montrer-que-o-oo-sin-2-t-t-2-e-xt-dt-est-continu-sur-R-

Question Number 171608 by SANOGO last updated on 18/Jun/22

m o n t r e r q u e

\int_{o}^{+ o o} \frac{{s i n}^{2} t}{t^{2}} e^{- x t} d t

e s t c o n t i n u s u r R^{+}

Answered by aleks041103 last updated on 18/Jun/22

I (x) = \int_{0}^{\infty} \frac{{s i n}^{2} t}{t^{2}} e^{- x t} d t

I^{″} = \int_{0}^{\infty} \frac{{s i n}^{2} t}{t^{2}} {(- t)}^{2} e^{- x t} d t =

= \int_{0}^{\infty} \frac{1 - c o s (2 t)}{2} e^{- x t} d t =

= \frac{1}{2} \int_{0}^{\infty} e^{- x t} d t - \frac{1}{4} \int_{0}^{\infty} c o s (u) e^{- \frac{x}{2} u} d u

= \frac{1}{2 x} (1) - \frac{1}{4} \int_{0}^{\infty} R e (e^{(- \frac{x}{2} + i) u}) d u

\int_{0}^{\infty} e^{s u} d u = \frac{1}{s} ({\underset{u \to \infty}{l i m e}}^{s u} - 1)

\Rightarrow \int_{0}^{\infty} R e (e^{(- \frac{x}{2} + i) u}) d u = R e (\frac{1}{- \frac{x}{2} + i} (0 - 1)) =

= R e (\frac{1}{\frac{x}{2} - i}) = R e (\frac{\frac{x}{2} + i}{{(\frac{x}{2})}^{2} + 1}) = \frac{2 x}{x^{2} + 4}

\Rightarrow I^{″} (x) = \frac{1}{2 x} - \frac{2 x}{4 (x^{2} + 4)}

\Rightarrow I^{'} (x) = \frac{1}{2} l n (x) - \frac{1}{4} l n (x^{2} + 4) + c_{1}

\dots