determine-L-1-cosx-x-2-

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 (\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 (\dots) = \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 \dots . be continued \dots .