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Question Number 206003 by mnjuly1970 last updated on 04/Apr/24

	Answered by Berbere last updated on 04/Apr/24

	$Ω = \int_{0}^{\infty} \frac{e^{- x}}{x^{2}} (1 - c o s (x)) d x$ $Ω (a) = \int_{0}^{\infty} \frac{e^{- x} (c o s (a x) - c o s (x))}{x^{2}} d x; Ω (1) = 0$ $Ω^{'} (a) = \int_{0}^{\infty} \frac{- e^{- x} s i n (a x)}{x} d x$ $Ω^{'} (0) = 0; Ω^{″} (a) = - \int_{0}^{\infty} e^{- x} c o s (a x) d x$ $= - R e \int_{0}^{\infty} e^{- x} e^{i a x} d x = \frac{1}{1 + a^{2}}$ $Ω^{'} (a) = \tan^{- 1} (a)$ $Ω (a) = \int_{1}^{a} \tan^{- 1} (x) d x = a \tan^{- 1} (a) - \frac{1}{2} l n (\frac{1 + a^{2}}{2}) - \frac{π}{4}$ $Ω = Ω (0) = \frac{1}{2} l n (2) - \frac{π}{4}$
	Answered by mathzup last updated on 05/Apr/24
	$I=_(by parts) [−(1/x)e^(−x) (1−cosx)]_0 ^(+∞) −∫_0 ^∞ (−(1/x)){−e^(−x) (1−cosx)+e^(−x) sinx}dx =0+∫_0 ^∞ (1/x)(e^(−x) (sinx+cosx−1))dx =∫_0 ^∞ (e^(−x) /x)(sinx+cosx−1)dx f(λ)=∫_0 ^∞ (e^(−λx) /x)(sinx+cosx−1)dx f^′ (λ)=−∫_0 ^∞ e^(−λx) (sinx+cosx−1)dx =∫_0 ^∞ e^(−λx) dx−∫_0 ^∞ e^(−λx) (cosx+sinx)dx but ∫_0 ^∞ e^(−λx) dx=[−(1/λ)e^(−λx) ]_0 ^∞ =(1/λ) ∫_0 ^∞ e^(−λx) (cosx +sinx)dx =Re(∫_0 ^∞ e^(−λx+ix) dx)+Im(∫_0 ^∞ e^(−λx+ix) dx) ∫_0 ^∞ e^((−λ+i)x) dx=[(1/(−λ+i))e^((−λ+i)x) ]_0 ^∞ =−(1/(−λ+i))=(1/(λ−i))=((λ+i)/(1+λ^2 )) ⇒ R(...)=(λ/(1+λ^2 )) and Im(...)=(1/(1+λ^2 )) ⇒f^′ (λ)=(1/λ)−(λ/(1+λ^2 ))−(1/(1+λ^2 )) ⇒ f(λ)=ln(λ)−(1/2)ln(λ^2 +1)−arctanλ +c lim_(λ→+∞) =0=−(π/2)+c ⇒c=(π/2) ⇒f(λ)=ln((λ/( (√(1+λ^2 )))))−arctanλ +(π/2) I=f(1)=ln((1/( (√2))))−(π/4)+(π/2) =−(1/2)ln(2)+(π/4) ⇒ I=(π/4)−(1/2)ln(2)$
	$I =_{b y p a r t s} {[- \frac{1}{x} e^{- x} (1 - c o s x)]}_{0}^{+ \infty}$ $- \int_{0}^{\infty} (- \frac{1}{x}) {- e^{- x} (1 - c o s x) + e^{- x} s i n x} d x$ $= 0 + \int_{0}^{\infty} \frac{1}{x} (e^{- x} (s i n x + c o s x - 1)) d x$ $= \int_{0}^{\infty} \frac{e^{- x}}{x} (s i n x + c o s x - 1) d x$ $f (λ) = \int_{0}^{\infty} \frac{e^{- λ x}}{x} (s i n x + c o s x - 1) d x$ $f^{^{'}} (λ) = - \int_{0}^{\infty} e^{- λ x} (s i n x + c o s x - 1) d x$ $= \int_{0}^{\infty} e^{- λ x} d x - \int_{0}^{\infty} e^{- λ x} (c o s x + s i n x) d x$ $b u t \int_{0}^{\infty} e^{- λ x} d x = {[- \frac{1}{λ} e^{- λ x}]}_{0}^{\infty} = \frac{1}{λ}$ $\int_{0}^{\infty} e^{- λ x} (c o s x + s i n x) d x$ $= R e (\int_{0}^{\infty} e^{- λ x + i x} d x) + I m (\int_{0}^{\infty} e^{- λ x + i x} d x)$ $\int_{0}^{\infty} e^{(- λ + i) x} d x = {[\frac{1}{- λ + i} e^{(- λ + i) x}]}_{0}^{\infty}$ $= - \frac{1}{- λ + i} = \frac{1}{λ - i} = \frac{λ + i}{1 + λ^{2}} \Rightarrow$ $R (. . .) = \frac{λ}{1 + λ^{2}} a n d I m (. . .) = \frac{1}{1 + λ^{2}}$ $\Rightarrow f^{^{'}} (λ) = \frac{1}{λ} - \frac{λ}{1 + λ^{2}} - \frac{1}{1 + λ^{2}} \Rightarrow$ $f (λ) = l n (λ) - \frac{1}{2} l n (λ^{2} + 1) - a r c t a n λ + c$ ${l i m}_{λ \to + \infty} = 0 = - \frac{π}{2} + c \Rightarrow c = \frac{π}{2}$ $\Rightarrow f (λ) = l n (\frac{λ}{\sqrt{1 + λ^{2}}}) - a r c t a n λ + \frac{π}{2}$ $I = f (1) = l n (\frac{1}{\sqrt{2}}) - \frac{π}{4} + \frac{π}{2}$ $= - \frac{1}{2} l n (2) + \frac{π}{4} \Rightarrow$ $I = \frac{π}{4} - \frac{1}{2} l n (2)$
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