∫_0 ^∞ ((e^(πx) −e^x )/(x(e^(πx) +1)(e^x +1)))dx


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Question Number 101828 by floor(10²Eta[1]) last updated on 05/Jul/20

$\underset{0}{\int^{\infty}} \frac{e^{π x} - e^{x}}{x (e^{π x} + 1) (e^{x} + 1)} dx$
	Answered by maths mind last updated on 05/Jul/20
	$=∫_0 ^(+∞) (1/x){(1/((e^x +1)))−(1/(e^(πx) +1))}dx let f bee lik t≥0 well defind C_∞ f(t)=∫_0 ^(+∞) (1/x){(1/(e^x +1))−(1/(e^(πx) +1))}e^(−xt) dx f′(t)=∫_0 ^(+∞) −(e^(−xt) /(e^x +1))dx+∫_0 ^(+∞) (e^(−xt) /(e^(πx) +1))dx =−∫_0 ^∞ e^(−x(t+1)) Σ_(k≥0) (−e^(−x) )^k dx+∫_0 ^(+∞) e^(−(t+π)) Σ(−e^(−πx) )^k dx =Σ_(k≥0) (−1)^(k+1) .(1/(k+t+1))+Σ_(k≥0) (((−1)^k )/(πk+t+π)) =Σ_(k≥0) ((((−1)^k )/(k+t+π))−(((−1)^k )/(k+t+1))) =Σ_(k≥0) ((1/(2k+t+π))−(1/(2k+t+1)))+Σ_(k≥0) ((1/(2k+2+t))−(1/(2k+t+π+1))) (1/(t+π))−(1/(t+1))+(1/(2+t))−(1/(t+π+1))+Σ_(k≥1) (−(1/(2k))+(1/(2k+t+π))+(1/(2k))−(1/(2k+t+1))) +Σ_(k≥1) ((1/(2k+2+t))−(1/(2k))+(1/(2k))−(1/(2k+t+π+1))) we have H(z)=Σ_(k≥1) ((1/k)−(1/(k+z)))^� ∀z∈C−{Z_− } we get f′(t)=(1/(t+π))−(1/(t+1))+(1/(t+2))−(1/(t+π+1))−(1/2)H(((t+π)/2))+(1/2)H(((t+1)/2)) −(1/2)H(((t+2)/2))+(1/2)H(((t+π+1)/2)) H(z)=Ψ(z+1)+γ we get f′(t)=(1/(t+π))−(1/(t+1))+(1/(t+2))−(1/(t+π+1))+(1/2)(Ψ(((t+3)/2))−Ψ(((t+π+2)/2))+Ψ(((t+π+1)/2))−Ψ(((t+2)/2))) f(t)=ln(((t+π)/(t+π+1)))+ln(((t+2)/(t+1)))+log(((Γ(((t+3)/2)))/(Γ(((t+2)/2)))))+log(((Γ(((t+π+1)/2)))/(Γ(((t+π+2)/2)))))+c wr used∫Ψ(z)dz=log(Γ(z))+c lim_(t→∞) f(t)=0 we need too finde thislim_(x→∞) ((Γ(x+(1/2)))/(Γ(x))) Γ(z+a)∼(√(2πz)).z^(z+a−(1/2)) e^(−z) f(t)→0 f(0)=∫_0 ^(+∞) ((e^(πx) −e^x )/(x(e^x +1)(e^(πx) +1)))dx=ln((π/(π+1)))+ln(2)+log(((Γ((3/2))Γ(((π+1)/2)))/(Γ(1)Γ(((π+2)/2)))))$
	$= \int_{0}^{+ \infty} \frac{1}{x} {\frac{1}{(e^{x} + 1)} - \frac{1}{e^{π x} + 1}} d x$ $l e t f b e e l i k t ⩾ 0 w e l l d e f i n d C_{\infty}$ $f (t) = \int_{0}^{+ \infty} \frac{1}{x} {\frac{1}{e^{x} + 1} - \frac{1}{e^{π x} + 1}} e^{- x t} d x$ $f^{'} (t) = \int_{0}^{+ \infty} - \frac{e^{- x t}}{e^{x} + 1} d x + \int_{0}^{+ \infty} \frac{e^{- x t}}{e^{π x} + 1} d x$ $= - \int_{0}^{\infty} e^{- x (t + 1)} \sum_{k ⩾ 0} {(- e^{- x})}^{k} d x + \int_{0}^{+ \infty} e^{- (t + π)} Σ {(- e^{- π x})}^{k} d x$ $= \sum_{k ⩾ 0} {(- 1)}^{k + 1} . \frac{1}{k + t + 1} + \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{π k + t + π}$ $= \sum_{k ⩾ 0} (\frac{{(- 1)}^{k}}{k + t + π} - \frac{{(- 1)}^{k}}{k + t + 1})$ $= \sum_{k ⩾ 0} (\frac{1}{2 k + t + π} - \frac{1}{2 k + t + 1}) + \sum_{k ⩾ 0} (\frac{1}{2 k + 2 + t} - \frac{1}{2 k + t + π + 1})$ $\frac{1}{t + π} - \frac{1}{t + 1} + \frac{1}{2 + t} - \frac{1}{t + π + 1} + \sum_{k ⩾ 1} (- \frac{1}{2 k} + \frac{1}{2 k + t + π} + \frac{1}{2 k} - \frac{1}{2 k + t + 1})$ $+ \sum_{k ⩾ 1} (\frac{1}{2 k + 2 + t} - \frac{1}{2 k} + \frac{1}{2 k} - \frac{1}{2 k + t + π + 1})$ $Missing \left or extra \right$ $w e g e t f^{'} (t) = \frac{1}{t + π} - \frac{1}{t + 1} + \frac{1}{t + 2} - \frac{1}{t + π + 1} - \frac{1}{2} H (\frac{t + π}{2}) + \frac{1}{2} H (\frac{t + 1}{2})$ $- \frac{1}{2} H (\frac{t + 2}{2}) + \frac{1}{2} H (\frac{t + π + 1}{2})$ $H (z) = Ψ (z + 1) + γ$ $w e g e t$ $f^{'} (t) = \frac{1}{t + π} - \frac{1}{t + 1} + \frac{1}{t + 2} - \frac{1}{t + π + 1} + \frac{1}{2} (Ψ (\frac{t + 3}{2}) - Ψ (\frac{t + π + 2}{2}) + Ψ (\frac{t + π + 1}{2}) - Ψ (\frac{t + 2}{2}))$ $f (t) = l n (\frac{t + π}{t + π + 1}) + l n (\frac{t + 2}{t + 1}) + l o g (\frac{Γ (\frac{t + 3}{2})}{Γ (\frac{t + 2}{2})}) + l o g (\frac{Γ (\frac{t + π + 1}{2})}{Γ (\frac{t + π + 2}{2})}) + c$ $w r u s e d \int Ψ (z) d z = l o g (Γ (z)) + c$ $\lim_{t \to \infty} f (t) = 0$ $w e n e e d t o o f i n d e t h i s \lim_{x \to \infty} \frac{Γ (x + \frac{1}{2})}{Γ (x)}$ $Γ (z + a) \sim \sqrt{2 π z} . z^{z + a - \frac{1}{2}} e^{- z}$ $f (t) \to 0$ $f (0) = \int_{0}^{+ \infty} \frac{e^{π x} - e^{x}}{x (e^{x} + 1) (e^{π x} + 1)} d x = l n (\frac{π}{π + 1}) + l n (2) + l o g (\frac{Γ (\frac{3}{2}) Γ (\frac{π + 1}{2})}{Γ (1) Γ (\frac{π + 2}{2})})$
	Commented by floor(10²Eta[1]) last updated on 05/Jul/20

	$someone help me with the solution :$ $I (t) = \underset{0}{\int^{\infty}} \frac{e^{tx} - e^{x}}{x (e^{tx} + 1) (e^{x} + 1)} dx$ $now i^{'} m gonna derivate and integrate again :$ $I^{'} (t) = \underset{0}{\int^{\infty}} \frac{\partial}{\partial t} (\frac{e^{tx} - e^{x}}{x (e^{tx} + 1) (e^{x} + 1)}) dx$ $= \underset{0}{\int^{\infty}} \frac{e^{tx}}{{(e^{tx} + 1)}^{2}} dx, u = e^{tx} + 1 \Rightarrow \frac{du}{t} = e^{tx} dx$ $= \frac{1}{t} \underset{2}{\int^{\infty}} \frac{du}{u^{2}} = \frac{1}{2 t}$ $\Rightarrow I^{'} (t) = \frac{1}{2 t} \Rightarrow I (t) = \int \frac{1}{2 t} = \frac{1}{2} \ln ∣ t ∣ + C$ $I (1) = 0 = C \Rightarrow I (t) = \frac{1}{2} \ln ∣ t ∣$ $\Rightarrow I (π) = \ln (\sqrt{π})$
	Answered by mathmax by abdo last updated on 05/Jul/20

	$I = \int_{0}^{\infty} \frac{e^{π x} - e^{x}}{x (e^{π x} + 1) (e^{x} + 1)} dx \Rightarrow I = \int_{0}^{\infty} \frac{e^{π x} + 1 - (1 + e^{x})}{x (e^{π x} + 1) (e^{x} + 1)} dx$ $I = \int_{0}^{\infty} \frac{dx}{x (e^{x} + 1)} - \int_{0}^{\infty} \frac{dx}{x (e^{π x} + 1)} = I_{1} - I_{2} we have I_{1} = \int_{0}^{\infty} \frac{e^{- x}}{x (1 + e^{- x})} dx$ $I_{2} = \int_{0}^{\infty} \frac{e^{- π x}}{x (1 + e^{- π x})} dx =_{π x = t} \int_{0}^{\infty} \frac{e^{- t}}{\frac{t}{π} (1 + e^{- t})} (\frac{dt}{π}) = \int_{0}^{\infty} \frac{e^{- t}}{t (1 + e^{- t})} dt = I_{1} \Rightarrow$ $I = 0$
	Commented by maths mind last updated on 05/Jul/20

	$I_{1,} I_{2} d i d n t c v$
	Commented by mathmax by abdo last updated on 05/Jul/20

	$but I_{1} = I_{2} . . .! i dont see the convergence . . .$
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