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Question Number 137419 by mnjuly1970 last updated on 02/Apr/21

\dots \dots \dots m a t h e m a t i c a l \dots . a n a l y s i s \dots \dots . .

e v a l u a t e \dots .

ϕ = \int_{0}^{\infty} \frac{e^{2 π x} - e^{π x}}{x (1 + e^{2 π x}) (1 + e^{π x})} d x = λ \int_{0}^{1} l n (Γ (x) d x

λ = ? ? ?

Answered by Dwaipayan Shikari last updated on 02/Apr/21

\int_{0}^{\infty} \frac{f (a x) - f (b x)}{x} d x = \lim_{z \to \infty} (f (z) - f (0)) l o g (\frac{a}{b})

\int_{0}^{\infty} \frac{e^{2 π x} - e^{π x}}{x (1 + e^{2 π x}) (1 + e^{π x})} d x = \int_{0}^{\infty} \frac{\frac{1}{1 + e^{π x}} - \frac{1}{1 + e^{2 π x}}}{x} d x = (- \frac{1}{2}) l o g (\frac{1}{2})

= \frac{l o g (2)}{2}

λ \int_{0}^{1} l o g (Γ (x)) d x = λ \frac{l o g (2 π)}{2} (P r e v i o u s E x a m p l e s o n C o m m u n i t y)

λ = \frac{l o g (2)}{l o g (2 π)} = \frac{1}{1 + {l o g}_{2} (π)}

Commented by mnjuly1970 last updated on 02/Apr/21

m e r c e y m r p a y a n \dots

Commented by mnjuly1970 last updated on 02/Apr/21