advanced-calculus-prove-that-i-0-1-2-e-2x-1-1-e-x-1-x-dx-log-1-pi-ii-0-1-2-1-1-e-x-e-2x-x-dx-log-pi-2

Question Number 139524 by mnjuly1970 last updated on 28/Apr/21

\dots \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 :

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

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

Answered by mindispower last updated on 30/Apr/21

i . . i i a r e s i m i l a r i n w a y t h i s c a n s o l v e b y

l e t \emptyset = \int_{0}^{\infty} (\frac{e^{- 2 x}}{2} - \frac{e^{- x}}{1 + e^{- x}}) . \frac{d x}{x}

e^{- x} = t \Rightarrow ϕ = \int_{0}^{1} (\frac{1}{1 + t} - \frac{t}{2}) \frac{d t}{l n (t)}

l e t f (z) = \int_{0}^{1} (\frac{1}{1 + t} - \frac{t}{2}) \frac{t^{z}}{l n (t)} d t

f^{'} (z) = \int_{0}^{1} (\frac{t^{z}}{1 + t} - \frac{t^{1 + z}}{2}) d t = - \frac{1}{2 (2 + z)} + \int_{0}^{1} \frac{t^{z} - t^{1 + z}}{1 - t^{2}} d t

t^{2} \to t i n 2 n d \Leftrightarrow f^{'} (z) = - \frac{1}{2 (2 + z)} + \frac{1}{2} (- γ + \int_{0}^{1} \frac{1 - t^{\frac{z}{2}}}{1 - t} . d t)

- \frac{1}{2} (- γ + \int_{0}^{1} \frac{1 - t^{\frac{z - 1}{2}}}{1 - t} d t)

Ψ (x + 1) = - γ + \int_{0}^{1} \frac{1 - t^{x}}{1 - t} d t \Rightarrow

f^{'} (z) = - \frac{1}{2 (2 + z)} + \frac{1}{2} (Ψ (\frac{z}{2} + 1) - Ψ (\frac{z + 1}{2}))

f (z) = - \frac{1}{2} l n (2 + z) + l o g (\frac{Γ (\frac{z}{2} + 1)}{Γ (\frac{z + 1}{2})}) + c

Γ (z + a) \sim \sqrt{2 π} . z^{z + a - \frac{1}{2}} e^{- z}

\frac{Γ (\frac{z}{2} + 1)}{Γ (\frac{z + 1}{2})} \sim \frac{{(\frac{z}{2})}^{\frac{z}{2} + \frac{1}{2}}}{{(z)}^{\frac{z}{2}}}

f (z) \sim l n (\frac{1}{\sqrt{2 + z}} . \sqrt{\frac{z}{2}}) + c \Rightarrow \lim_{z \to \infty} f (z) = l n (\frac{1}{\sqrt{2}}) + c

b a c k t o f (z) = \int_{0}^{1} (\frac{1}{1 + t} - \frac{t}{2}) \frac{t^{z}}{l n (t)} d t \Rightarrow \lim_{z \to \infty} f (z) = 0

\Rightarrow c = l n (\sqrt{2}) \Leftrightarrow f (z) = \int_{0}^{1} (\frac{1}{1 + t} - \frac{t}{2}) . \frac{t^{z}}{l n (t)} = l n (\frac{1}{\sqrt{2 + z}})

+ l o g (\frac{Γ (\frac{z}{2} + 1)}{Γ (\frac{z + 1}{2})}) + l n (\sqrt{2})

f (0) = ϕ = l n (\frac{1}{\sqrt{2}}) + l n (\sqrt{2}) + l o g (\frac{Γ (1)}{Γ (\frac{1}{2})}) = l o g (\frac{1}{\sqrt{π}})

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