let f(x)=∫_0 ^∞ e^(−ax) ln(1+e^(−bx) )dx with a>0 and b>0 1) calculate (∂f/∂a)(x) 2) calculate (∂f/∂b)(x) 3)find the value of ∫_0 ^∞ e^(−2x) ln(1+e^(−x) )dx and ∫


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Question Number 41301 by math khazana by abdo last updated on 05/Aug/18

$l e t f (x) = \int_{0}^{\infty} e^{- a x} l n (1 + e^{- b x}) d x w i t h a > 0 a n d$ $b > 0$ $1) c a l c u l a t e \frac{\partial f}{\partial a} (x)$ $2) c a l c u l a t e \frac{\partial f}{\partial b} (x)$ $3) f i n d t h e v a l u e o f \int_{0}^{\infty} e^{- 2 x} l n (1 + e^{- x}) d x a n d$ $\int_{0}^{\infty} e^{- x} l n (1 + e^{- 2 x}) d x .$
Commented byprof Abdo imad last updated on 05/Aug/18

$1) w e h a v e \frac{\partial f}{\partial a} (x) = - \int_{0}^{\infty} {x e}^{- a x} l n (1 + e^{- b x}) d x$ $b u t f o r ∣ u ∣< 1 l n (1 + u) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1} u^{n}}{n} \Rightarrow$ $\frac{\partial f}{\partial a} (x) = - \int_{0}^{\infty} x^{} e^{- a x} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} e^{- n b x}) d x$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} \int_{0}^{\infty} {x e}^{- (a + n b) x} d x b y p a r t s$ $A_{n} = \int_{0}^{\infty} x e^{- (a + n b) x} d x = {[\frac{- x}{a + n b} e^{- (a + n b) x}]}_{0}^{+ \infty}$ $- \int_{0}^{\infty} - \frac{1}{a + n b} e^{- (a + n b) x} d x$ $= \frac{1}{a + n b} \int_{0}^{\infty} e^{- (a + n b) x} d x = - \frac{1}{a + n b} {[\frac{1}{a + n b}]}_{0}^{+ \infty}$ $= \frac{1}{{(a + n b)}^{2}} \Rightarrow \frac{\partial f}{\partial a} (x) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n {(a + n b)}^{2}} . . .$
Commented byprof Abdo imad last updated on 05/Aug/18

$2) w e h a v e \frac{\partial f}{\partial b} (x) = - \int_{0}^{\infty} e^{- a x} \frac{x}{1 + e^{- b x}} d x$ $= - \int_{0}^{\infty} x e^{- a x} (\sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- n b x}) d x$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n + 1} \int_{0}^{\infty} x e^{- (a + n b) x} d x b u t w e h a v e$ $p r o v e d t h a t \int_{0}^{\infty} x e^{- (a + n b) x} d x = \frac{1}{{(a + n b)}^{2}} \Rightarrow$ $\frac{\partial f}{\partial b} (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{{(a + n b)}^{2}} . . .$
Commented byprof Abdo imad last updated on 05/Aug/18
$3)let I = ∫_0 ^∞ e^(−2x) ln(1+e^(−x)) dx I = ∫_0 ^∞ e^(−2x) (Σ_(n=1) ^∞ (((−1)^(n−1) )/n) e^(−nx) )dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) ∫_0 ^∞ e^(−(n+2)x) dx but ∫_0 ^∞ e^(−(n+2)x) dx=[((−1)/(n+2)) e^(−(n+2)x) ]_0 ^(+∞) =(1/(n+2)) ⇒ I = Σ_(n=1) ^∞ (((−1)^(n−1) )/(n(n+2))) =(1/2)Σ_(n=1) ^∞ (−1)^(n−1) {(1/n) −(1/(n+2))} =(1/2) Σ_(n=1) ^∞ (((−1)^(n−1) )/n) −(1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/(n+2)) but Σ_(n=1) ^∞ (((−1)^(n−1) )/n) =ln(2) Σ_(n=1) ^∞ (((−1)^(n−1) )/(n+2)) =Σ_(n=3) ^∞ (((−1)^(n−3) )/n) =Σ_(n=3) ^∞ (((−1)^(n−1) )/n) =ln(2)−{1 −(1/2)}=ln(2)−(1/2) ⇒ I =(1/2)ln(2)−(1/2){ln(2)−(1/2)} ⇒I =(1/4)$
$3) l e t I = \int_{0}^{\infty} e^{- 2 x} l n (1 + e^{- x)} d x$ $I = \int_{0}^{\infty} e^{- 2 x} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} e^{- n x}) d x$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{\infty} e^{- (n + 2) x} d x b u t$ $\int_{0}^{\infty} e^{- (n + 2) x} d x = {[\frac{- 1}{n + 2} e^{- (n + 2) x}]}_{0}^{+ \infty} = \frac{1}{n + 2} \Rightarrow$ $I = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n (n + 2)} = \frac{1}{2} \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} {\frac{1}{n} - \frac{1}{n + 2}}$ $= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} - \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n + 2} b u t$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = l n (2)$ $\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n + 2} = \sum_{n = 3}^{\infty} \frac{{(- 1)}^{n - 3}}{n}$ $= \sum_{n = 3}^{\infty} \frac{{(- 1)}^{n - 1}}{n} = l n (2) - {1 - \frac{1}{2}} = l n (2) - \frac{1}{2}$ $\Rightarrow I = \frac{1}{2} l n (2) - \frac{1}{2} {l n (2) - \frac{1}{2}} \Rightarrow I = \frac{1}{4}$
Commented byprof Abdo imad last updated on 05/Aug/18
$let J = ∫_0 ^∞ e^(−x) ln(1+e^(−2x) )dx we have J = ∫_0 ^∞ e^(−x) (Σ_(n=1) ^∞ (((−1)^(n−1) )/n) e^(−2nx) )dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) ∫_0 ^∞ e^(−(2n+1)x) dx =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)[((−1)/(2n+1)) e^(−(2n+1)x) ]_0 ^(+∞) =Σ_(n=1) ^∞ (((−1)^(n−1) )/(n(2n+1))) ⇒ (1/2)J =Σ_(n=1) ^∞ (−1)^(n−1) { (1/(2n))−(1/(2n+1))} =(1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/n) +Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) =((ln(2))/2) +(π/4) −1 ⇒J =ln(2)+(π/2) −2 .$
$l e t J = \int_{0}^{\infty} e^{- x} l n (1 + e^{- 2 x}) d x w e h a v e$ $J = \int_{0}^{\infty} e^{- x} (\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} e^{- 2 n x}) d x$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{\infty} e^{- (2 n + 1) x} d x$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} {[\frac{- 1}{2 n + 1} e^{- (2 n + 1) x}]}_{0}^{+ \infty}$ $= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n (2 n + 1)} \Rightarrow$ $\frac{1}{2} J = \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} {\frac{1}{2 n} - \frac{1}{2 n + 1}}$ $= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} + \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1}$ $= \frac{l n (2)}{2} + \frac{π}{4} - 1 \Rightarrow J = l n (2) + \frac{π}{2} - 2 .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18
	$f(a,b)=∫_0 ^∞ e^(−ax) ln(1+e^(−bx) )dx after intregation the results will be in terms of a and b so we can write df=((∂f/∂a))_b da +((∂f/∂b))_a db (df/da)=∫_0 ^∞ ln(1+e^(−bx) ).(∂/∂a)(e^(−ax) ) dx (df/da)=∫_0 ^∞ ln(1+e^(−bx) ).(−ae^(−ax) )dx (df/da)=−af (df/f)=−ada lnf=−(a^2 /2)+c_1 f(a,b) =∫_0 ^∞ ln(1+e^(−bx) )e^(−ax) dx (df/db)=∫_0 ^∞ e^(−ax) .(∂/∂b){ln(1+e^(−bx) ) dx (df/db)=∫_0 ^∞ e^(−ax) .((−be^(−bx) )/(1+e^(−bx) )) dx (df/db)=∫_0 ^∞ (−b)(e^(−ax−bx) /(e^(bx) +1)).(e^(bx) /)dx (df/db)=(−b)∫_0 ^∞ (e^(−ax) /(1+e^(bx) )) dx let t=1+e^(bx) dt=e^(bx) .b.dx dx=(dt/(b(t−1))) e^(bx) =t−1 e^x =(t−1)^(1/b) so e^(−ax) =(t−1)^((−a)/b) (df/db)=(−b)∫_2 ^∞ (((t−1)^((−a)/b) )/t)×(dt/(b(t−1))) (df/db)=(−1)∫_2 ^∞ (((t−1)^(((−a)/b)−1) )/t) contd...$
	$f (a, b) = \int_{0}^{\infty} e^{- a x} l n (1 + e^{- b x}) d x$ $a f t e r i n t r e g a t i o n t h e r e s u l t s w i l l b e i n t e r m s o f a a n d b$ $s o w e c a n w r i t e$ $d f = {(\frac{\partial f}{\partial a})}_{b} d a + {(\frac{\partial f}{\partial b})}_{a} d b$ $\frac{d f}{d a} = \int_{0}^{\infty} l n (1 + e^{- b x}) . \frac{\partial}{\partial a} (e^{- a x}) d x$ $\frac{d f}{d a} = \int_{0}^{\infty} l n (1 + e^{- b x}) . (- {a e}^{- a x}) d x$ $\frac{d f}{d a} = - a f$ $\frac{d f}{f} = - a d a$ $l n f = - \frac{a^{2}}{2} + c_{1}$ $f (a, b) = \int_{0}^{\infty} l n (1 + e^{- b x}) e^{- a x} d x$ $\frac{d f}{d b} = \int_{0}^{\infty} e^{- a x} . \frac{\partial}{\partial b} {l n (1 + e^{- b x}) d x$ $\frac{d f}{d b} = \int_{0}^{\infty} e^{- a x} . \frac{- {b e}^{- b x}}{1 + e^{- b x}} d x$ $\frac{d f}{d b} = \int_{0}^{\infty} (- b) \frac{e^{- a x - b x}}{e^{b x} + 1} . \frac{e^{b x}}{} d x$ $\frac{d f}{d b} = (- b) \int_{0}^{\infty} \frac{e^{- a x}}{1 + e^{b x}} d x$ $l e t t = 1 + e^{b x} d t = e^{b x} . b . d x$ $d x = \frac{d t}{b (t - 1)}$ $e^{b x} = t - 1 e^{x} = {(t - 1)}^{\frac{1}{b}} s o e^{- a x} = {(t - 1)}^{\frac{- a}{b}}$ $\frac{d f}{d b} = (- b) \int_{2}^{\infty} \frac{{(t - 1)}^{\frac{- a}{b}}}{t} \times \frac{d t}{b (t - 1)}$ $\frac{d f}{d b} = (- 1) \int_{2}^{\infty} \frac{{(t - 1)}^{\frac{- a}{b} - 1}}{t}$ $c o n t d . . .$
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