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Question Number 38453 by maxmathsup by imad last updated on 25/Jun/18

f i n d f (x) = \int_{0}^{\infty} \frac{1 - c o s (x t)}{t} e^{- x t} d t w i t h x > 0

1) f i n d a s i m p l e f o r m o f f (x)

2) c a l c u l a t e \int_{0}^{\infty} \frac{1 - c o s (π t)}{t} e^{- t} d t

3) c a l c u l a t e \int_{0}^{\infty} \frac{1 - c o s (3 t)}{t} e^{- 2 t} d t

Commented by math khazana by abdo last updated on 26/Jun/18

t h e Q i s f i n d f (x) = \int_{0}^{\infty} \frac{1 - c o s (a t)}{t} e^{- x t} d t

Commented by math khazana by abdo last updated on 29/Jun/18

1) w e h a v e f (x) = \int_{0}^{\infty} \frac{1 - c o s (a t)}{t} e^{- x t} d t \Rightarrow

f^{^{'}} (x) = - \int_{0}^{\infty} (1 - c o s (a t)) e^{- x t} d t

= \int_{0}^{\infty} c o s (a t) e^{- x t} d t - \int_{0}^{\infty} e^{- x t} d t b u t

\int_{0}^{\infty} e^{- x t} d t = {[- \frac{1}{x} e^{- x t}]}_{0}^{+ \infty} = \frac{1}{x}

\int_{0}^{\infty} c o s (a t) e^{- x t} d t = R e (\int_{0}^{\infty} e^{i a t - x t} d t)

\int_{0}^{\infty} e^{(- x + i a) t} d t = {[\frac{1}{- x + i a} e^{(- x + i a) t}]}_{0}^{+ \infty}

= \frac{- 1}{- x + i a} = \frac{1}{x - i a} = \frac{x + i a}{x^{2} + a^{2}} \Rightarrow

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

f^{^{'}} (x) = \frac{x}{x^{2} + a^{2}} - \frac{1}{x} \Rightarrow

f (x) = \frac{1}{2} l n (x^{2} + a^{2}) - l n (x) + c b u t \exists m > 0 <

∣ \int_{0}^{\infty} \frac{1 - c o s (a t)}{t} e^{- x t} d t ∣ ⩽ m \int_{0}^{\infty} e^{- x t} d t = \frac{m}{x} \to 0

w h e n x \to + \infty a l s o w e h a v e

f (x) = \frac{1}{2} l n (\frac{x^{2} + a^{2}}{x^{2}}) + c \Rightarrow

c = {l i m}_{x \to + \infty} f (x) - \frac{1}{2} l n (\frac{x^{2} + a^{2}}{x^{2}}) = 0 \Rightarrow

f (x) = \frac{1}{2} l n (x^{2} + a^{2}) - l n (x) w i t h x > 0

Commented by math khazana by abdo last updated on 29/Jun/18

2) w e h a v e p r o v e d t h a t

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

\int_{0}^{\infty} \frac{1 - c o s (π t)}{t} e^{- t} = \frac{1}{2} l n (1 + π^{2}) .

Commented by math khazana by abdo last updated on 29/Jun/18

3) ? \int_{0}^{\infty} \frac{1 - c o s (3 t)}{t} e^{- 2 t} d t = \frac{1}{2} l n (2^{2} + 3^{2}) - l n (2)

= \frac{1}{2} l n (13) - l n (2) .