advanced-calculus-prove-that-0-1-ln-x-cos-2-pix-dx-ln-2pi-4-1-8-m-n-1970-

Question Number 117280 by mnjuly1970 last updated on 10/Oct/20

\dots a d v a n c e d c a l c u l u s \dots

p r o v e t h a t ::

\int_{0}^{1} l n (Γ (x)) . {c o s}^{2} (π x) d x

= \frac{l n (2 π)}{4} + \frac{1}{8}

m . n . 1970

Answered by AbduraufKodiriy last updated on 10/Oct/20

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

I = \int_{0}^{1} l n (Γ (x)) {c o s}^{2} (π x) d x = \int_{0}^{1} l n (Γ (1 - x)) {c o s}^{2} (π (1 - x)) d x =

= \int_{0}^{1} l n (Γ (1 - x)) {c o s}^{2} (π x) d x

T h e r e f o r e : 2 I = \int_{0}^{1} (l n (Γ (x)) + l n (Γ (1 - x))) {c o s}^{2} (π x) d x \Rightarrow

\Rightarrow I = \frac{1}{2} \int_{0}^{1} l n (Γ (x) Γ (1 - x)) {c o s}^{2} (π x) d x = \frac{1}{2} \int_{0}^{1} l n (\frac{π}{s i n (π x)}) {c o s}^{2} (π x) d x =

= \frac{l n (π)}{2} \int_{0}^{1} {c o s}^{2} (π x) d x - \frac{1}{2} \int {c o s}^{2} (π x) l n (s i n (π x)) d x;

I_{1} = \int_{0}^{1} {c o s}^{2} (π x) l n (s i n (π x)) d x \Rightarrow I = \frac{l n (π)}{4} - \frac{1}{2} I_{1}

I_{1} = | \begin{matrix} π x = t \Rightarrow x = \frac{t}{π} \\ d x = \frac{1}{π} d t \end{matrix} | = \frac{1}{π} \int_{0}^{π} {c o s}^{2} (t) l n (s i n (t)) d t =

= \frac{1}{2 π} \int_{0}^{π} l n (s i n (t)) d t + \frac{1}{2 π} \int_{0}^{π} c o s (2 t) l n (s i n (t)) d t; W e k n o w : \int_{0}^{π} l n (s i n (t)) d t = - π l n 2

I_{1} = - \frac{1}{2} l n (2) + \frac{1}{2} \int_{0}^{π} c o s (2 t) l n (s i n (t)) d t = - \frac{1}{2} l n 2 + (\frac{s i n (2 t) l n (s i n (t))}{4 π} - \frac{t}{4 π} - \frac{s i n (2 t)}{8 π}) ∣_{0}^{π} =

= - \frac{1}{2} l n (2) - \frac{1}{4} \Rightarrow I = \frac{l n (π)}{4} + \frac{l n (2)}{4} + \frac{1}{8} = \frac{l n (2 π)}{4} + \frac{1}{8}

Commented by AbduraufKodiriy last updated on 10/Oct/20

Commented by mnjuly1970 last updated on 10/Oct/20

t h a n y o u m r a b d u r a u f k . .

Commented by mnjuly1970 last updated on 10/Oct/20

j a v o b = j a v a b = a n s w e r \dots

Commented by mnjuly1970 last updated on 10/Oct/20

m e r c e y a g h a y e a b d u r a u f k