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Question Number 39368 by maxmathsup by imad last updated on 05/Jul/18

l e t F (t) = \int_{0}^{+ \infty} \frac{s i n x}{x (1 + x^{2})} e^{- t x (1 + x^{2})} d x w i t h t ⩾ 0

1) c a c u l a t e \frac{d F}{d t} (t)

2) f i n d a s i m p l e f o r m o f F (t)

3) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{s i n x}{x (1 + x^{2}) d x} .

Commented by math khazana by abdo last updated on 08/Jul/18

1) w e h a v e \frac{d F}{d t} (t) = - \int_{0}^{+ \infty} s i n x e^{- t x - {t x}^{3}} d x

2) l e t I = \int_{0}^{\infty} \frac{s i n x}{x (1 + x^{2})} d x

2 I = \int_{- \infty}^{+ \infty} \frac{s i n x}{x (1 + x^{2})} d x = I m (\int_{- \infty}^{+ \infty} \frac{e^{i x}}{x (1 + x^{2})} d x)

l e t φ (z) = \frac{e^{i z}}{z (1 + z^{2})} t h e p o l e s o f φ a r e 0, i, - i

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, 0) + R e s (φ, i)}

R e s (φ, 0) = {l i m}_{z \to 0} z φ (z) = 1

R e s (φ, i) = {l i m}_{z \to i} (z - i) φ (z) = \frac{e^{- 1}}{i (2 i)} = - \frac{e^{- 1}}{2}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {1 - \frac{e^{- 1}}{2}} \Rightarrow

2 I = I m (\int_{- \infty}^{+ \infty} φ (z) d z) = 2 π {1 - \frac{e^{- 1}}{2}}

\Rightarrow I = π {1 - \frac{e^{- 1}}{2}} .