Compute lim_(A→+∞) {(1/A)∫_1 ^A A^(1/x) dx}. (IMC 2015)


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Question Number 7501 by Yozzia last updated on 01/Sep/16
$Compute lim_(A→+∞) {(1/A)∫_1 ^A A^(1/x) dx}. (IMC 2015)$
$C o m p u t e \lim_{A \to + \infty} {\frac{1}{A} \int_{1}^{A} A^{\frac{1}{x}} d x} .$ $(I M C 2015)$
	Answered by FilupSmith last updated on 01/Sep/16
	$L =lim_(A→∞) {(1/A)∫_1 ^A A^(1/x) dx} unsure if math is correct. L =lim_(A→∞) {(1/A)∫_1 ^A e^((1/x)ln(A)) dx} u=e^((1/x)ln(A)) ⇒ du=−x^(−2) ln(A)e^(x^(−1) ln(A)) dx x=((ln(A))/(ln(u))) du=−((ln(u))/x)udx du=−((ln^2 (u))/(ln(A)))udx dx=−((ln(A))/(uln^2 (u)))du L =lim_(A→∞) {(1/A)∫_(x=1) ^( x=A) u(−((ln(A))/(uln^2 (u)))du)} L =lim_(A→∞) {−(1/A)∫_(x=1) ^( x=A) u((ln(A))/(uln^2 (u)))du} L =lim_(A→∞) {−((ln(A))/A)∫_(x=1) ^( x=A) (1/(ln^2 (u)))du} limit of ((ln(A))/A)=(1/A) as A→∞ = 0 0×constant=0 continue$
	$L = \lim_{A \to \infty} {\frac{1}{A} \int_{1}^{A} A^{\frac{1}{x}} d x}$ $u n s u r e i f m a t h i s c o r r e c t .$ $L = \lim_{A \to \infty} {\frac{1}{A} \int_{1}^{A} e^{\frac{1}{x} \ln (A)} d x}$ $u = e^{\frac{1}{x} \ln (A)} \Rightarrow d u = - x^{- 2} \ln (A) e^{x^{- 1} \ln (A)} d x$ $x = \frac{\ln (A)}{\ln (u)}$ $d u = - \frac{\ln (u)}{x} u d x$ $d u = - \frac{\ln^{2} (u)}{\ln (A)} u d x$ $d x = - \frac{\ln (A)}{u \ln^{2} (u)} d u$ $L = \lim_{A \to \infty} {\frac{1}{A} \int_{x = 1}^{x = A} u (- \frac{\ln (A)}{u \ln^{2} (u)} d u)}$ $L = \lim_{A \to \infty} {- \frac{1}{A} \int_{x = 1}^{x = A} u \frac{\ln (A)}{u \ln^{2} (u)} d u}$ $L = \lim_{A \to \infty} {- \frac{\ln (A)}{A} \int_{x = 1}^{x = A} \frac{1}{\ln^{2} (u)} d u}$ $l i m i t o f \frac{\ln (A)}{A} = \frac{1}{A} a s A \to \infty = 0$ $0 \times c o n s t a n t = 0$ $c o n t i n u e$
	Commented by FilupSmith last updated on 01/Sep/16
	$∫(1/(ln^2 (u)))du t=ln^2 (u) u=e^t^(1/2) dt=2ln(u)(1/u)du (u/(2ln(u)))dt=du (e^t^(1/2) /(2t^(1/2) ))dt=du L =lim_(A→∞) {−((ln(A))/A)∫_(x=1) ^( x=A) (1/(ln^2 (u)))du} L =lim_(A→∞) {−((ln(A))/A)∫_(x=1) ^( x=A) (e^t^(1/2) /(2t^(3/2) ))dt} correct???$
	$\int \frac{1}{\ln^{2} (u)} d u$ $t = \ln^{2} (u)$ $u = e^{t^{1 / 2}}$ $d t = 2 \ln (u) \frac{1}{u} d u$ $\frac{u}{2 \ln (u)} d t = d u$ $\frac{e^{t^{1 / 2}}}{2 t^{1 / 2}} d t = d u$ $L = \lim_{A \to \infty} {- \frac{\ln (A)}{A} \int_{x = 1}^{x = A} \frac{1}{\ln^{2} (u)} d u}$ $L = \lim_{A \to \infty} {- \frac{\ln (A)}{A} \int_{x = 1}^{x = A} \frac{e^{t^{1 / 2}}}{2 t^{3 / 2}} d t}$ $c o r r e c t ? ? ?$
	Commented by Yozzia last updated on 01/Sep/16

	$S o f a r i t s e e m s o k . I c a n a c c e s s t h e$ $s o l u t i o n o n l i n e, b u t I {d o n}^{'} t k n o w i t .$ ${I t}^{'} s a q u e s t i o n f r o m t h e 2015 I n t e r n a t i o n a l$ $M a t h e m a t i c s C o m p e t i t i o n f o r U n i v e r s i t i e s .$ $J u s t r e c e n t l y I g o t w i n d o f t h i s c o m p e t i t i o n$ $a n d n o w I h a v e a c c e s s t o a v a r i e t y o f$ $i n t r i g u i n g p r o b l e m s w h i c h I^{'} l l h o p e f u l l y$ $b e a b l e t o s o l v e a s I p r o g r e s s i n m y$ $u n d e r g r a d u a t e s t u d i e s . I^{'} m h o p i n g$ $t o s h a r e s o m e o f t h o s e p r o b l e m s h e r e$ $w i t h t h e f o r u m!$
	Commented by FilupSmith last updated on 02/Sep/16

	$That is awsome . I have no formal education .$ $I am completely self taught and you have$ $taught me lot!$
	Commented by Alimurtaza last updated on 05/Jan/17


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