Mixing
Evaluate definite integral with variable limits [closed]
$begingroup$
Consider the function $f: (0, infty) to R $
$$ f(x)= begin{cases}
int_{x}^{x^2} frac{dt}{log(t)} & xneq 1 \
l & x=1 \
end{cases}
$$
Determine $l$ such that the function $f$ should be continuous at 1. For the determined value of $l$, is the function differentiable?
I started out by trying to calculate the following limit, however, I had trouble continuing.
$$lim_{xto1}int_{x}^{x^2}frac{mathrm dt}{log(t)}$$
Thank you!
real-analysis calculus integration definite-integrals
asked Jan 4 at 13:06
math1945math1945
102
$endgroup$
closed as off-topic by amWhy, RRL, Abcd, jgon, rtybase Jan 4 at 21:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, RRL, Abcd, jgon, rtybase
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider the function $f: (0, infty) to R $
$$ f(x)= begin{cases}
int_{x}^{x^2} frac{dt}{log(t)} & xneq 1 \
l & x=1 \
end{cases}
$$
Determine $l$ such that the function $f$ should be continuous at 1. For the determined value of $l$, is the function differentiable?
I started out by trying to calculate the following limit, however, I had trouble continuing.
$$lim_{xto1}int_{x}^{x^2}frac{mathrm dt}{log(t)}$$
Thank you!
real-analysis calculus integration definite-integrals
asked Jan 4 at 13:06
math1945math1945
102
$endgroup$
closed as off-topic by amWhy, RRL, Abcd, jgon, rtybase Jan 4 at 21:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, RRL, Abcd, jgon, rtybase
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Your integral itself is a very special integral called the Logarithmic Integral. If you were to evaluate the given integral, you would get $ lim_{x to 1 } (Gamma{(0, -ln x)} - Gamma{(0, -2 ln x)} )$
$endgroup$
– Tusky
Jan 4 at 13:32
$begingroup$
The3 result should be $$ln(2)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 4 at 13:36
add a comment |
$begingroup$
Consider the function $f: (0, infty) to R $
$$ f(x)= begin{cases}
int_{x}^{x^2} frac{dt}{log(t)} & xneq 1 \
l & x=1 \
end{cases}
$$
Determine $l$ such that the function $f$ should be continuous at 1. For the determined value of $l$, is the function differentiable?
I started out by trying to calculate the following limit, however, I had trouble continuing.
$$lim_{xto1}int_{x}^{x^2}frac{mathrm dt}{log(t)}$$
Thank you!
real-analysis calculus integration definite-integrals
asked Jan 4 at 13:06
math1945math1945
102
$endgroup$
Consider the function $f: (0, infty) to R $
$$ f(x)= begin{cases}
int_{x}^{x^2} frac{dt}{log(t)} & xneq 1 \
l & x=1 \
end{cases}
$$
Determine $l$ such that the function $f$ should be continuous at 1. For the determined value of $l$, is the function differentiable?
I started out by trying to calculate the following limit, however, I had trouble continuing.
$$lim_{xto1}int_{x}^{x^2}frac{mathrm dt}{log(t)}$$
Thank you!
real-analysis calculus integration definite-integrals
real-analysis calculus integration definite-integrals
asked Jan 4 at 13:06
math1945math1945
102
asked Jan 4 at 13:06
math1945math1945
102
edited Jan 4 at 14:05
math1945
asked Jan 4 at 13:06
math1945math1945
102
asked Jan 4 at 13:06
math1945math1945
102
asked Jan 4 at 13:06
math1945math1945
102
102
closed as off-topic by amWhy, RRL, Abcd, jgon, rtybase Jan 4 at 21:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, RRL, Abcd, jgon, rtybase
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, RRL, Abcd, jgon, rtybase Jan 4 at 21:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, RRL, Abcd, jgon, rtybase
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Your integral itself is a very special integral called the Logarithmic Integral. If you were to evaluate the given integral, you would get $ lim_{x to 1 } (Gamma{(0, -ln x)} - Gamma{(0, -2 ln x)} )$
$endgroup$
– Tusky
Jan 4 at 13:32
$begingroup$
The3 result should be $$ln(2)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 4 at 13:36
add a comment |
1
$begingroup$
Your integral itself is a very special integral called the Logarithmic Integral. If you were to evaluate the given integral, you would get $ lim_{x to 1 } (Gamma{(0, -ln x)} - Gamma{(0, -2 ln x)} )$
$endgroup$
– Tusky
Jan 4 at 13:32
$begingroup$
The3 result should be $$ln(2)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 4 at 13:36
1
1
$begingroup$
Your integral itself is a very special integral called the Logarithmic Integral. If you were to evaluate the given integral, you would get $ lim_{x to 1 } (Gamma{(0, -ln x)} - Gamma{(0, -2 ln x)} )$
$endgroup$
– Tusky
Jan 4 at 13:32
$begingroup$
Your integral itself is a very special integral called the Logarithmic Integral. If you were to evaluate the given integral, you would get $ lim_{x to 1 } (Gamma{(0, -ln x)} - Gamma{(0, -2 ln x)} )$
$endgroup$
– Tusky
Jan 4 at 13:32
$begingroup$
The3 result should be $$ln(2)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 4 at 13:36
$begingroup$
The3 result should be $$ln(2)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 4 at 13:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To calculate the limit itself (and not address the differentiability question)...
Below, consider $x>1$ as fixed, and make (as Tommaso does) the substitution $t = e^s$. Then
$$f(x) = int_x^{x^2} {dt over ln t} = int_{ln x}^{2ln x} {e^sover s}, ds.$$
Make the substitution $s = u ln x$:
$$ f(x) = int^2_1 e^{uln x} {duover u}.$$
As $xdownarrow 1$, we see that
$$ f(x) to int^2_1{duover u}= ln 2.$$
(Bringing the limit inside the integral is valid - by dominated convergence, for instance.)
Addendum/Edit:
1) The same argument works starting with $0<x<1$, and taking $xuparrow 1$.
2) If one sets $f(1) = ln 2$, one has that
$$ f(x) = int^2_1 e^{u ln x} {duover u}$$
identically, for $xin (0,infty)$. So, on condition that you can show that to calculate limits, one can perform the same corresponding calculations on the integrand, you can show that $f$ is differentiable.
answered Jan 4 at 15:27
peter a gpeter a g
3,1001614
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add a comment |
$begingroup$
First change variables $t = e^s$ and get
$$ int_{ln(x)}^{ln(x^2)} e^s s^{-1}ds. $$
Now expand the exponential around zero $e^s = sum_{ngeq 0} s^n/n!$ and pass the series outside the integral as $s^{-1} s^n/n! |_{(ln(x),ln(x^2)}$ converges uniformly to $0$. Pay attention to the first term, as $s^0/0!equiv 1$
begin{align*} int_{ln(x)}^{ln(x^2)} (1 + sum_{ngeq 1}s^n/n!) s^{-1}ds &= left. ln(s) + sum_{ngeq 1} s^n/(n!n) right|_{ln(x)}^{ln(x^2)} \
& = ln(ln(x^2)/ln(x)) + sum_{ngeq 1} (2^n-1)ln(x)^n/(n!n).
end{align*}
The term on the right goes to zero as $xto 0$ because the series converges absolutely in a right neighborhood of $1$. Therefore
$$ lim_{xto 1^+} f(x) = lim_{xto 1^+} ln(ln(x^2)/ln(x)) = ln(lim_{xto 1^+} ln(x^2)/ln(x)) = ln(lim_{xto 1^+} x^{-2}2x/(x^{-1}) = ln(2). $$
answered Jan 4 at 15:16
Tommaso SeneciTommaso Seneci
99428
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To calculate the limit itself (and not address the differentiability question)...
Below, consider $x>1$ as fixed, and make (as Tommaso does) the substitution $t = e^s$. Then
$$f(x) = int_x^{x^2} {dt over ln t} = int_{ln x}^{2ln x} {e^sover s}, ds.$$
Make the substitution $s = u ln x$:
$$ f(x) = int^2_1 e^{uln x} {duover u}.$$
As $xdownarrow 1$, we see that
$$ f(x) to int^2_1{duover u}= ln 2.$$
(Bringing the limit inside the integral is valid - by dominated convergence, for instance.)
Addendum/Edit:
1) The same argument works starting with $0<x<1$, and taking $xuparrow 1$.
2) If one sets $f(1) = ln 2$, one has that
$$ f(x) = int^2_1 e^{u ln x} {duover u}$$
identically, for $xin (0,infty)$. So, on condition that you can show that to calculate limits, one can perform the same corresponding calculations on the integrand, you can show that $f$ is differentiable.
answered Jan 4 at 15:27
peter a gpeter a g
3,1001614
$endgroup$
add a comment |
$begingroup$
To calculate the limit itself (and not address the differentiability question)...
Below, consider $x>1$ as fixed, and make (as Tommaso does) the substitution $t = e^s$. Then
$$f(x) = int_x^{x^2} {dt over ln t} = int_{ln x}^{2ln x} {e^sover s}, ds.$$
Make the substitution $s = u ln x$:
$$ f(x) = int^2_1 e^{uln x} {duover u}.$$
As $xdownarrow 1$, we see that
$$ f(x) to int^2_1{duover u}= ln 2.$$
(Bringing the limit inside the integral is valid - by dominated convergence, for instance.)
Addendum/Edit:
1) The same argument works starting with $0<x<1$, and taking $xuparrow 1$.
2) If one sets $f(1) = ln 2$, one has that
$$ f(x) = int^2_1 e^{u ln x} {duover u}$$
identically, for $xin (0,infty)$. So, on condition that you can show that to calculate limits, one can perform the same corresponding calculations on the integrand, you can show that $f$ is differentiable.
answered Jan 4 at 15:27
peter a gpeter a g
3,1001614
$endgroup$
add a comment |
$begingroup$
To calculate the limit itself (and not address the differentiability question)...
Below, consider $x>1$ as fixed, and make (as Tommaso does) the substitution $t = e^s$. Then
$$f(x) = int_x^{x^2} {dt over ln t} = int_{ln x}^{2ln x} {e^sover s}, ds.$$
Make the substitution $s = u ln x$:
$$ f(x) = int^2_1 e^{uln x} {duover u}.$$
As $xdownarrow 1$, we see that
$$ f(x) to int^2_1{duover u}= ln 2.$$
(Bringing the limit inside the integral is valid - by dominated convergence, for instance.)
Addendum/Edit:
1) The same argument works starting with $0<x<1$, and taking $xuparrow 1$.
2) If one sets $f(1) = ln 2$, one has that
$$ f(x) = int^2_1 e^{u ln x} {duover u}$$
identically, for $xin (0,infty)$. So, on condition that you can show that to calculate limits, one can perform the same corresponding calculations on the integrand, you can show that $f$ is differentiable.
answered Jan 4 at 15:27
peter a gpeter a g
3,1001614
$endgroup$
To calculate the limit itself (and not address the differentiability question)...
Below, consider $x>1$ as fixed, and make (as Tommaso does) the substitution $t = e^s$. Then
$$f(x) = int_x^{x^2} {dt over ln t} = int_{ln x}^{2ln x} {e^sover s}, ds.$$
Make the substitution $s = u ln x$:
$$ f(x) = int^2_1 e^{uln x} {duover u}.$$
As $xdownarrow 1$, we see that
$$ f(x) to int^2_1{duover u}= ln 2.$$
(Bringing the limit inside the integral is valid - by dominated convergence, for instance.)
Addendum/Edit:
1) The same argument works starting with $0<x<1$, and taking $xuparrow 1$.
2) If one sets $f(1) = ln 2$, one has that
$$ f(x) = int^2_1 e^{u ln x} {duover u}$$
identically, for $xin (0,infty)$. So, on condition that you can show that to calculate limits, one can perform the same corresponding calculations on the integrand, you can show that $f$ is differentiable.
answered Jan 4 at 15:27
peter a gpeter a g
3,1001614
edited Jan 4 at 18:23
answered Jan 4 at 15:27
peter a gpeter a g
3,1001614
answered Jan 4 at 15:27
peter a gpeter a g
3,1001614
answered Jan 4 at 15:27
peter a gpeter a g
3,1001614
3,1001614
add a comment |
add a comment |
$begingroup$
First change variables $t = e^s$ and get
$$ int_{ln(x)}^{ln(x^2)} e^s s^{-1}ds. $$
Now expand the exponential around zero $e^s = sum_{ngeq 0} s^n/n!$ and pass the series outside the integral as $s^{-1} s^n/n! |_{(ln(x),ln(x^2)}$ converges uniformly to $0$. Pay attention to the first term, as $s^0/0!equiv 1$
begin{align*} int_{ln(x)}^{ln(x^2)} (1 + sum_{ngeq 1}s^n/n!) s^{-1}ds &= left. ln(s) + sum_{ngeq 1} s^n/(n!n) right|_{ln(x)}^{ln(x^2)} \
& = ln(ln(x^2)/ln(x)) + sum_{ngeq 1} (2^n-1)ln(x)^n/(n!n).
end{align*}
The term on the right goes to zero as $xto 0$ because the series converges absolutely in a right neighborhood of $1$. Therefore
$$ lim_{xto 1^+} f(x) = lim_{xto 1^+} ln(ln(x^2)/ln(x)) = ln(lim_{xto 1^+} ln(x^2)/ln(x)) = ln(lim_{xto 1^+} x^{-2}2x/(x^{-1}) = ln(2). $$
answered Jan 4 at 15:16
Tommaso SeneciTommaso Seneci
99428
$endgroup$
add a comment |
$begingroup$
First change variables $t = e^s$ and get
$$ int_{ln(x)}^{ln(x^2)} e^s s^{-1}ds. $$
Now expand the exponential around zero $e^s = sum_{ngeq 0} s^n/n!$ and pass the series outside the integral as $s^{-1} s^n/n! |_{(ln(x),ln(x^2)}$ converges uniformly to $0$. Pay attention to the first term, as $s^0/0!equiv 1$
begin{align*} int_{ln(x)}^{ln(x^2)} (1 + sum_{ngeq 1}s^n/n!) s^{-1}ds &= left. ln(s) + sum_{ngeq 1} s^n/(n!n) right|_{ln(x)}^{ln(x^2)} \
& = ln(ln(x^2)/ln(x)) + sum_{ngeq 1} (2^n-1)ln(x)^n/(n!n).
end{align*}
The term on the right goes to zero as $xto 0$ because the series converges absolutely in a right neighborhood of $1$. Therefore
$$ lim_{xto 1^+} f(x) = lim_{xto 1^+} ln(ln(x^2)/ln(x)) = ln(lim_{xto 1^+} ln(x^2)/ln(x)) = ln(lim_{xto 1^+} x^{-2}2x/(x^{-1}) = ln(2). $$
answered Jan 4 at 15:16
Tommaso SeneciTommaso Seneci
99428
$endgroup$
add a comment |
$begingroup$
First change variables $t = e^s$ and get
$$ int_{ln(x)}^{ln(x^2)} e^s s^{-1}ds. $$
Now expand the exponential around zero $e^s = sum_{ngeq 0} s^n/n!$ and pass the series outside the integral as $s^{-1} s^n/n! |_{(ln(x),ln(x^2)}$ converges uniformly to $0$. Pay attention to the first term, as $s^0/0!equiv 1$
begin{align*} int_{ln(x)}^{ln(x^2)} (1 + sum_{ngeq 1}s^n/n!) s^{-1}ds &= left. ln(s) + sum_{ngeq 1} s^n/(n!n) right|_{ln(x)}^{ln(x^2)} \
& = ln(ln(x^2)/ln(x)) + sum_{ngeq 1} (2^n-1)ln(x)^n/(n!n).
end{align*}
The term on the right goes to zero as $xto 0$ because the series converges absolutely in a right neighborhood of $1$. Therefore
$$ lim_{xto 1^+} f(x) = lim_{xto 1^+} ln(ln(x^2)/ln(x)) = ln(lim_{xto 1^+} ln(x^2)/ln(x)) = ln(lim_{xto 1^+} x^{-2}2x/(x^{-1}) = ln(2). $$
answered Jan 4 at 15:16
Tommaso SeneciTommaso Seneci
99428
$endgroup$
First change variables $t = e^s$ and get
$$ int_{ln(x)}^{ln(x^2)} e^s s^{-1}ds. $$
Now expand the exponential around zero $e^s = sum_{ngeq 0} s^n/n!$ and pass the series outside the integral as $s^{-1} s^n/n! |_{(ln(x),ln(x^2)}$ converges uniformly to $0$. Pay attention to the first term, as $s^0/0!equiv 1$
begin{align*} int_{ln(x)}^{ln(x^2)} (1 + sum_{ngeq 1}s^n/n!) s^{-1}ds &= left. ln(s) + sum_{ngeq 1} s^n/(n!n) right|_{ln(x)}^{ln(x^2)} \
& = ln(ln(x^2)/ln(x)) + sum_{ngeq 1} (2^n-1)ln(x)^n/(n!n).
end{align*}
The term on the right goes to zero as $xto 0$ because the series converges absolutely in a right neighborhood of $1$. Therefore
$$ lim_{xto 1^+} f(x) = lim_{xto 1^+} ln(ln(x^2)/ln(x)) = ln(lim_{xto 1^+} ln(x^2)/ln(x)) = ln(lim_{xto 1^+} x^{-2}2x/(x^{-1}) = ln(2). $$
answered Jan 4 at 15:16
Tommaso SeneciTommaso Seneci
99428
answered Jan 4 at 15:16
Tommaso SeneciTommaso Seneci
99428
answered Jan 4 at 15:16
Tommaso SeneciTommaso Seneci
99428
answered Jan 4 at 15:16
Tommaso SeneciTommaso Seneci
99428
99428
add a comment |
add a comment |
1
$begingroup$
Your integral itself is a very special integral called the Logarithmic Integral. If you were to evaluate the given integral, you would get $ lim_{x to 1 } (Gamma{(0, -ln x)} - Gamma{(0, -2 ln x)} )$
$endgroup$
– Tusky
Jan 4 at 13:32
$begingroup$
The3 result should be $$ln(2)$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 4 at 13:36