nice-integral-please-prove-0-pi-2-ln-2-cot-x-dx-pi-3-8-m-n-1070-

Question Number 117953 by mnjuly1970 last updated on 14/Oct/20

\dots ◂ n i c e i n t e g r a l ▸ \dots

p l e a s e p r o v e :

Ω = \int_{0}^{\frac{π}{2}} {l n}^{2} (c o t (x)) d x = \frac{π^{3}}{8}

\dots ♠ m . n . 1070 ♠ \dots

Answered by mindispower last updated on 14/Oct/20

Ω = \int_{0}^{\frac{π}{2}} {l n}^{2} (c o t (x)) d x; c o t (x) = s \Rightarrow

Ω = \int_{0}^{\infty} \frac{{l n}^{2} (s)}{1 + s^{2}} d s = 2 \int_{0}^{1} \frac{{l n}^{2} (s)}{1 + s^{2}} d s = 2 \int_{0}^{1} Σ {(- s^{2})}^{k} {l n}^{2} (s) d s

= 2 \sum_{n ⩾ 0} {(- 1)}^{n} \int_{0}^{1} s^{2 k} {l n}^{2} (s) d s

\int_{0}^{1} s^{n} {l n}^{2} (s) d s = - \frac{2}{n + 1} \int_{0}^{1} s^{n} l n (s) d s

= \frac{2}{{(n + 1)}^{2}} \int_{0}^{1} s^{n} d s = \frac{2}{{(n + 1)}^{3}}

w e g e t \sum_{n ⩾ 0} \frac{4 {(- 1)}^{n}}{{(2 n + 1)}^{3}} = \sum_{n ⩾ 0} \frac{4}{{(4 n + 1)}^{3}} - \sum_{n ⩾ 0} \frac{4}{{(4 n + 3)}^{3}}

= \frac{1}{16} [\sum_{n ⩾ 0} \frac{1}{{(n + \frac{1}{4})}^{3}} - \sum_{n ⩾ 0} \frac{1}{{(n + \frac{3}{4})}^{3}}]

= \frac{1}{16} [Ψ^{3} (\frac{1}{4}) - Ψ^{3} (\frac{3}{4})]

Ψ (1 - x) - Ψ (x) = π c o t (π x)

\Rightarrow Ψ^{3} (1 - x) - Ψ^{3} (x) = \frac{d^{2}}{{d x}^{2}} (π c o t (π x))

= \frac{d}{d x} (- π^{2} (1 + {c o t}^{2} (π x)) = 2 π^{3} (1 + {c o t}^{2} (π x)) c o t (π x))

\frac{1}{16} [Ψ^{3} (\frac{1}{4}) - Ψ^{3} (\frac{3}{4})] = \frac{1}{16} (2 π^{3} (1 + {c o t}^{2} (\frac{π}{4})) c o t (\frac{π}{4})

= \frac{π^{3}}{4} m a y b e e f o r g e t s o m t h i n g

Commented by mnjuly1970 last updated on 14/Oct/20

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i s f a b u l o u s . g r a t e f u l .

Commented by mindispower last updated on 19/Oct/20

t h a n k y o u w i t h e p l e s u r

s i r d o y o u h a v e s o m m e l e c t u r a b o u t p o l y g a r i t h m s

f u n c t i o n i d e n t i t i e s

t h a n k y o u f o r y o u a n s w e r

Answered by mathmax by abdo last updated on 14/Oct/20

A = \int_{0}^{\frac{π}{2}} \ln^{2} (cotanx) dx \Rightarrow A = \int_{0}^{\frac{π}{2}} \ln^{2} (\frac{1}{tanx}) dx

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

= \int_{0}^{1} \frac{\ln^{2} t}{1 + t^{2}} dt + \int_{1}^{\infty} \frac{\ln^{2} t}{1 + t^{2}} dt (\to t = \frac{1}{u})

= \int_{0}^{1} \frac{\ln^{2} t}{1 + t^{2}} dt + \int_{0}^{1} \frac{\ln^{2} u}{(1 + \frac{1}{u^{2}})} \frac{du}{u^{2}} = 2 \int_{0}^{1} \frac{\ln^{2} x}{1 + x^{2}} dx

= 2 \int_{0}^{1} \ln^{2} x (\sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n}) dx

= 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{2 n} \ln^{2} x dx = 2 \sum_{n = 0}^{\infty} {(- 1)}^{n} A_{n} by parts

A_{n} = \int_{0}^{1} x^{2 n} \ln^{2} xdx = {[\frac{x^{2 n + 1}}{2 n + 1} \ln^{2} x]}_{0}^{1} - \int_{0}^{1} \frac{x^{2 n + 1}}{2 n + 1} \frac{2 lnx}{x} dx

= - \frac{2}{(2 n + 1)} \int_{0}^{1} x^{2 n} lnx dx = - \frac{2}{(2 n + 1)} {{[\frac{x^{2 n + 1}}{2 n + 1} lnx]}_{0}^{1} - \int_{0}^{1} \frac{x^{2 n}}{2 n + 1}}

= \frac{2}{{(2 n + 1)}^{3}} \Rightarrow A = 4 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} rest to find the value of

this serie by sir fourier \dots be continued \dots .

Commented by mnjuly1970 last updated on 15/Oct/20

t h a n k y o u s o m u c h s i r .

g r a t e f u l \dots