Math Question and Answers


Question and Answers Forum

All Questions Topic List
Differential Equation Questions
Previous in All Question Next in All Question
Previous in Differential Equation Next in Differential Equation
Question Number 83565 by Jidda28 last updated on 03/Mar/20

Commented by abdomathmax last updated on 04/Mar/20
$I=∫_(−∞) ^(+∞) (e^(2x) /(e^(3x) +1))dx changement e^x =t give I =∫_0 ^(+∞) (t^2 /(t^3 +1))×(dt/t) =∫_0 ^∞ ((tdt)/(t^3 +1)) let decompose F(t)=(t/(t^3 +1)) ⇒F(t)=(t/((t+1)(t^2 −t+1))) =(a/(t+1)) +((bt +c)/(t^2 −t +1)) a=(t+1)F(t)∣_(t=−1) =((−1)/3) lim_(t→+∞) tF(t)=0=a+b ⇒b=(1/3) F(0)=0=a+c ⇒c=(1/3)⇒F(t)=((−1)/(3(t+1))) +(1/3)((t+1)/(t^2 −t+1)) ⇒∫ F(t)dt =−(1/3)∫ (dt/(t+1)) +(1/6)∫ ((2t+2)/(t^2 −t+1))dt =−(1/3)∫ (dt/(t+1)) +(1/6)∫ ((2t−1)/(t^2 −t+1))dt +(1/2)∫ (dt/(t^2 −t+1)) =−(1/3)ln∣t+1∣+(1/6)ln(t^2 −t+1) +(1/2)∫ (dt/((t−(1/2))^2 +(3/4))) (t−(1/2)=((√3)/2)u) =ln((((t^2 −t+1)^(1/6) )/(∣t+1∣^(1/3) )))+(1/2)×(4/3)∫ (1/(u^2 +1))×((√3)/2)du =ln(....)+(1/(√3)) arctan(((2t−1)/(√3))) ⇒ I =∫_0 ^∞ F(t)dt =[ln((((t^2 −t+1)^(1/6) )/((t+1)^(1/3) )))]_0 ^(+∞) +(1/(√3))[arctan(((2t−1)/(√3)))]_0 ^(+∞) =(1/(√3)){(π/2) +arctan((1/(√3)))} =(1/(√3)){ (π/2)+(π/6)} =(1/(√3)){((4π)/6)} =((2π)/(3(√3)))$
$I = \int_{- \infty}^{+ \infty} \frac{e^{2 x}}{e^{3 x} + 1} d x c h a n g e m e n t e^{x} = t g i v e$ $I = \int_{0}^{+ \infty} \frac{t^{2}}{t^{3} + 1} \times \frac{d t}{t} = \int_{0}^{\infty} \frac{t d t}{t^{3} + 1}$ $l e t d e c o m p o s e F (t) = \frac{t}{t^{3} + 1} \Rightarrow F (t) = \frac{t}{(t + 1) (t^{2} - t + 1)}$ $= \frac{a}{t + 1} + \frac{b t + c}{t^{2} - t + 1}$ $a = (t + 1) F (t) ∣_{t = - 1} = \frac{- 1}{3}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + b \Rightarrow b = \frac{1}{3}$ $F (0) = 0 = a + c \Rightarrow c = \frac{1}{3} \Rightarrow F (t) = \frac{- 1}{3 (t + 1)} + \frac{1}{3} \frac{t + 1}{t^{2} - t + 1}$ $\Rightarrow \int F (t) d t = - \frac{1}{3} \int \frac{d t}{t + 1} + \frac{1}{6} \int \frac{2 t + 2}{t^{2} - t + 1} d t$ $= - \frac{1}{3} \int \frac{d t}{t + 1} + \frac{1}{6} \int \frac{2 t - 1}{t^{2} - t + 1} d t + \frac{1}{2} \int \frac{d t}{t^{2} - t + 1}$ $= - \frac{1}{3} l n ∣ t + 1 ∣ + \frac{1}{6} l n (t^{2} - t + 1)$ $+ \frac{1}{2} \int \frac{d t}{{(t - \frac{1}{2})}^{2} + \frac{3}{4}} (t - \frac{1}{2} = \frac{\sqrt{3}}{2} u)$ $= l n (\frac{{(t^{2} - t + 1)}^{\frac{1}{6}}}{∣ t + 1 ∣^{\frac{1}{3}}}) + \frac{1}{2} \times \frac{4}{3} \int \frac{1}{u^{2} + 1} \times \frac{\sqrt{3}}{2} d u$ $= l n (. . . .) + \frac{1}{\sqrt{3}} a r c t a n (\frac{2 t - 1}{\sqrt{3}}) \Rightarrow$ $I = \int_{0}^{\infty} F (t) d t = {[l n (\frac{{(t^{2} - t + 1)}^{\frac{1}{6}}}{{(t + 1)}^{\frac{1}{3}}})]}_{0}^{+ \infty}$ $+ \frac{1}{\sqrt{3}} {[a r c t a n (\frac{2 t - 1}{\sqrt{3}})]}_{0}^{+ \infty}$ $= \frac{1}{\sqrt{3}} {\frac{π}{2} + a r c t a n (\frac{1}{\sqrt{3}})} = \frac{1}{\sqrt{3}} {\frac{π}{2} + \frac{π}{6}}$ $= \frac{1}{\sqrt{3}} {\frac{4 π}{6}} = \frac{2 π}{3 \sqrt{3}}$
Commented by abdomathmax last updated on 04/Mar/20

$\Rightarrow I = \frac{2 π \sqrt{3}}{9} (e r r o r i n t h e Q! . . .)$
Commented by Jidda28 last updated on 04/Mar/20

$t h a n k y o u s i r$
Commented by msup trace by abdo last updated on 04/Mar/20

$y o u a r e w e l c o m e$
	Answered by mind is power last updated on 03/Mar/20
	$i found just this near to gamma Ψ(x)=((Γ′(x))/(Γ(x))) and ∫_0 ^(+∞) e^(−ax) dx=(1/a)∫_0 ^(+∞) x^(1−1) e^(−x) =((Γ(1))/a),lign (5→6)bellow =∫_0 ^(+∞) (e^(−2x) /(1+e^(−3x) ))dx=∫_0 ^(+∞) (e^x /(1+e^(3x) ))dx ⇒∫_(−∞) ^(+∞) (e^(2x) /(1+e^(3x) ))dx=∫_0 ^(+∞) (e^x /(1+e^(3x) ))dx+∫_0 ^(+∞) (e^(2x) /(1+e^(3x) )) =∫_0 ^(+∞) (e^(−2x) /(e^(−3x) +1))dx+∫_0 ^(+∞) (e^(−x) /(1+e^(−3x) ))dx =∫_0 ^(+∞) e^(−2x) {Σ_(k≥0) (−e^(−3x) )^k )dx+∫_0 ^(+∞) e^(−x) (Σ_(k≥0) (−e^(−3x) )^k )dx =Σ_(k≥0) (−1)^k ∫_0 ^(+∞) (e^(−(3k+2)x) +e^(−(3k+1)x) )dx =Σ_(k≥0) (((−1)^k )/(3k+2))+Σ_(k≥0) (((−1)^k )/(3k+1)) =Σ_(k≥0) ((1/(6k+2))−(1/(6k+5)))+Σ_(k≥0) ((1/(6k+1))−(1/(6k+4))) =Σ_(k≥0) ((3/((6k+2)(6k+5))))+Σ_(k≥0) ((3/((6k+1)(6k+4)))) =(1/(12))Σ_(k≥0) ((1/((k+(2/6))(k+(5/6)))))+(1/(12))Σ_(k≥0) (1/((k+(1/6))(k+(4/6)))) =(1/(12))((Ψ((5/6))−Ψ((2/6)))/((5−2)/6))+(1/(12))((Ψ((4/6))−Ψ((1/6)))/((4−1)/6)) =(1/6)[Ψ((5/6))−Ψ((1/6))+Ψ((4/6))−Ψ((2/6))] =(1/6)[−(π/2)cot(((5π)/6))+(π/2)cot((π/6))−(π/2)cot(((4π)/6))+(π/2)cot(((2π)/6))] =(1/6)[πcot((π/6))+πcot((π/3))] =(π/6)[(1/(√3))+(√3)]=(π/6).((4/(√3)))=((2π)/(3(√3)))=((2π(√3))/9) 2 nd way =∫_0 ^(+∞) (e^x /(e^(3x) +1))dx+∫_0 ^(+∞) (e^(2x) /(1+e^(3x) ))dx=∫_0 ^(+∞) ((e^x (1+e^x )dx)/((e^x +1)(e^(2x) −e^x +1))) =∫_0 ^(+∞) (e^x /(e^(2x) −e^x +1))dx =∫_1 ^(+∞) (du/(u^2 −u+1))=∫_1 ^(+∞) (du/((u−(1/2))^2 +(3/4))) =(2/(√3))[arctan(((2u)/(√3))−(1/(√3)))]_1 ^(+∞) (2/(√3))[(π/2)−(π/6)]=((4π)/(6(√3)))=((2π(√3))/9)$
	$i f o u n d j u s t t h i s n e a r t o g a m m a Ψ (x) = \frac{Γ^{'} (x)}{Γ (x)}$ $a n d \int_{0}^{+ \infty} e^{- a x} d x = \frac{1}{a} \int_{0}^{+ \infty} x^{1 - 1} e^{- x} = \frac{Γ (1)}{a}, l i g n (5 \to 6) b e l l o w$ $= \int_{0}^{+ \infty} \frac{e^{- 2 x}}{1 + e^{- 3 x}} d x = \int_{0}^{+ \infty} \frac{e^{x}}{1 + e^{3 x}} d x$ $\Rightarrow \int_{- \infty}^{+ \infty} \frac{e^{2 x}}{1 + e^{3 x}} d x = \int_{0}^{+ \infty} \frac{e^{x}}{1 + e^{3 x}} d x + \int_{0}^{+ \infty} \frac{e^{2 x}}{1 + e^{3 x}}$ $= \int_{0}^{+ \infty} \frac{e^{- 2 x}}{e^{- 3 x} + 1} d x + \int_{0}^{+ \infty} \frac{e^{- x}}{1 + e^{- 3 x}} d x$ $= \int_{0}^{+ \infty} e^{- 2 x} {\sum_{k ⩾ 0} {(- e^{- 3 x})}^{k}) d x + \int_{0}^{+ \infty} e^{- x} (\sum_{k ⩾ 0} {(- e^{- 3 x})}^{k}) d x$ $= \sum_{k ⩾ 0} {(- 1)}^{k} \int_{0}^{+ \infty} (e^{- (3 k + 2) x} + e^{- (3 k + 1) x}) d x$ $= \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{3 k + 2} + \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{3 k + 1}$ $= \sum_{k ⩾ 0} (\frac{1}{6 k + 2} - \frac{1}{6 k + 5}) + \sum_{k ⩾ 0} (\frac{1}{6 k + 1} - \frac{1}{6 k + 4})$ $= \sum_{k ⩾ 0} (\frac{3}{(6 k + 2) (6 k + 5)}) + \sum_{k ⩾ 0} (\frac{3}{(6 k + 1) (6 k + 4)})$ $= \frac{1}{12} \sum_{k ⩾ 0} (\frac{1}{(k + \frac{2}{6}) (k + \frac{5}{6})}) + \frac{1}{12} \sum_{k ⩾ 0} \frac{1}{(k + \frac{1}{6}) (k + \frac{4}{6})}$ $= \frac{1}{12} \frac{Ψ (\frac{5}{6}) - Ψ (\frac{2}{6})}{\frac{5 - 2}{6}} + \frac{1}{12} \frac{Ψ (\frac{4}{6}) - Ψ (\frac{1}{6})}{\frac{4 - 1}{6}}$ $= \frac{1}{6} [Ψ (\frac{5}{6}) - Ψ (\frac{1}{6}) + Ψ (\frac{4}{6}) - Ψ (\frac{2}{6})]$ $= \frac{1}{6} [- \frac{π}{2} c o t (\frac{5 π}{6}) + \frac{π}{2} c o t (\frac{π}{6}) - \frac{π}{2} c o t (\frac{4 π}{6}) + \frac{π}{2} c o t (\frac{2 π}{6})]$ $= \frac{1}{6} [π c o t (\frac{π}{6}) + π c o t (\frac{π}{3})]$ $= \frac{π}{6} [\frac{1}{\sqrt{3}} + \sqrt{3}] = \frac{π}{6} . (\frac{4}{\sqrt{3}}) = \frac{2 π}{3 \sqrt{3}} = \frac{2 π \sqrt{3}}{9}$ $2 n d w a y$ $= \int_{0}^{+ \infty} \frac{e^{x}}{e^{3 x} + 1} d x + \int_{0}^{+ \infty} \frac{e^{2 x}}{1 + e^{3 x}} d x = \int_{0}^{+ \infty} \frac{e^{x} (1 + e^{x}) d x}{(e^{x} + 1) (e^{2 x} - e^{x} + 1)}$ $= \int_{0}^{+ \infty} \frac{e^{x}}{e^{2 x} - e^{x} + 1} d x$ $= \int_{1}^{+ \infty} \frac{d u}{u^{2} - u + 1} = \int_{1}^{+ \infty} \frac{d u}{{(u - \frac{1}{2})}^{2} + \frac{3}{4}}$ $= \frac{2}{\sqrt{3}} {[a r c t a n (\frac{2 u}{\sqrt{3}} - \frac{1}{\sqrt{3}})]}_{1}^{+ \infty}$ $\frac{2}{\sqrt{3}} [\frac{π}{2} - \frac{π}{6}] = \frac{4 π}{6 \sqrt{3}} = \frac{2 π \sqrt{3}}{9}$
	Commented by Jidda28 last updated on 04/Mar/20

	$t h a n k y o u s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum