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Question Number 62826 by Cheyboy last updated on 25/Jun/19

Commented by mathmax by abdo last updated on 25/Jun/19
$b) I =∫_0 ^π ((sinθ)/(√(π−θ))) changement(√(π−θ)) =x give I =∫_(√π) ^0 ((sin(π−x^2 ))/x) (−2x)dx = 2 ∫_0 ^(√π) sin(x^2 )dx the function x→sin(x^2 ) is continous on[0,(√π)] so the integral converges. 4b) i think the Q .here is find lim_(n→∞) (1/n){ f((1/n))+f((2/n))+....f((n/n))} =lim_(n→+∞) ((1−o)/n) Σ_(k=0) ^n f(0 +((k(1−o))/n)) =∫_0 ^1 f(x)dx .$
$b) I = \int_{0}^{π} \frac{s i n θ}{\sqrt{π - θ}} c h a n g e m e n t \sqrt{π - θ} = x g i v e I = \int_{\sqrt{π}}^{0} \frac{s i n (π - x^{2})}{x} (- 2 x) d x$ $= 2 \int_{0}^{\sqrt{π}} s i n (x^{2}) d x t h e f u n c t i o n x \to s i n (x^{2}) i s c o n t i n o u s o n [0, \sqrt{π}] s o t h e$ $i n t e g r a l c o n v e r g e s .$ $4 b) i t h i n k t h e Q . h e r e i s f i n d {l i m}_{n \to \infty} \frac{1}{n} {f (\frac{1}{n}) + f (\frac{2}{n}) + . . . . f (\frac{n}{n})}$ $= {l i m}_{n \to + \infty} \frac{1 - o}{n} \sum_{k = 0}^{n} f (0 + \frac{k (1 - o)}{n}) = \int_{0}^{1} f (x) d x .$
Commented by mathmax by abdo last updated on 25/Jun/19

$5) l e t A_{p} = \int_{2}^{\infty} \frac{d x}{x {(l n x)}^{p}} c h a n g e m e n t l n (x) = t g i v e x = e^{t}$ $A_{p} = \int_{l n (2)}^{+ \infty} \frac{e^{t} d t}{e^{t} t^{p}} = \int_{l n (2)}^{+ \infty} \frac{d t}{t^{p}} a n d A_{p} c o n v e r g e s \Leftrightarrow p > 1 .$
Commented by Cheyboy last updated on 26/Jun/19

$G o d b l e s s y o u S i r$
Commented by mathmax by abdo last updated on 26/Jun/19

$y o u a r e w e l c o m e s i r .$
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