let-f-x-0-tsin-xt-x-2-t-2-2-dt-x-gt-0-calculate-f-x-

Question Number 131050 by mathmax by abdo last updated on 31/Jan/21

let f (x) = \int_{0}^{\infty} \frac{tsin (xt)}{{(x^{2} + t^{2})}^{2}} dt (x > 0)

calculate f^{^{'}} (x)

Answered by mathmax by abdo last updated on 02/Feb/21

changement t = xz give f (x) = \int_{0}^{\infty} \frac{xzsin (x^{2} z)}{x^{4} {(z^{2} + 1)}^{2}} xdz

= \frac{1}{x^{2}} \int_{0}^{\infty} \frac{zsin (x^{2} z)}{{(z^{2} + 1)}^{2}} dz \Rightarrow f (x) = \frac{1}{{2 x}^{2}} \int_{- \infty}^{+ \infty} \frac{zsin (x^{2} z)}{{(z^{2} + 1)}^{2}} dz = \frac{1}{{2 x}^{2}} Im (\int_{- \infty}^{+ \infty} \frac{{ze}^{{ix}^{2} z}}{{(z^{2} + 1)}^{2}} dz)

\Rightarrow φ (z) = \frac{z e^{{ix}^{2} z}}{{(z^{2} + 1)}^{2}} \Rightarrow φ (z) = \frac{{ze}^{{ix}^{2} z}}{{(z - i)}^{2} {(z + i)}^{2}} residus theorem give

\int_{R} φ (z) dz = 2 i π Res (φ, i)

Res (φ, i) = \lim_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}

= \lim_{z \to i} {\frac{{ze}^{{ix}^{2} z}}{{(z + i)}^{2}}}^{(1)} = \lim_{z \to i} \frac{(e^{{ix}^{2} z} + {ix}^{2} z e^{{ix}^{2} z}) {(z + i)}^{2} - 2 (z + i) {ze}^{{ix}^{2} z}}{{(z + i)}^{4}}

= \lim_{z \to i} \frac{e^{{ix}^{2} z} (1 + {ix}^{2} z) (z + i) - {2 ze}^{{ix}^{2} z}}{{(z + i)}^{3}}

= \frac{e^{- x^{2}} (1 - x^{2}) (2 i) - 2 i e^{- x^{2}}}{{(2 i)}^{3}} = \frac{2 i {(1 - x^{2}) e^{- x^{2}} - e^{- x^{2}}}}{- 8 i}

= - \frac{1}{4} (- x^{2}) e^{- x^{2}} = \frac{x^{2}}{4} e^{- x^{2}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) dz = 2 i π \times \frac{x^{2}}{4} e^{- x^{2}}

= \frac{i π x^{2}}{2} e^{- x^{2}} \Rightarrow f (x) = \frac{1}{{2 x}^{2}} . \frac{π}{2} x^{2} e^{- x^{2}} = \frac{π}{4} e^{- x^{2}} \Rightarrow

f^{^{'}} (x) = \frac{π}{4} (- 2 x) e^{- x^{2}} = - \frac{π x}{2} e^{- x^{2}}