Question-159552

Question Number 159552 by Ar Brandon last updated on 18/Nov/21

Commented by Ar Brandon last updated on 18/Nov/21

Prove the above results

Commented by mindispower last updated on 19/Nov/21

b o n j o u r t e s E t u d i a n t ?

Commented by Ar Brandon last updated on 19/Nov/21

Oui monsieur . \overset{´}{E t u d i a n t} en 1^{er} ann \overset{´}{e e} IUT .

Commented by puissant last updated on 20/Nov/21

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Answered by mindispower last updated on 19/Nov/21

(1)

l e t f (b) = \int_{0}^{\infty} \frac{c o s (a x) l n (x^{2} + {b z}^{2})}{x^{2} + z^{2}}, b ⩾ 0

f^{'} (b) = \int_{0}^{\infty} \frac{z^{2} c o s (a x)}{(x^{2} + z^{2}) (x^{2} + {b z}^{2})} d x

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

A = \frac{z^{2}}{2} \int_{- \infty}^{\infty} \frac{e^{i a x}}{(x^{2} + z^{2}) (x^{2} + {b z}^{2})} d x =

= z^{2} . i π . \frac{e^{- a z}}{2 i z} . \frac{1}{(1 - b) z^{2}} + i π . \frac{e^{- a z \sqrt{b}}}{z^{2} (1 - b) . (2 i z \sqrt{b})}

= \frac{π}{2 z} (\frac{e^{- a z}}{(1 - b)} + \frac{e^{- a z \sqrt{b}}}{(1 - b) \sqrt{b}})

f (b) = \frac{π}{2 z} \int \frac{e^{- a z}}{1 - b} + \frac{e^{- a z \sqrt{b}}}{(1 - b) \sqrt{b}} d b

= \frac{π}{2 z} (- l n (1 - b) e^{- a z} + \int \frac{e^{- a z \sqrt{b}}}{(1 - b) \sqrt{b}} d b ∣_{\sqrt{b} = u_{}}^{}

\int \frac{e^{- a u z}}{(1 - u^{2})} . 2 d u

= \int \frac{e^{- a u z}}{1 - u} + \frac{e^{- a u z}}{1 + u} d u

= - \int \frac{e^{- a z (1 - x)}}{x} d x + \int \frac{e^{- a z (y - 1)}}{y} d y

= e^{- a z} . - \int \frac{e^{a z x}}{a z x} d a z x + e^{a z} \int \frac{e^{- a z y}}{a z y} d (a z y)

= e^{- a z} - E_{1} (- a z x) + e^{a z} . E_{1} (a z y)

= - e^{- a z} E_{1} (- a z (1 - \sqrt{b})) + e^{a z} (a z (1 + \sqrt{b}))

- e^{a z} . E_{i} (- a z (1 + \sqrt{b}) + e^{- a z} E_{i} (a z (1 - \sqrt{b}))

w e g e t

\frac{π}{2 z} (e^{- a z} (E_{i} (a z (1 - \sqrt{b}) - l n (1 - b)) - e^{a z} E_{i} (- a z (1 + \sqrt{b}))) + C

f (0) = \int_{0}^{\infty} \frac{c o s (a x) l n (x^{2})}{x^{2} + z^{2}} d x = 2 \int_{0}^{\infty} \frac{c o s (a x) l n (x)}{x^{2} + z^{2}} d x ∣_{z \neq 0}

A = \int_{- \infty}^{0} \frac{c o s (a x) l n (x)}{x^{2} + z^{2}} + \int_{0}^{\infty} \frac{c o s (a x) l n (x)}{x^{2} + z^{2}} = R e (2 i π . \frac{e^{- a z}}{2 i z} l n (i z))

= R e π \frac{e^{- a z}}{z} (l n (z) + \frac{i π}{2}) = π \frac{e^{- a z}}{z} l n (z)

l n (- x) = l n (x) + i π, \forall x > 0

A 2 \int_{0}^{a n a \infty d} \frac{c o s (a x) l n (x)}{x^{2} + z^{2}} d x + i π \int_{0}^{\infty} \frac{c o s (a x)}{x^{2} + z^{2}} x g o o d

\int_{0}^{\infty} \frac{c o s (a x) l n (x^{2})}{x^{2} + z^{2}} = f (0) = \frac{π}{z} e^{- a z} l n (z), \forall z, R e (z) > 0

C = \frac{π}{z} e^{- a z} l n (z) - \frac{π}{2 z} e^{- a z} E_{i} (a z) + \frac{π}{2 z} e^{a z} E_{i} (- a z)

E_{i} (x) = γ + l n (x) + o (x), x \to 1 t o o b e e c o n t i n u e d

n o t ? m a n y t i m e s n o w

i t h i n k i t s a g o o d p a t h E_{i} \dots r e l a t e d t o c h i . .

a n d S h i

Commented by Ar Brandon last updated on 19/Nov/21

Belle d \overset{´}{e m o n s t r a t i o n}, monsieur!