nice-calculus-prove-that-0-pi-tan-1-tan-2-x-tan-2-x-dx-pi-2-pi-4-

Question Number 128225 by mnjuly1970 last updated on 05/Jan/21

\dots n i c e c a l c u l u s \dots

p r o v e t h a t :

ϕ = \int_{0}^{π} \frac{{t a n}^{- 1} ({t a n}^{2} (x))}{{t a n}^{2} (x} d x \overset{? ? ?}{=} π (\sqrt{2} - \frac{π}{4})

Answered by mathmax by abdo last updated on 05/Jan/21

Φ = \int_{0}^{π} \frac{\arctan (\tan^{2} x)}{\tan^{2} x} dx let f (a) = \int_{0}^{π} \frac{\arctan ({atan}^{2} x)}{\tan^{2} x} dx (a > 0) \Rightarrow

f^{^{'}} (a) = \int_{0}^{π} \frac{1}{(1 + a^{2} \tan^{4} x)} dx = \int_{0}^{\frac{π}{2}} (\dots) dx + \int_{\frac{π}{2}}^{π} (\dots) dx

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

= \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dt}{(t^{2} + 1) (a^{2} t^{4} + 1)} let φ (z) = \frac{1}{(z^{2} + 1) (a^{2} z^{4} + 1)}

\Rightarrow φ (z) = \frac{1}{a^{2} (z^{2} + 1) (z^{4} + \frac{1}{a^{2}})} = \frac{1}{a^{2} (z - i) (z + i) (z^{2} - \frac{i}{a}) (z^{2} + \frac{i}{a})}

= \frac{1}{a^{2} (z - i) (z + i) (z - \frac{1}{\sqrt{a}} e^{\frac{i π}{4}}) (z + \frac{1}{\sqrt{a}} e^{\frac{i π}{4}}) (z - \frac{1}{\sqrt{a}} e^{- \frac{i π}{4}}) (z + \frac{1}{\sqrt{a}} e^{- \frac{i π}{4}})}

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, i) + Res (φ, \frac{1}{\sqrt{a}} e^{\frac{i π}{4}}) + Res (φ, - \frac{1}{\sqrt{a}} e^{- \frac{i π}{4}})}

\dots be continued \dots

Answered by mindispower last updated on 06/Jan/21

t a n (x) = t

= \int_{- \infty}^{\infty} \frac{{t a n}^{-} (t^{2})}{t^{2} (1 + t^{2})} d t

\int_{- \infty}^{\infty} \frac{{t a n}^{-} (t^{2})}{t^{2}} - \int_{- \infty}^{\infty} \frac{{t a n}^{-} (t^{2})}{1 + t^{2}} d t

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

= 2 A - 2 B

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

= \frac{1}{2} \int_{0}^{\infty} \frac{u^{- \frac{3}{4}}}{1 + u} = \frac{1}{2} β (\frac{1}{4}, \frac{3}{4}) = \frac{π}{2 s i n (\frac{π}{4})} = \frac{π}{\sqrt{2}}

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

= \frac{π^{2}}{4} - \int_{0}^{\infty} t^{4} \frac{\frac{2}{t} (\frac{π}{2} - {t a n}^{-} (t))}{1 + t^{4}} . \frac{d t}{t^{2}}_{= \frac{π^{2}}{4} - C}

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

{\Rightarrow 2 C = \int_{0}^{\infty} \frac{π t}{1 + t^{4}} = \frac{π}{2} \int_{0}^{\infty} . \frac{d (t^{2})}{1 + {(t^{2})}^{2}} = \frac{π}{2 .} . {t a n}^{-} (t^{2})]}_{0}^{\infty}

= \frac{π^{2}}{4} \Rightarrow C = \frac{π^{2}}{8}

B = \frac{π^{2}}{8}

2 A - 2 B = π \sqrt{2} - \frac{π^{2}}{4} = π (\sqrt{2} - \frac{π}{4}) = \int_{0}^{\frac{π}{2}} \frac{{t a n}^{-} ({t g}^{2} (t))}{{t g}^{2} (t)} d t