nice-calculus-prove-that-0-pi-2-log-1-tan-x-tan-x-dx-5pi-2-48-

Question Number 124827 by mnjuly1970 last updated on 06/Dec/20

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

p r o v e t h a t ::

\int_{0}^{\frac{π}{2}} \frac{l o g (1 + t a n (x))}{t a n (x)} d x = \frac{5 π^{2}}{48} ✓

Answered by mindispower last updated on 06/Dec/20

f (t) = \int_{0}^{\infty} \frac{l n (1 + t . t a n (x))}{t a n (x)} d x

q = f (1), q

f^{'} (t) = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + t . t a n (x)} = \int_{0}^{\frac{π}{2}} \frac{c o s (x)}{c o s (x) + t s i n (x)} d x

c o s (x) = α (c o s (x) + t s i n (x)) + β (- s i n (x) + t c o s (x)

\Rightarrow α + β t = 1

α t - β = 0

α = \frac{1}{1 + t^{2}}, β = \frac{t}{1 + t^{2}}

f^{'} (t) = \frac{1}{1 + t^{2}} \int_{0}^{\frac{π}{2}} d x + \frac{t}{1 + t^{2}} \int_{0}^{\frac{π}{2}} \frac{- s i n (x) + t c o s (x)}{c o s (x) + t s i n (x)} d x

= \frac{π}{2 (1 + t^{2})} + {[\frac{t}{1 + t^{2}} l n (c o s (x) + t s i n (x))]}_{0}^{\frac{π}{2}}

= \frac{π}{2 (1 + t^{2})} + \frac{t l n (t)}{1 + t^{2}}

f (0) = \int_{0}^{\frac{π}{2}} \frac{l n (1 + o t a n (x))}{t a n (x)} d x = 0

q = \int_{0}^{1} \frac{π}{2 (1 + t^{2})} + \frac{t l n (t)}{1 + t^{2}} d t

= \frac{π}{2} \tan^{- 1} (1) + \sum_{k ⩾ 0} \int_{0}^{1} {(- 1)}^{k} t^{2 k + 1} l n (t) d t

= \frac{π}{2} . \frac{π}{4} + \sum_{k ⩾ 0} \frac{{(- 1)}^{k + 1}}{{(2 k + 2)}^{2}} = \frac{π}{8} - \frac{1}{4} \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{{(k + 1)}^{2}}

= \frac{π}{8} + \frac{1}{4} (\sum_{k ⩾ 0} \frac{1}{{(2 k + 2)}^{2}} - \frac{1}{4} \sum_{k ⩾ 0} \frac{1}{{(2 k + 1)}^{2}})

= \frac{π^{2}}{8} + \frac{1}{16} ζ (2) - \frac{3}{16} ζ (2)

= \frac{π^{2}}{8} - \frac{1}{8} . \frac{π^{2}}{6} = \frac{5 π^{2}}{48} = \int_{0}^{\frac{π}{2}} \frac{l n (1 + t a n (x))}{t a n (x)} d x

Commented by mnjuly1970 last updated on 06/Dec/20

a g a i n t h a n k y o u

s i r m i n d s p o w e r

Commented by mindispower last updated on 06/Dec/20

a l w a y s p l e a s u r s i r

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

let f (a) = \int_{0}^{\frac{π}{2}} \frac{\log (1 + atanx)}{tanx} dx with a > 0 \Rightarrow f^{^{'}} (a) = \int_{0}^{\frac{π}{2}} \frac{tanx}{(1 + atanx) tanx} dx

= \int_{0}^{\frac{π}{2}} \frac{dx}{1 + atanx} =_{tanx = t} \int_{0}^{\infty} \frac{dt}{(1 + t^{2}) (1 + at)} =_{at = z} \int_{0}^{\infty} \frac{dz}{a (1 + \frac{z^{2}}{a^{2}}) (z + 1)}

= a \int_{0}^{\infty} \frac{dz}{(z^{2} + a^{2}) (z + 1)} let decompose F (z) = \frac{1}{(z + 1) (z^{2} + a^{2})}

F (z) = \frac{α}{z + 1} + \frac{β z + n}{z^{2} + a^{2}}

α = \frac{1}{a^{2} + 1}, \lim_{z \to + \infty} zF (z) = 0 = α + β \Rightarrow β = - \frac{1}{a^{2} + 1}

F (0) = \frac{1}{a^{2}} = α + \frac{n}{a^{2}} \Rightarrow 1 = a^{2} α + n \Rightarrow n = 1 - \frac{a^{2}}{a^{2} + 1} = \frac{1}{a^{2} + 1} \Rightarrow

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

\int_{0}^{\infty} F (z) dz = \frac{1}{a^{2} + 1} \int_{0}^{\infty} \frac{dz}{z + 1} - \frac{1}{2 (a^{2} + 1)} \int_{0}^{\infty} \frac{2 z - 2}{z^{2} + a^{2}} dz

= \frac{1}{a^{2} + 1} \int_{0}^{\infty} (\frac{1}{z + 1} - \frac{1}{2} \times \frac{2 z}{z^{2} + 1}) dz + \frac{1}{a^{2} + 1} \int_{0}^{\infty} \frac{dz}{z^{2} + a^{2}} (\to z = au)

= \frac{1}{a^{2} + 1} {[\ln ∣ \frac{z + 1}{\sqrt{z^{2} + 1}} ∣]}_{0}^{\infty} + \frac{1}{a^{2} + 1} \int_{0}^{\infty} \frac{adu}{a^{2} (u^{2} + 1)}

= \frac{1}{a (a^{2} + 1)} \times \frac{π}{2} \Rightarrow f^{^{'}} (a) = \frac{π}{2 (1 + a^{2})} \Rightarrow f (a) = \frac{π}{2} \arctan (a) + C

f (0) = C \Rightarrow f (a) = \frac{π}{2} \arctan (a) and

\int_{0}^{\frac{π}{2}} \frac{\log (1 + tanx)}{tanx} dx = \frac{π}{2} \arctan (1) = \frac{π}{2} . \frac{π}{4} = \frac{π^{2}}{8}

Commented by mnjuly1970 last updated on 06/Dec/20

t h a n k y o u s o m u c h s i r m a x \dots

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

you are welcome sir