determine L(((1−cosx)/x^2 ))


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Question Number 99460 by mathmax by abdo last updated on 21/Jun/20

$determine L (\frac{1 - cosx}{x^{2}})$
	Answered by mathmax by abdo last updated on 23/Jun/20
	$L(((1−cosx)/x^2 )) =∫_0 ^∞ ((1−cost)/t^2 )e^(−xt) dt =f(x) ⇒ f^′ (x) =−∫_0 ^∞ (((1−cost)e^(−xt) )/t)dt ⇒f^((2)) (x) =∫_0 ^∞ (1−cost)e^(−xt) dt =∫_0 ^∞ e^(−xt) dt −∫_0 ^∞ e^(−xt) cost dt but ∫_0 ^∞ e^(−xt) dt =[−(1/x)e^(−xt) ]_0 ^∞ =(1/x) and ∫_0 ^∞ e^(−xt) cost dt =Re(∫_0 ^∞ e^(−xt+it) dt) =Re(∫_0 ^∞ e^((−x+i)t) dt) ∫_0 ^∞ e^((−x+i)t) dt =[(1/(−x+i))e^((−x+i)t) ]_0 ^∞ =−(1/(−x+i)) =(1/(x−i)) =((x+i)/(x^2 +1)) ⇒ Re(...) =(x/(x^2 +1)) ⇒f^((2)) (x) =(1/x)−(x/(x^2 +1)) ⇒ f^′ (x) =lnx −(1/2)ln(x^2 +1) +k ⇒ f(x) =∫ ln(x)dx −(1/2)∫ ln(x^2 +1)dx+kx +c we have ∫ ln(x)dx =xln(x)−x ∫ ln(x^2 +1)dx =xln(x^2 +1)−∫ x×((2x)/(1+x^2 ))dx =xln(x^2 +1)−2 ∫ ((x^2 +1−1)/(1+x^2 ))dx =xln(1+x^2 )−2x +2arctanx ⇒ f(x) =xlnx−x−(x/2)ln(1+x^2 )+x −arctanx +kx +c c =lim_(x→0) f(x) =∫_0 ^∞ ((1−cost)/t^2 ) =2 ∫_0 ^∞ ((sin^2 t)/t^2 )dt =2{ [−((sin^2 t)/t)]_0 ^∞ +∫_0 ^∞ ((2sintcost)/t) dt} =2 ∫_0 ^∞ ((sin(2t))/t) dt =_(2t=u) 2∫_0 ^∞ ((sinu)/(u/2))×(du/2) =2∫_0 ^∞ ((sinu)/u)du =π ⇒c =π rest to find k ....be continued....$
	$L (\frac{1 - cosx}{x^{2}}) = \int_{0}^{\infty} \frac{1 - cost}{t^{2}} e^{- xt} dt = f (x) \Rightarrow$ $f^{^{'}} (x) = - \int_{0}^{\infty} \frac{(1 - cost) e^{- xt}}{t} dt \Rightarrow f^{(2)} (x) = \int_{0}^{\infty} (1 - cost) e^{- xt} dt$ $= \int_{0}^{\infty} e^{- xt} dt - \int_{0}^{\infty} e^{- xt} cost dt but \int_{0}^{\infty} e^{- xt} dt = {[- \frac{1}{x} e^{- xt}]}_{0}^{\infty}$ $= \frac{1}{x} and \int_{0}^{\infty} e^{- xt} cost dt = Re (\int_{0}^{\infty} e^{- xt + it} dt) = Re (\int_{0}^{\infty} e^{(- x + i) t} dt)$ $\int_{0}^{\infty} e^{(- x + i) t} dt = {[\frac{1}{- x + i} e^{(- x + i) t}]}_{0}^{\infty} = - \frac{1}{- x + i} = \frac{1}{x - i} = \frac{x + i}{x^{2} + 1} \Rightarrow$ $Re (. . .) = \frac{x}{x^{2} + 1} \Rightarrow f^{(2)} (x) = \frac{1}{x} - \frac{x}{x^{2} + 1} \Rightarrow$ $f^{^{'}} (x) = lnx - \frac{1}{2} \ln (x^{2} + 1) + k \Rightarrow$ $f (x) = \int \ln (x) dx - \frac{1}{2} \int \ln (x^{2} + 1) dx + kx + c we have$ $\int \ln (x) dx = xln (x) - x$ $\int \ln (x^{2} + 1) dx = xln (x^{2} + 1) - \int x \times \frac{2 x}{1 + x^{2}} dx = xln (x^{2} + 1) - 2 \int \frac{x^{2} + 1 - 1}{1 + x^{2}} dx$ $= xln (1 + x^{2}) - 2 x + 2 arctanx \Rightarrow$ $f (x) = xlnx - x - \frac{x}{2} \ln (1 + x^{2}) + x - arctanx + kx + c$ $c = \lim_{x \to 0} f (x) = \int_{0}^{\infty} \frac{1 - cost}{t^{2}} = 2 \int_{0}^{\infty} \frac{\sin^{2} t}{t^{2}} dt$ $= 2 {{[- \frac{\sin^{2} t}{t}]}_{0}^{\infty} + \int_{0}^{\infty} \frac{2 sintcost}{t} dt} = 2 \int_{0}^{\infty} \frac{\sin (2 t)}{t} dt =_{2 t = u} 2 \int_{0}^{\infty} \frac{sinu}{\frac{u}{2}} \times \frac{du}{2}$ $= 2 \int_{0}^{\infty} \frac{sinu}{u} du = π \Rightarrow c = π rest to find k . . . . be continued . . . .$
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