0-pi-2-x-tan-x-dx-

Question Number 89052 by M±th+et£s last updated on 15/Apr/20

\int_{0}^{\frac{π}{2}} \frac{x}{t a n (x)} d x

Commented by abdomathmax last updated on 15/Apr/20

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

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

f^{^{'}} (a) = \int_{0}^{\infty} \frac{d t}{(1 + a^{2} t^{2}) (1 + t^{2})}

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

= \int_{0}^{\infty} \frac{d u}{(1 + u^{2}) (a^{2} + u^{2})} d e c o m p o s i t i o n o f

F (u) = \frac{1}{(u^{2} + 1) (u^{2} + a^{2})} = \frac{α u + β}{u^{2} + 1} + \frac{λ u + ρ}{u^{2} + a^{2}}

F (- u) = F (u) \Rightarrow

\frac{- α u + β}{u^{2} + 1} + \frac{- λ u + ρ}{u^{2} + a^{2}} = F (u) \Rightarrow α = λ = 0 a n d ρ = β \Rightarrow

F (u) = \frac{β}{u^{2} + 1} + \frac{ρ}{u^{2} + a^{2}}

{l i m}_{u \to + \infty} u^{2} F (u) = 0 = β + ρ \Rightarrow ρ = - β \Rightarrow

F (u) = \frac{β}{u^{2} + 1} - \frac{β}{u^{2} + a^{2}}

F (0) = \frac{1}{a^{2}} = β - \frac{β}{a^{2}} \Rightarrow 1 = a^{2} β - β = (a^{2} - 1) β \Rightarrow

β = \frac{1}{a^{2} - 1} \Rightarrow F (u) = \frac{1}{a^{2} - 1} (\frac{1}{u^{2} + 1} - \frac{1}{u^{2} + a^{2}}) \Rightarrow

f^{^{'}} (a) = \frac{1}{a^{2} - 1} \int_{0}^{\infty} (\frac{1}{u^{2} + 1} - \frac{1}{u^{2} + a^{2}}) d u

\int \frac{d u}{u^{2} + a^{2}} =_{u = a z} \int \frac{a d z}{a^{2} (1 + z^{2})} = \frac{1}{a} a r c t a n (\frac{u}{a}) \Rightarrow

f^{^{'}} (a) = \frac{1}{a^{2} - 1} {[a r c t a n u - \frac{1}{a} a r c t a n (\frac{u}{a})]}_{0}^{+ \infty}

= \frac{1}{a^{2} - 1} {\frac{π}{2} - \frac{π}{2 a}} = \frac{π}{2} \times \frac{1}{a^{2} - 1} (1 - \frac{1}{a})

= \frac{π}{2 (a^{2} - 1)} (\frac{a - 1}{a}) = \frac{π}{2 a (a + 1)} \Rightarrow

f (a) = \frac{π}{2} \int \frac{d a}{a (a + 1)} + c

= \frac{π}{2} \int (\frac{1}{a} - \frac{1}{a + 1}) d a + c = \frac{π}{2} l n ∣ \frac{a}{a + 1} ∣ + c

c = {l i m}_{a \to + \infty} f (a)

I = f (1) = \frac{π}{2} l n (\frac{1}{2}) + c = c - \frac{π}{2} l n (2)

r e s t t o f i n d c \dots

Commented by abdomathmax last updated on 15/Apr/20

l e t t r y r e s i d u s m e t h o d w e ? h s v e

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

l e t φ (z) = \frac{a r c t a n (z)}{z (1 + z^{2})} \Rightarrow φ (z) = \frac{a r c t a n z}{z (z - i) (z + i)}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, i) + R e s (φ, o)}

= 2 i π \times \frac{a r c t a n (i)}{i (2 i)} = - i π a r c t a n (i)

b u t t h e f r m u l a a r c t a n (z) = \frac{1}{2 i} l n (\frac{1 + i z}{1 - i z}) i s n o t

v a l i d e f o r z = i \dots b e c o n t i n u e d \dots

Commented by M±th+et£s last updated on 15/Apr/20

t h a n x s i r

Commented by mathmax by abdo last updated on 15/Apr/20

y o u a r e w e l c o m e