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

Question Number 123159 by Lordose last updated on 23/Nov/20

\int_{0}^{\frac{π}{2}} \log^{2} (\tan (x)) dx

Answered by mnjuly1970 last updated on 23/Nov/20

a n s := \frac{π^{3}}{8}

Commented by Lordose last updated on 23/Nov/20

{It}^{'} s correct sir

Commented by mnjuly1970 last updated on 23/Nov/20

t h a n k y o u

s o l u t i o n

I = \int_{0}^{\frac{π}{2}} {l o g}^{2} (t a n (x)) d x \overset{t a n (x) = y}{=}

= \int_{0}^{\infty} {l o g}^{2} (y) \frac{d y}{1 + y^{2}} \overset{y^{2} = t}{=} \frac{1}{8} \int_{0}^{\infty} \frac{t^{\frac{- 1}{2}} {l o g}^{2} (t)}{1 + t} d t

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

g o a l :: I = \frac{1}{8} I^{″} (0)

b u t : I (a) = β (\frac{1}{2} + a, \frac{1}{2} - a)

= Γ (\frac{1}{2} + a) Γ (\frac{1}{2} - a)

\underset{r e f l e c t i o n f o r m u l a}{\overset{e u l e r}{=}} \frac{π}{s i n (\frac{π}{2} + π a)} = \frac{π}{c o s (π a)}

I^{'} (a) = \frac{π^{2} s i n (π a)}{{c o s}^{2} (π a)}

I^{″} (a) = π^{2} (\frac{π {c o s}^{3} (π a) + 2 π {s i n}^{2} (π a) c o s (π a)}{{c o s}^{4} (π a)}) ∣_{a = 0}

I^{″} (0) = π^{3} \Rightarrow I = \frac{π^{3}}{8} ✓ ✓

Commented by Dwaipayan Shikari last updated on 23/Nov/20

G r e a t s i r! i {h a v e n}^{'} t t h o u g h t o f t h i s

Commented by mnjuly1970 last updated on 23/Nov/20

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

t o m e y o u r s o l u t i o n i s v e r y v e r y

n i c e a n d u n i q u e \dots

Answered by Dwaipayan Shikari last updated on 23/Nov/20

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

= \int_{0}^{\infty} \frac{{l o g}^{2} t}{t^{2} + 1} d t

= \underset{n ⩾ 0}{\sum^{\infty}} \int_{0}^{\infty} {l o g}^{2} t {(- 1)}^{n} t^{2 n} d t l o g t = u \Rightarrow \frac{1}{t} = \frac{d u}{d t}

= \underset{n ⩾ 0}{\sum^{\infty}} {(- 1)}^{n} \int_{- \infty}^{\infty} u^{2} e^{2 n u + u} d t u (2 n + 1) = Λ

= 2 \underset{n ⩾ 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} \int_{0}^{\infty} Λ^{2} e^{- Λ} d Λ

= 2 \underset{n ⩾ 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} Γ (3)

= 4 \underset{n ⩾ 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} = 4 . \frac{π^{3}}{32} = \frac{π^{3}}{8}

Commented by mnjuly1970 last updated on 23/Nov/20

v e r y g o o d \dots

Commented by Lordose last updated on 23/Nov/20

Can you explain from 4 th line

Commented by Dwaipayan Shikari last updated on 24/Nov/20

\underset{n ⩾ 0}{\sum^{\infty}} {(- 1)}^{n} \int_{- \infty}^{\infty} \frac{Λ^{2}}{{(2 n + 1)}^{2} (2 n + 1)} e^{Λ} d Λ u (2 n + 1) = Λ

= \underset{n ⩾ 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} \int_{\infty}^{- \infty} - Φ^{2} e^{- Φ} d Φ Λ = - Φ

= 2 \underset{n ⩾ 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} \int_{0}^{\infty} Φ^{2} e^{- Φ} d Φ = 2 . Γ (3) \frac{π^{3}}{32} = \frac{π^{3}}{8}

Answered by mathmax by abdo last updated on 23/Nov/20

A = \int_{0}^{\frac{π}{2}} \ln^{2} (tanx) dx changementtanx = t give

A = \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}{x})

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

u_{n} = \int_{0}^{1} x^{2 n} \ln^{2} (x) dx = {[\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} \ln (x) 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)} dx}

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

serie by fourier \dots .