0-pi-ln-sinx-dx-piln2-0-1-ln-x-dx-ln-2pi-

Question Number 207582 by sniper237 last updated on 19/May/24

\int_{0}^{π} l n (s i n x) d x = - π l n 2

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

Answered by mathzup last updated on 21/May/24

I = \int_{0}^{π} l n (s i n x) d x = \int_{0}^{\frac{π}{2}} l n (s i n x) d x

+ \int_{\frac{π}{2}}^{π} l n (s i n x) d x (\to x = \frac{π}{2} + t)

= - \frac{π}{2} l n (2) + \int_{0}^{\frac{π}{2}} l n (c o s t) d t

= - \frac{π}{2} l n 2 - \frac{π}{2} l n 2 = - π l n 2

Answered by mathzup last updated on 21/May/24

Γ (x) . Γ (1 - x) = \frac{π}{s i n (π x)} \Rightarrow

l n (Γ (x)) + l n (Γ (1 - x)) = l n π - l n (s i n (π x))

\Rightarrow \int_{0}^{1} l n (Γ (x)) d x + \int_{0}^{1} l n (Γ (1 - x)) d x

= l n (π) - \int_{0}^{1} l n (s i n (π x)) d x (\to π x = t)

= l n (π) - \int_{0}^{π} l n (s i n t) \frac{d t}{π}

= l n (π) - \frac{1}{π} (- π l n 2)

= l n (π) + l n (2) = l n (2 π)

b u t \int_{0}^{1} l n (Γ (1 - x)) d x =_{1 - x = t} \int_{1}^{0} l n (Γ (t)) (- d t)

= \int_{0}^{1} l n (Γ (x) d x \Rightarrow

\int_{0}^{1} l n (Γ (x)) d x = \frac{1}{2} l n (2 π)

i l y a e r r e u r d a n s l e n o n c e! \dots

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