Question Number 124888 by mnjuly1970 last updated on 06/Dec/20

::::: p r o v e t h a t

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

Answered by mnjuly1970 last updated on 06/Dec/20

s o l u t i o n

ϕ \overset{x = \frac{1}{t} \Rightarrow}{=} \int_{0}^{\infty} a r c c o t (t^{2}) d t = {[t a r c c o t (t^{2})]}_{0}^{\infty} + 2 \int_{0}^{\infty} \frac{t^{2}}{1 + t^{4}} d t

\overset{t^{4} = u}{=} \frac{1}{2} \int_{0}^{\infty} \frac{u^{\frac{1}{2}} u^{\frac{- 3}{4}}}{1 + u} d u = \frac{1}{2} Γ (\frac{3}{4}) Γ (\frac{1}{4})

= \frac{1}{2} \frac{π}{s i n (\frac{π}{4})} = \frac{1}{2} \frac{2 π}{\sqrt{2}} = \frac{π}{\sqrt{2}} ✓

Commented by Dwaipayan Shikari last updated on 06/Dec/20

G r e a t s i r!

Commented by mnjuly1970 last updated on 06/Dec/20

t h a n k s a l o t . .

Answered by Dwaipayan Shikari last updated on 06/Dec/20

\int_{0}^{\infty} \frac{{t a n}^{- 1} (x^{2})}{x^{2}} d x = {[- \frac{{t a n}^{- 1} x^{2}}{x}]}_{0}^{\infty} + \int_{0}^{\infty} \frac{2 x}{x (x^{4} + 1)} d x

= 2 \int_{0}^{\infty} \frac{d x}{(x^{4} + 1)} = \int_{0}^{\frac{π}{2}} \frac{d θ}{\sqrt{t a n θ} ({t a n}^{2} θ + 1)} {s e c}^{2} θ x^{2} = t a n θ

= 2 \int_{0}^{\frac{π}{2}} \frac{1}{\sqrt{t a n θ}} d θ = 2 \int_{0}^{\frac{π}{2}} \frac{1}{\sqrt{c o t θ}} d θ = I

\Rightarrow 2 I = 2 \int_{0}^{\frac{π}{2}} \frac{s i n θ + c o s θ}{\sqrt{s i n θ c o s θ}} d θ

\Rightarrow I = \sqrt{2} \int_{0}^{\frac{π}{2}} \frac{s i n θ + c o s θ}{\sqrt{1 - {(s i n θ - c o s θ)}^{2}}} d θ = \sqrt{2} \int_{0}^{1} \frac{d t}{\sqrt{1 - t^{2}}}

\Rightarrow I = \sqrt{2} {[{s i n}^{- 1} t]}_{0}^{1} = \frac{π}{\sqrt{2}}

Commented by mnjuly1970 last updated on 06/Dec/20

t h a n k y o u m r

d w a i p a y a n . .

Answered by mathmax by abdo last updated on 06/Dec/20

I = \int_{0}^{\infty} \frac{\arctan (x^{2})}{x^{2}} dx by psrts I = {[- \frac{1}{x} \arctan (x^{2})]}_{0}^{\infty} + \int_{0}^{\infty} \frac{1}{x} \times \frac{2 x}{1 + x^{4}} dx

= 2 \int_{0}^{\infty} \frac{dx}{1 + x^{4}} =_{x = t^{\frac{1}{4}}} 2 \int_{0}^{\infty} \frac{1}{4 (1 + t)} t^{\frac{1}{4} - 1} dt = \frac{1}{2} \int_{0}^{\infty} \frac{t^{\frac{1}{4} - 1}}{1 + t} dt

= \frac{1}{2} \times \frac{π}{\sin (\frac{π}{4})} = \frac{π}{2 \times \frac{1}{\sqrt{2}}} = \frac{π}{\sqrt{2}} \Rightarrow I = \frac{π}{\sqrt{2}}