Limit of a Definite Integral 4

Calculus Level 5

lim ⁡ n → ∞ ∫ e π ( x − e ) n ( π − x ) n d x n = ? \large{\lim_{n \to \infty} \sqrt[n]{\int_e^\pi (x-e)^n (\pi - x)^n \, dx} = \ ?}

( π − e ) 4 \dfrac{(\pi - e)}{4} ( π − e ) 2 \dfrac{(\pi - e)}{2} 0 \large{0} ( π − e ) \large{(\pi - e)} ( π − e ) 2 2 \dfrac{(\pi - e)^2}{2} ( π − e ) 4 4 \dfrac{(\pi - e)^4}{4} ( π − e ) 2 4 \dfrac{(\pi - e)^2}{4} ( π − e ) 4 2 \dfrac{(\pi - e)^4}{2}

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Satyajit Mohanty by Satyajit Mohanty , 24, India

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1 solution

Ayush Verma
Jul 26, 2015

B is beta function.

I = ∫ e π ( x − e ) n ( π − x ) n d x l e t t = x − e ⇒ d x = d t , π − x = ( π − e − t ) ⇒ I = ∫ 0 π − e t n ( π − e − t ) n d t = ∫ 0 π − e ( π − e ) n ( t π − e ) n ( π − e ) n ( 1 − t π − e ) n d t L e t t π − e = y ⇒ d t = ( π − e ) d y ∴ I = ∫ 0 1 ( π − e ) n y n ( π − e ) n ( 1 − y ) n ( π − e ) d y = ( π − e ) 2 n + 1 ∫ 0 1 y n + 1 − 1 ( 1 − y ) n + 1 − 1 d y = ( π − e ) 2 n + 1 B ( n + 1 , n + 1 ) = ( π − e ) 2 n + 1 ( n ! ) 2 ( 2 n + 1 ) ! ⇒ L = l i m n → ∞ { ( π − e ) 2 n + 1 ( n ! ) 2 ( 2 n + 1 ) ! } 1 n = l i m n → ∞ ( π − e ) 2 + 1 n { ( n ! ) 2 ( 2 n + 1 ) ! } 1 n = ( π − e ) 2 4 I=\int _{ e }^{ \pi }{ { \left( x-e \right) }^{ n }{ \left( \pi -x \right) }^{ n } } dx\\ \\ let\quad t=x-e\Rightarrow dx=dt,\pi -x=\left( \pi -e-t \right) \\ \\ \Rightarrow I=\int _{ 0 }^{ \pi -e }{ { t }^{ n }{ \left( \pi -e-t \right) }^{ n } } dt\\ \\ =\int _{ 0 }^{ \pi -e }{ { { \left( \pi -e \right) }^{ n }{ \left( \cfrac { t }{ \pi -e } \right) }^{ n } }{ { \left( \pi -e \right) }^{ n }\left( 1-\cfrac { t }{ \pi -e } \right) }^{ n } } dt\\ \\ Let\quad \cfrac { t }{ \pi -e } =y\Rightarrow dt=\left( \pi -e \right) dy\\ \\ \therefore \quad I=\int _{ 0 }^{ 1 }{ { { \left( \pi -e \right) }^{ n }{ y }^{ n } }{ { \left( \pi -e \right) }^{ n }\left( 1-y \right) }^{ n } } \left( \pi -e \right) dy\\ \\ ={ \left( \pi -e \right) }^{ 2n+1 }\int _{ 0 }^{ 1 }{ { y }^{ n+1-1 } } { \left( 1-y \right) }^{ n+1-1 }dy\\ \\ ={ \left( \pi -e \right) }^{ 2n+1 }B\left( n+1,n+1 \right) \\ \\ ={ \left( \pi -e \right) }^{ 2n+1 }\cfrac { { \left( n! \right) }^{ 2 } }{ \left( 2n+1 \right) ! } \\ \\ \Rightarrow L=\underset { n\rightarrow \infty }{ lim } { \left\{ { \left( \pi -e \right) }^{ 2n+1 }\cfrac { { \left( n! \right) }^{ 2 } }{ \left( 2n+1 \right) ! } \right\} }^{ \cfrac { 1 }{ n } }\\ \\ =\underset { n\rightarrow \infty }{ lim } { \left( \pi -e \right) }^{ 2+\cfrac { 1 }{ n } }{ \left\{ \cfrac { { \left( n! \right) }^{ 2 } }{ \left( 2n+1 \right) ! } \right\} }^{ \cfrac { 1 }{ n } }\\ \\ =\cfrac { { \left( \pi -e \right) }^{ 2 } }{ 4 }

6 Helpful 0 Interesting 1 Brilliant 0 Confused

Hmmmmm could someone correct me where I went wrong.

lim ⁡ n → ∞ [ ∫ e π ( x − e ) n ( π − x ) n d x ] 1 n \lim _{ n\rightarrow \infty }{ { \left[ \int _{ e }^{ \pi }{ {(x-e)}^{n}{(\pi -x)}^{n}dx } \right] }^{ \frac { 1 }{ n } } }

Setting u = x-e makes du = dx yielding:

lim ⁡ n → ∞ [ ∫ 0 α ( u ) n ( α − u ) n d u ] 1 n ∣ α = π − e \lim _{ n\rightarrow \infty }{ { \left[ \int _{ 0 }^{ \alpha }{ {(u)}^{n}{(\alpha -u)}^{n}du } \right] }^{ \frac { 1 }{ n } } } |\alpha =\pi -e

Setting the integral equal to S n {S}_{n} and repeatedly applying integration by parts, we get:

S n = u n + 1 ( α − u ) n n + 1 + u n + 2 ( α − u ) n − 1 n + 2 + u n + 3 ( α − u ) n − 2 n + 3 + u n + 4 ( α − u ) n − 3 n + 4 . . . + ∫ 0 α u n + n ( α − u ) n − n n + n d u { S }_{ n }=\frac { { u }^{ n+1 }{ (\alpha -u) }^{ n } }{ n+1 } +\frac { { u }^{ n+2 }{ (\alpha -u) }^{ n-1 } }{ n+2 } +\frac { { u }^{ n+3 }{ (\alpha -u) }^{ n-2 } }{ n+3 } +\frac { { u }^{ n+4 }{ (\alpha -u) }^{ n-3 } }{ n+4 } ...+\int _{ 0 }^{ \alpha }{ \frac { { u }^{ n+n }{ (\alpha -u) }^{ n-n } }{ n+n } } du

Each of the terms outside of the integral, when evaluated, yield 0. Thus we are left with:

∫ 0 α u n + n ( α − u ) n − n n + n d u = ∫ 0 α u 2 n 2 n d u \int _{ 0 }^{ \alpha }{ \frac { { u }^{ n+n }{ (\alpha -u) }^{ n-n } }{ n+n } } du = \int _{ 0 }^{ \alpha }{ \frac { { u }^{ 2n } }{ 2n } } du

After evaluation, we are left with the following limit:

lim ⁡ n → ∞ u 2 n + 1 4 n 2 + 2 n 1 n \lim _{ n\rightarrow \infty }{ { \frac { { u }^{ 2n+1 } }{ 4{ n }^{ 2 }+2n } ^{ \frac { 1 }{ n } } } }

And it is at this point that my answer then disagrees with the answer given, as the numerator agrees with that as given in the answer, but the denominator disagrees. Where have I made an error?

Seth Lovelace - 5 years, 4 months ago

Same method but answer not matching

Akash aggrawal - 4 years, 11 months ago

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