Show: $lim_{nrightarrow infty} left|int_{1}^{e}left[ln(x)right]^n:dx right|= 0 $
$begingroup$
This is probably an unremarkable result, but I was curious about exploring the following family of definite integrals to see what asymptotic behaviour it as as $n rightarrow infty$
$$I_n = int_{1}^{e}left[ln(x)right]^n:dx mbox{ for } n in mathbb{N}$$
The method I took:
Employ integration by parts:
begin{align}
v'(x) &= 1 & u(x) &= left[ln(x)right]^n \
v(x) &= x & u'(x) &= nleft[ln(x)right]^{n - 1}frac{1}{x}
end{align}
Thus,
begin{align}
I_n &= int_{1}^{e}left[ln(x)right]^n:dx = x cdot left[ln(x)right]^n Big|_{1}^{e} - int_{1}^{e} x cdot nleft[ln(x)right]^{n - 1}frac{1}{x} \
&= e - nint_{1}^{e}left[ln(x)right]^{n - 1}:dx
end{align}
Thus,
begin{equation}
I_n = e - nI_{n - 1}
end{equation}
Where
begin{equation}
I_1 = int_{1}^{e} ln(x):dx = xleft(ln(x) - 1right)Big|_{1}^{e} = 1
end{equation}
This yields the solution:
begin{align}
I_n &= n!left[(-1)^{n - 1} + esum_{i = 0}^{n - 2} frac{(-1)^i}{(n - i)!} right] = n!left(-1right)^nleft[-1 + esum_{i = 2}^{n } frac{(-1)^i}{i!} right] \
&= n!left(-1right)^nleft[-1 + esum_{i = 0}^{n } frac{(-1)^i}{i!} right]
end{align}
We see that:
begin{equation}
lim_{nrightarrow infty} I_n = n! (-1)^n left( -1 + ecdot e^{-1} right) = 0
end{equation}
Or
begin{equation}
lim_{nrightarrow infty}int_{1}^{e}left[ln(x)right]^n:dx = 0
end{equation}
I was wondering if I could have a more technical eye have a look over this to see if the result I've obtained is correct.
Edit - Final result was incorrect and now corrected as per Song's pickup.
integration limits proof-verification definite-integrals
asked Dec 10 '18 at 23:17
DavidGDavidG
2,2641721
$endgroup$
add a comment |
$begingroup$
This is probably an unremarkable result, but I was curious about exploring the following family of definite integrals to see what asymptotic behaviour it as as $n rightarrow infty$
$$I_n = int_{1}^{e}left[ln(x)right]^n:dx mbox{ for } n in mathbb{N}$$
The method I took:
Employ integration by parts:
begin{align}
v'(x) &= 1 & u(x) &= left[ln(x)right]^n \
v(x) &= x & u'(x) &= nleft[ln(x)right]^{n - 1}frac{1}{x}
end{align}
Thus,
begin{align}
I_n &= int_{1}^{e}left[ln(x)right]^n:dx = x cdot left[ln(x)right]^n Big|_{1}^{e} - int_{1}^{e} x cdot nleft[ln(x)right]^{n - 1}frac{1}{x} \
&= e - nint_{1}^{e}left[ln(x)right]^{n - 1}:dx
end{align}
Thus,
begin{equation}
I_n = e - nI_{n - 1}
end{equation}
Where
begin{equation}
I_1 = int_{1}^{e} ln(x):dx = xleft(ln(x) - 1right)Big|_{1}^{e} = 1
end{equation}
This yields the solution:
begin{align}
I_n &= n!left[(-1)^{n - 1} + esum_{i = 0}^{n - 2} frac{(-1)^i}{(n - i)!} right] = n!left(-1right)^nleft[-1 + esum_{i = 2}^{n } frac{(-1)^i}{i!} right] \
&= n!left(-1right)^nleft[-1 + esum_{i = 0}^{n } frac{(-1)^i}{i!} right]
end{align}
We see that:
begin{equation}
lim_{nrightarrow infty} I_n = n! (-1)^n left( -1 + ecdot e^{-1} right) = 0
end{equation}
Or
begin{equation}
lim_{nrightarrow infty}int_{1}^{e}left[ln(x)right]^n:dx = 0
end{equation}
I was wondering if I could have a more technical eye have a look over this to see if the result I've obtained is correct.
Edit - Final result was incorrect and now corrected as per Song's pickup.
integration limits proof-verification definite-integrals
asked Dec 10 '18 at 23:17
DavidGDavidG
2,2641721
$endgroup$
1
$begingroup$
Your mistake is $-1+ecdot e^{-1} = 1$. It should be $0$. Actually, by dominated convergence theorem, $int_1^e (log x)^n dx to 0$. In the same spirit, we can show $int_1^A (log x)^n dx = mathcal{o}((log A)^n)$.
$endgroup$
– Song
Dec 10 '18 at 23:22
$begingroup$
Hi @Song - Yes, indeed- Thank you very much. I will edit my post.
$endgroup$
– DavidG
Dec 10 '18 at 23:23
2
$begingroup$
For a more direct approach leading to the same optimal result, note that, by concavity of the logarithm, for every $1<x<e$, $ln x<e^{-1}x$ hence $$I_n<e^{-n}int_1^ex^ndx<e^{-n}int_0^ex^ndx=frac e{n+1}$$
$endgroup$
– Did
Dec 10 '18 at 23:39
add a comment |
$begingroup$
This is probably an unremarkable result, but I was curious about exploring the following family of definite integrals to see what asymptotic behaviour it as as $n rightarrow infty$
$$I_n = int_{1}^{e}left[ln(x)right]^n:dx mbox{ for } n in mathbb{N}$$
The method I took:
Employ integration by parts:
begin{align}
v'(x) &= 1 & u(x) &= left[ln(x)right]^n \
v(x) &= x & u'(x) &= nleft[ln(x)right]^{n - 1}frac{1}{x}
end{align}
Thus,
begin{align}
I_n &= int_{1}^{e}left[ln(x)right]^n:dx = x cdot left[ln(x)right]^n Big|_{1}^{e} - int_{1}^{e} x cdot nleft[ln(x)right]^{n - 1}frac{1}{x} \
&= e - nint_{1}^{e}left[ln(x)right]^{n - 1}:dx
end{align}
Thus,
begin{equation}
I_n = e - nI_{n - 1}
end{equation}
Where
begin{equation}
I_1 = int_{1}^{e} ln(x):dx = xleft(ln(x) - 1right)Big|_{1}^{e} = 1
end{equation}
This yields the solution:
begin{align}
I_n &= n!left[(-1)^{n - 1} + esum_{i = 0}^{n - 2} frac{(-1)^i}{(n - i)!} right] = n!left(-1right)^nleft[-1 + esum_{i = 2}^{n } frac{(-1)^i}{i!} right] \
&= n!left(-1right)^nleft[-1 + esum_{i = 0}^{n } frac{(-1)^i}{i!} right]
end{align}
We see that:
begin{equation}
lim_{nrightarrow infty} I_n = n! (-1)^n left( -1 + ecdot e^{-1} right) = 0
end{equation}
Or
begin{equation}
lim_{nrightarrow infty}int_{1}^{e}left[ln(x)right]^n:dx = 0
end{equation}
I was wondering if I could have a more technical eye have a look over this to see if the result I've obtained is correct.
Edit - Final result was incorrect and now corrected as per Song's pickup.
integration limits proof-verification definite-integrals
asked Dec 10 '18 at 23:17
DavidGDavidG
2,2641721
$endgroup$
This is probably an unremarkable result, but I was curious about exploring the following family of definite integrals to see what asymptotic behaviour it as as $n rightarrow infty$
$$I_n = int_{1}^{e}left[ln(x)right]^n:dx mbox{ for } n in mathbb{N}$$
The method I took:
Employ integration by parts:
begin{align}
v'(x) &= 1 & u(x) &= left[ln(x)right]^n \
v(x) &= x & u'(x) &= nleft[ln(x)right]^{n - 1}frac{1}{x}
end{align}
Thus,
begin{align}
I_n &= int_{1}^{e}left[ln(x)right]^n:dx = x cdot left[ln(x)right]^n Big|_{1}^{e} - int_{1}^{e} x cdot nleft[ln(x)right]^{n - 1}frac{1}{x} \
&= e - nint_{1}^{e}left[ln(x)right]^{n - 1}:dx
end{align}
Thus,
begin{equation}
I_n = e - nI_{n - 1}
end{equation}
Where
begin{equation}
I_1 = int_{1}^{e} ln(x):dx = xleft(ln(x) - 1right)Big|_{1}^{e} = 1
end{equation}
This yields the solution:
begin{align}
I_n &= n!left[(-1)^{n - 1} + esum_{i = 0}^{n - 2} frac{(-1)^i}{(n - i)!} right] = n!left(-1right)^nleft[-1 + esum_{i = 2}^{n } frac{(-1)^i}{i!} right] \
&= n!left(-1right)^nleft[-1 + esum_{i = 0}^{n } frac{(-1)^i}{i!} right]
end{align}
We see that:
begin{equation}
lim_{nrightarrow infty} I_n = n! (-1)^n left( -1 + ecdot e^{-1} right) = 0
end{equation}
Or
begin{equation}
lim_{nrightarrow infty}int_{1}^{e}left[ln(x)right]^n:dx = 0
end{equation}
I was wondering if I could have a more technical eye have a look over this to see if the result I've obtained is correct.
Edit - Final result was incorrect and now corrected as per Song's pickup.
integration limits proof-verification definite-integrals
integration limits proof-verification definite-integrals
asked Dec 10 '18 at 23:17
DavidGDavidG
2,2641721
asked Dec 10 '18 at 23:17
DavidGDavidG
2,2641721
edited Dec 10 '18 at 23:24
DavidG
asked Dec 10 '18 at 23:17
DavidGDavidG
2,2641721
asked Dec 10 '18 at 23:17
DavidGDavidG
2,2641721
asked Dec 10 '18 at 23:17
DavidGDavidG
2,2641721
2,2641721
1
$begingroup$
Your mistake is $-1+ecdot e^{-1} = 1$. It should be $0$. Actually, by dominated convergence theorem, $int_1^e (log x)^n dx to 0$. In the same spirit, we can show $int_1^A (log x)^n dx = mathcal{o}((log A)^n)$.
$endgroup$
– Song
Dec 10 '18 at 23:22
$begingroup$
Hi @Song - Yes, indeed- Thank you very much. I will edit my post.
$endgroup$
– DavidG
Dec 10 '18 at 23:23
2
$begingroup$
For a more direct approach leading to the same optimal result, note that, by concavity of the logarithm, for every $1<x<e$, $ln x<e^{-1}x$ hence $$I_n<e^{-n}int_1^ex^ndx<e^{-n}int_0^ex^ndx=frac e{n+1}$$
$endgroup$
– Did
Dec 10 '18 at 23:39
add a comment |
1
$begingroup$
Your mistake is $-1+ecdot e^{-1} = 1$. It should be $0$. Actually, by dominated convergence theorem, $int_1^e (log x)^n dx to 0$. In the same spirit, we can show $int_1^A (log x)^n dx = mathcal{o}((log A)^n)$.
$endgroup$
– Song
Dec 10 '18 at 23:22
$begingroup$
Hi @Song - Yes, indeed- Thank you very much. I will edit my post.
$endgroup$
– DavidG
Dec 10 '18 at 23:23
2
$begingroup$
For a more direct approach leading to the same optimal result, note that, by concavity of the logarithm, for every $1<x<e$, $ln x<e^{-1}x$ hence $$I_n<e^{-n}int_1^ex^ndx<e^{-n}int_0^ex^ndx=frac e{n+1}$$
$endgroup$
– Did
Dec 10 '18 at 23:39
1
1
$begingroup$
Your mistake is $-1+ecdot e^{-1} = 1$. It should be $0$. Actually, by dominated convergence theorem, $int_1^e (log x)^n dx to 0$. In the same spirit, we can show $int_1^A (log x)^n dx = mathcal{o}((log A)^n)$.
$endgroup$
– Song
Dec 10 '18 at 23:22
$begingroup$
Your mistake is $-1+ecdot e^{-1} = 1$. It should be $0$. Actually, by dominated convergence theorem, $int_1^e (log x)^n dx to 0$. In the same spirit, we can show $int_1^A (log x)^n dx = mathcal{o}((log A)^n)$.
$endgroup$
– Song
Dec 10 '18 at 23:22
$begingroup$
Hi @Song - Yes, indeed- Thank you very much. I will edit my post.
$endgroup$
– DavidG
Dec 10 '18 at 23:23
$begingroup$
Hi @Song - Yes, indeed- Thank you very much. I will edit my post.
$endgroup$
– DavidG
Dec 10 '18 at 23:23
2
2
$begingroup$
For a more direct approach leading to the same optimal result, note that, by concavity of the logarithm, for every $1<x<e$, $ln x<e^{-1}x$ hence $$I_n<e^{-n}int_1^ex^ndx<e^{-n}int_0^ex^ndx=frac e{n+1}$$
$endgroup$
– Did
Dec 10 '18 at 23:39
$begingroup$
For a more direct approach leading to the same optimal result, note that, by concavity of the logarithm, for every $1<x<e$, $ln x<e^{-1}x$ hence $$I_n<e^{-n}int_1^ex^ndx<e^{-n}int_0^ex^ndx=frac e{n+1}$$
$endgroup$
– Did
Dec 10 '18 at 23:39
add a comment |
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$begingroup$
Your mistake is $-1+ecdot e^{-1} = 1$. It should be $0$. Actually, by dominated convergence theorem, $int_1^e (log x)^n dx to 0$. In the same spirit, we can show $int_1^A (log x)^n dx = mathcal{o}((log A)^n)$.
$endgroup$
– Song
Dec 10 '18 at 23:22
$begingroup$
Hi @Song - Yes, indeed- Thank you very much. I will edit my post.
$endgroup$
– DavidG
Dec 10 '18 at 23:23
2
$begingroup$
For a more direct approach leading to the same optimal result, note that, by concavity of the logarithm, for every $1<x<e$, $ln x<e^{-1}x$ hence $$I_n<e^{-n}int_1^ex^ndx<e^{-n}int_0^ex^ndx=frac e{n+1}$$
$endgroup$
– Did
Dec 10 '18 at 23:39