let f(x)=∫_0 ^∞ (t^(a−1) /(x+t^n )) dt with 0<a<1 and x>0 and n≥2 1) determine a explicit form of f(x) 2) calculate g(x) =∫_0 ^∞ (t^(a−1) /((x+t^n )^2 )) dt 3) find f^((k)) (x) at form of integrals 4) calculate ∫_0 ^∞ (t^(a−1) /(9+t^2 )) dt and ∫_0 ^∞ (t^(a−1) /((9+t^2 )^2 )) 5) calculate U_n =∫_0 ^∞ (t^((1/n)−1) /(2^n +t^n )) dt and study the convergence of Σ U


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Question Number 63664 by mathmax by abdo last updated on 07/Jul/19

$l e t f (x) = \int_{0}^{\infty} \frac{t^{a - 1}}{x + t^{n}} d t w i t h 0 < a < 1 a n d x > 0 a n d n ⩾ 2$ $1) d e t e r m i n e a e x p l i c i t f o r m o f f (x)$ $2) c a l c u l a t e g (x) = \int_{0}^{\infty} \frac{t^{a - 1}}{{(x + t^{n})}^{2}} d t$ $3) f i n d f^{(k)} (x) a t f o r m o f i n t e g r a l s$ $4) c a l c u l a t e \int_{0}^{\infty} \frac{t^{a - 1}}{9 + t^{2}} d t a n d \int_{0}^{\infty} \frac{t^{a - 1}}{{(9 + t^{2})}^{2}}$ $5) c a l c u l a t e U_{n} = \int_{0}^{\infty} \frac{t^{\frac{1}{n} - 1}}{2^{n} + t^{n}} d t a n d s t u d y t h e c o n v e r g e n c e o f Σ U_{n}$
Commented bymathmax by abdo last updated on 09/Jul/19

$1) w e h a v e f (x) = \int_{0}^{\infty} \frac{t^{a - 1}}{x + t^{n}} d t c h a n g e m e n t t =^{n} \sqrt{x} u g i v e$ $f (x) = \int_{0}^{\infty} \frac{{(^{n} \sqrt{x} u)}^{a - 1}}{x + {x u}^{n}}^{n} \sqrt{x} d u$ $= \frac{x^{\frac{a - 1}{n} + \frac{1}{n}}}{x} \int_{0}^{\infty} \frac{u^{a - 1}}{1 + u^{n}} d u = x^{\frac{a}{n} - 1} \int_{0}^{\infty} \frac{u^{a - 1}}{1 + u^{n}}$ $=_{u = α^{\frac{1}{n}}} x^{\frac{a}{n} - 1} \int_{0}^{\infty} \frac{α^{\frac{a - 1}{n}}}{1 + α} \frac{1}{n} α^{\frac{1}{n} - 1} d α$ $= x^{\frac{a}{n} - 1} \int_{0}^{\infty} \frac{α^{\frac{a - 1}{n} + \frac{1}{n} - 1}}{1 + α} d α = x^{\frac{a}{n} - 1} \int_{0}^{\infty} \frac{α^{\frac{a}{n} - 1}}{1 + α} d α \Rightarrow$ $f (x) = x^{\frac{a}{n} - 1} \frac{π}{s i n (\frac{π a}{n})}$
Commented bymathmax by abdo last updated on 09/Jul/19

$2) w e h a v e b y d e r i v a t i o n f^{^{'}} (x) = \int_{0}^{\infty} \frac{\partial}{\partial x} (\frac{t^{a - 1}}{x + t^{n}}) d t$ $= - \int_{0}^{\infty} \frac{t^{a - 1}}{{(x + t^{n})}^{2}} d t = - g (x) \Rightarrow g (x) = - f^{^{'}} (x)$ $f (x) = \frac{π}{s i n (\frac{π a}{n})} x^{\frac{a}{n} - 1} \Rightarrow f^{^{'}} (x) = \frac{π (\frac{a}{n} - 1)}{s i n (\frac{π a}{n})} x^{\frac{a}{n} - 2} \Rightarrow$ $g (x) = \int_{0}^{\infty} \frac{t^{a - 1}}{{(x + t^{n})}^{2}} d t = \frac{π (1 - \frac{a}{n})}{s i n (\frac{π a}{n})} x^{\frac{a}{n} - 2}$
Commented bymathmax by abdo last updated on 09/Jul/19

$3) w e h a v e f (x) = \int_{0}^{\infty} \frac{t^{a - 1}}{x + t^{n}} d t \Rightarrow f^{(k)} (x) = \int_{0}^{\infty} \frac{\partial^{k}}{\partial x^{k}} (\frac{t^{a - 1}}{x + t^{n}}) d t$ $= \int_{0}^{\infty} \frac{{(- 1)}^{k} k!}{{(x + t^{n})}^{k + 1}} t^{a - 1} d t \Rightarrow f^{(k)} (x) = {(- 1)}^{k} k! \int_{0}^{\infty} \frac{t^{a - 1}}{{(x + t^{n})}^{k + 1}} d t$
Commented bymathmax by abdo last updated on 09/Jul/19

$4) w e h a v e p r o v e d t h a t \int_{0}^{\infty} \frac{t^{a - 1}}{x + t^{n}} d t = \frac{π}{s i n (\frac{π a}{n})} x^{\frac{a}{n} - 1} a n d$ $\int_{0}^{\infty} \frac{t^{a - 1}}{{(x + t^{n})}^{2}} d t = \frac{π (1 - \frac{a}{n})}{s i n (\frac{π a}{n})} x^{\frac{a}{n} - 2}$ $x = 9 a n d n = 2 \Rightarrow \int_{0}^{\infty} \frac{t^{a - 1}}{9 + t^{2}} d t = \frac{π}{s i n (\frac{π a}{2})} \times 9^{\frac{a}{2} - 1}$ $\int_{0}^{\infty} \frac{t^{a - 1}}{{(9 + t^{2})}^{2}} d t = \frac{π (1 - \frac{a}{2})}{s i n (\frac{π a}{2})} \times 9^{\frac{a}{2} - 2}$
Commented bymathmax by abdo last updated on 09/Jul/19

$5) w e h a v e \int_{0}^{\infty} \frac{t^{a - 1}}{x + t^{n}} d t = \frac{π}{s i n (\frac{π a}{n})} x^{\frac{a}{n} - 1}$ $a = \frac{1}{n} a n d x = 2^{n} \Rightarrow \int_{0}^{\infty} \frac{t^{\frac{1}{n} - 1}}{2^{n} + t^{n}} d t = \frac{π}{s i n (\frac{π}{n^{2}})} {(2^{n})}^{\frac{1}{n^{2}} - 1}$ $= \frac{π}{s i n (\frac{π}{n^{2}})} \times 2^{\frac{1}{n} - n} = \frac{π}{2^{n - \frac{1}{n}} s i n (\frac{π}{n^{2}})} .$
Commented bymathmax by abdo last updated on 09/Jul/19

$\Rightarrow U_{n} = \frac{π}{2^{n - \frac{1}{n}} s i n (\frac{π}{n^{2}})}$
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