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Question Number 108786 by 150505R last updated on 19/Aug/20

	Answered by mathmax by abdo last updated on 19/Aug/20
	$first we study the convergence A_n =∫_0 ^∞ x^(n−1 ) cos(ax)dx A_n =∫_0 ^1 x^(n−1) cos(ax)dx +∫_1 ^(+∞) x^(n−1) cos(ax)dx ∣x^(n−1) cos(ax)∣≤x^(n−1) and ∫_0 ^1 (dx/x^(1−n) ) converges ⇔1−n<1 ⇒n>0 letξ>0 lim_(x→+∞) x^ξ x^(n−1) cos(ax) =0 ⇒lim_(x→+∞) x^(ξ+n−1) cos(ax) =0 ⇒ξ +n−1<0 ∀ξ>0 ⇒n<1−ξ ∀ξ>0 ⇒0<n<1 (so n is not natural) A_n =Re(∫_0 ^∞ x^(n−1) e^(−iax) dx) and ∫_0 ^∞ x^(n−1) e^(−iax) dx =_(iax =t) ∫_0 ^∞ ((t/(ia)))^(n−1) e^(−t) (dt/(ia)) =(1/((ia)^n )) ∫_0 ^∞ t^(n−1) e^(−t) dt =(1/a^n )e^(−i((nπ)/2)) Γ(n) =((Γ(n))/a^n ){cos(((nπ)/2))−isin(((nπ)/2))} ⇒ A_n =((Γ(n))/a^n )cos(((nπ)/2)) with n ∈]0,1[$
	$first we study the convergence A_{n} = \int_{0}^{\infty} x^{n - 1} \cos (ax) dx$ $A_{n} = \int_{0}^{1} x^{n - 1} \cos (ax) dx + \int_{1}^{+ \infty} x^{n - 1} \cos (ax) dx$ $∣ x^{n - 1} \cos (ax) ∣⩽ x^{n - 1} and \int_{0}^{1} \frac{dx}{x^{1 - n}} converges \Leftrightarrow 1 - n < 1 \Rightarrow n > 0$ $let ξ > 0 \lim_{x \to + \infty} x^{ξ} x^{n - 1} \cos (ax) = 0 \Rightarrow \lim_{x \to + \infty} x^{ξ + n - 1} \cos (ax) = 0$ $\Rightarrow ξ + n - 1 < 0 \forall ξ > 0 \Rightarrow n < 1 - ξ \forall ξ > 0 \Rightarrow 0 < n < 1$ $(so n is not natural)$ $A_{n} = Re (\int_{0}^{\infty} x^{n - 1} e^{- iax} dx) and \int_{0}^{\infty} x^{n - 1} e^{- iax} dx$ $=_{iax = t} \int_{0}^{\infty} {(\frac{t}{ia})}^{n - 1} e^{- t} \frac{dt}{ia} = \frac{1}{{(ia)}^{n}} \int_{0}^{\infty} t^{n - 1} e^{- t} dt$ $= \frac{1}{a^{n}} e^{- i \frac{n π}{2}} Γ (n) = \frac{Γ (n)}{a^{n}} {\cos (\frac{n π}{2}) - isin (\frac{n π}{2})} \Rightarrow$ $A_{n} = \frac{Γ (n)}{a^{n}} \cos (\frac{n π}{2}) with n \in] 0, 1 [$
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