Finding convergence (or divergence) of $int_0^{1/e} frac{dx}{x ln^2 x}$.
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Compute or prove the divergence of $$int_0^{1/e} frac{dx}{x ln^2 x}$$
I have tried to find the usual derivative taking integration from α..(0<α<1/e)to 1/e..it turned out be log0..but the given solution is 1.
calculus integration convergence definite-integrals
edited Nov 26 at 22:24
Key Flex
7,29941231
asked Nov 26 at 16:11
Kashmira
123
add a comment |
up vote
1
down vote
favorite
Compute or prove the divergence of $$int_0^{1/e} frac{dx}{x ln^2 x}$$
I have tried to find the usual derivative taking integration from α..(0<α<1/e)to 1/e..it turned out be log0..but the given solution is 1.
calculus integration convergence definite-integrals
edited Nov 26 at 22:24
Key Flex
7,29941231
asked Nov 26 at 16:11
Kashmira
123
Please use MathJax in future. It is a type of $LaTeX$. A tutorial on it from this site is readily available online.
– Shaun
Nov 26 at 16:20
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Compute or prove the divergence of $$int_0^{1/e} frac{dx}{x ln^2 x}$$
I have tried to find the usual derivative taking integration from α..(0<α<1/e)to 1/e..it turned out be log0..but the given solution is 1.
calculus integration convergence definite-integrals
edited Nov 26 at 22:24
Key Flex
7,29941231
asked Nov 26 at 16:11
Kashmira
123
Compute or prove the divergence of $$int_0^{1/e} frac{dx}{x ln^2 x}$$
I have tried to find the usual derivative taking integration from α..(0<α<1/e)to 1/e..it turned out be log0..but the given solution is 1.
calculus integration convergence definite-integrals
calculus integration convergence definite-integrals
edited Nov 26 at 22:24
Key Flex
7,29941231
asked Nov 26 at 16:11
Kashmira
123
edited Nov 26 at 22:24
Key Flex
7,29941231
asked Nov 26 at 16:11
Kashmira
123
edited Nov 26 at 22:24
Key Flex
7,29941231
edited Nov 26 at 22:24
Key Flex
7,29941231
edited Nov 26 at 22:24
Key Flex
7,29941231
7,29941231
asked Nov 26 at 16:11
Kashmira
123
asked Nov 26 at 16:11
Kashmira
123
asked Nov 26 at 16:11
Kashmira
123
123
Please use MathJax in future. It is a type of $LaTeX$. A tutorial on it from this site is readily available online.
– Shaun
Nov 26 at 16:20
add a comment |
Please use MathJax in future. It is a type of $LaTeX$. A tutorial on it from this site is readily available online.
– Shaun
Nov 26 at 16:20
Please use MathJax in future. It is a type of $LaTeX$. A tutorial on it from this site is readily available online.
– Shaun
Nov 26 at 16:20
Please use MathJax in future. It is a type of $LaTeX$. A tutorial on it from this site is readily available online.
– Shaun
Nov 26 at 16:20
add a comment |
2 Answers
2
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up vote
2
down vote
$$int_0^{frac1e}dfrac{1}{xln^2x} dx$$
Use u-substitution: $u=ln ximplies du=dfrac{dx}{x}$
first compute without boundaries
$$=intdfrac{1}{u^2} du=dfrac{u^{-2+1}}{-2+1}=-left[dfrac{1}{ln x}right]_0^{frac1e}=1$$
answered Nov 26 at 16:17
Key Flex
7,29941231
1
Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
– clathratus
Nov 26 at 17:13
@clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
– Key Flex
Nov 26 at 17:16
So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
– clathratus
Nov 26 at 17:19
@clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
– Key Flex
Nov 26 at 17:22
1
Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
– zhw.
Nov 26 at 17:46
add a comment |
up vote
0
down vote
HINT
Substitute $u = ln x$ with $du = dx/x$ to get $int u^{-2}du$...
answered Nov 26 at 16:13
gt6989b
32.7k22351
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$$int_0^{frac1e}dfrac{1}{xln^2x} dx$$
Use u-substitution: $u=ln ximplies du=dfrac{dx}{x}$
first compute without boundaries
$$=intdfrac{1}{u^2} du=dfrac{u^{-2+1}}{-2+1}=-left[dfrac{1}{ln x}right]_0^{frac1e}=1$$
answered Nov 26 at 16:17
Key Flex
7,29941231
1
Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
– clathratus
Nov 26 at 17:13
@clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
– Key Flex
Nov 26 at 17:16
So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
– clathratus
Nov 26 at 17:19
@clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
– Key Flex
Nov 26 at 17:22
1
Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
– zhw.
Nov 26 at 17:46
add a comment |
up vote
2
down vote
$$int_0^{frac1e}dfrac{1}{xln^2x} dx$$
Use u-substitution: $u=ln ximplies du=dfrac{dx}{x}$
first compute without boundaries
$$=intdfrac{1}{u^2} du=dfrac{u^{-2+1}}{-2+1}=-left[dfrac{1}{ln x}right]_0^{frac1e}=1$$
answered Nov 26 at 16:17
Key Flex
7,29941231
1
Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
– clathratus
Nov 26 at 17:13
@clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
– Key Flex
Nov 26 at 17:16
So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
– clathratus
Nov 26 at 17:19
@clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
– Key Flex
Nov 26 at 17:22
1
Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
– zhw.
Nov 26 at 17:46
add a comment |
up vote
2
down vote
up vote
2
down vote
$$int_0^{frac1e}dfrac{1}{xln^2x} dx$$
Use u-substitution: $u=ln ximplies du=dfrac{dx}{x}$
first compute without boundaries
$$=intdfrac{1}{u^2} du=dfrac{u^{-2+1}}{-2+1}=-left[dfrac{1}{ln x}right]_0^{frac1e}=1$$
answered Nov 26 at 16:17
Key Flex
7,29941231
$$int_0^{frac1e}dfrac{1}{xln^2x} dx$$
Use u-substitution: $u=ln ximplies du=dfrac{dx}{x}$
first compute without boundaries
$$=intdfrac{1}{u^2} du=dfrac{u^{-2+1}}{-2+1}=-left[dfrac{1}{ln x}right]_0^{frac1e}=1$$
answered Nov 26 at 16:17
Key Flex
7,29941231
answered Nov 26 at 16:17
Key Flex
7,29941231
answered Nov 26 at 16:17
Key Flex
7,29941231
answered Nov 26 at 16:17
Key Flex
7,29941231
7,29941231
1
Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
– clathratus
Nov 26 at 17:13
@clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
– Key Flex
Nov 26 at 17:16
So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
– clathratus
Nov 26 at 17:19
@clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
– Key Flex
Nov 26 at 17:22
1
Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
– zhw.
Nov 26 at 17:46
add a comment |
1
Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
– clathratus
Nov 26 at 17:13
@clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
– Key Flex
Nov 26 at 17:16
So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
– clathratus
Nov 26 at 17:19
@clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
– Key Flex
Nov 26 at 17:22
1
Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
– zhw.
Nov 26 at 17:46
1
1
Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
– clathratus
Nov 26 at 17:13
Pardon me if this is a very stupid question, but wouldn't the integral diverge because of the fact that $$lim_{xto0^+}frac1{xlog^2x}=+infty$$
– clathratus
Nov 26 at 17:13
@clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
– Key Flex
Nov 26 at 17:16
@clathratus Yes, it diverges because it has discontinuity at $x=0$ which gives improper bound
– Key Flex
Nov 26 at 17:16
So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
– clathratus
Nov 26 at 17:19
So why does your answer say that $$int_0^{frac1e}frac{dx}{xlog^2x}=1$$
– clathratus
Nov 26 at 17:19
@clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
– Key Flex
Nov 26 at 17:22
@clathratus I am evaluating the limit for $dfrac{1}{ln x}$ but not for $dfrac{1}{xln^2x}$
– Key Flex
Nov 26 at 17:22
1
1
Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
– zhw.
Nov 26 at 17:46
Your answer to the first comment is not correct. The question in the comment is: doesn't the integral diverge since the integrand blows up at $0$? Answer: No
– zhw.
Nov 26 at 17:46
add a comment |
up vote
0
down vote
HINT
Substitute $u = ln x$ with $du = dx/x$ to get $int u^{-2}du$...
answered Nov 26 at 16:13
gt6989b
32.7k22351
add a comment |
up vote
0
down vote
HINT
Substitute $u = ln x$ with $du = dx/x$ to get $int u^{-2}du$...
answered Nov 26 at 16:13
gt6989b
32.7k22351
add a comment |
up vote
0
down vote
up vote
0
down vote
HINT
Substitute $u = ln x$ with $du = dx/x$ to get $int u^{-2}du$...
answered Nov 26 at 16:13
gt6989b
32.7k22351
HINT
Substitute $u = ln x$ with $du = dx/x$ to get $int u^{-2}du$...
answered Nov 26 at 16:13
gt6989b
32.7k22351
answered Nov 26 at 16:13
gt6989b
32.7k22351
answered Nov 26 at 16:13
gt6989b
32.7k22351
answered Nov 26 at 16:13
gt6989b
32.7k22351
32.7k22351
add a comment |
add a comment |
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Please use MathJax in future. It is a type of $LaTeX$. A tutorial on it from this site is readily available online.
– Shaun
Nov 26 at 16:20