Solution to $int_0^1 left(frac{ln(x)}{1-x}right)^2dx$ in a closed form. [duplicate]
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This question already has an answer here:
A proof of $int_{0}^{1}left( frac{ln t}{1-t}right)^2,mathrm{d}t=frac{pi^2}{3}$
8 answers
I'm looking for the solution to the integral $$int_0^1 left(frac{ln(x)}{1-x}right)^2dx$$ I solved and know that the solution to $$-int_0^1 frac{ln(x)}{1-x}dx = frac{pi^2}{6}$$ through a taylor series argument, and am wondering if a similar approach is the best way to go.
definite-integrals taylor-expansion improper-integrals zeta-functions
edited Dec 21 '18 at 1:22
DavidG
2,1851720
asked Dec 20 '18 at 19:32
Suchetan DonthaSuchetan Dontha
14312
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marked as duplicate by Martin R, Community♦ Dec 20 '18 at 19:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
A proof of $int_{0}^{1}left( frac{ln t}{1-t}right)^2,mathrm{d}t=frac{pi^2}{3}$
8 answers
I'm looking for the solution to the integral $$int_0^1 left(frac{ln(x)}{1-x}right)^2dx$$ I solved and know that the solution to $$-int_0^1 frac{ln(x)}{1-x}dx = frac{pi^2}{6}$$ through a taylor series argument, and am wondering if a similar approach is the best way to go.
definite-integrals taylor-expansion improper-integrals zeta-functions
edited Dec 21 '18 at 1:22
DavidG
2,1851720
asked Dec 20 '18 at 19:32
Suchetan DonthaSuchetan Dontha
14312
$endgroup$
marked as duplicate by Martin R, Community♦ Dec 20 '18 at 19:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
A proof of $int_{0}^{1}left( frac{ln t}{1-t}right)^2,mathrm{d}t=frac{pi^2}{3}$
8 answers
I'm looking for the solution to the integral $$int_0^1 left(frac{ln(x)}{1-x}right)^2dx$$ I solved and know that the solution to $$-int_0^1 frac{ln(x)}{1-x}dx = frac{pi^2}{6}$$ through a taylor series argument, and am wondering if a similar approach is the best way to go.
definite-integrals taylor-expansion improper-integrals zeta-functions
edited Dec 21 '18 at 1:22
DavidG
2,1851720
asked Dec 20 '18 at 19:32
Suchetan DonthaSuchetan Dontha
14312
$endgroup$
This question already has an answer here:
A proof of $int_{0}^{1}left( frac{ln t}{1-t}right)^2,mathrm{d}t=frac{pi^2}{3}$
8 answers
I'm looking for the solution to the integral $$int_0^1 left(frac{ln(x)}{1-x}right)^2dx$$ I solved and know that the solution to $$-int_0^1 frac{ln(x)}{1-x}dx = frac{pi^2}{6}$$ through a taylor series argument, and am wondering if a similar approach is the best way to go.
This question already has an answer here:
A proof of $int_{0}^{1}left( frac{ln t}{1-t}right)^2,mathrm{d}t=frac{pi^2}{3}$
8 answers
definite-integrals taylor-expansion improper-integrals zeta-functions
definite-integrals taylor-expansion improper-integrals zeta-functions
edited Dec 21 '18 at 1:22
DavidG
2,1851720
asked Dec 20 '18 at 19:32
Suchetan DonthaSuchetan Dontha
14312
edited Dec 21 '18 at 1:22
DavidG
2,1851720
asked Dec 20 '18 at 19:32
Suchetan DonthaSuchetan Dontha
14312
edited Dec 21 '18 at 1:22
DavidG
2,1851720
edited Dec 21 '18 at 1:22
DavidG
2,1851720
edited Dec 21 '18 at 1:22
DavidG
2,1851720
2,1851720
asked Dec 20 '18 at 19:32
Suchetan DonthaSuchetan Dontha
14312
asked Dec 20 '18 at 19:32
Suchetan DonthaSuchetan Dontha
14312
asked Dec 20 '18 at 19:32
Suchetan DonthaSuchetan Dontha
14312
14312
marked as duplicate by Martin R, Community♦ Dec 20 '18 at 19:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Community♦ Dec 20 '18 at 19:38
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
hints
By parts
$$u=frac{-1}{1-x}$$
$$v=(ln(x))^2$$
the integral becomes
$$[uv]+2intfrac{ln(x)}{x(1-x)}$$
with
$$frac{1}{x(1-x)}=frac 1x+frac{1}{1-x}$$
answered Dec 20 '18 at 19:42
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
hints
By parts
$$u=frac{-1}{1-x}$$
$$v=(ln(x))^2$$
the integral becomes
$$[uv]+2intfrac{ln(x)}{x(1-x)}$$
with
$$frac{1}{x(1-x)}=frac 1x+frac{1}{1-x}$$
answered Dec 20 '18 at 19:42
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
hints
By parts
$$u=frac{-1}{1-x}$$
$$v=(ln(x))^2$$
the integral becomes
$$[uv]+2intfrac{ln(x)}{x(1-x)}$$
with
$$frac{1}{x(1-x)}=frac 1x+frac{1}{1-x}$$
answered Dec 20 '18 at 19:42
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
hints
By parts
$$u=frac{-1}{1-x}$$
$$v=(ln(x))^2$$
the integral becomes
$$[uv]+2intfrac{ln(x)}{x(1-x)}$$
with
$$frac{1}{x(1-x)}=frac 1x+frac{1}{1-x}$$
answered Dec 20 '18 at 19:42
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
hints
By parts
$$u=frac{-1}{1-x}$$
$$v=(ln(x))^2$$
the integral becomes
$$[uv]+2intfrac{ln(x)}{x(1-x)}$$
with
$$frac{1}{x(1-x)}=frac 1x+frac{1}{1-x}$$
answered Dec 20 '18 at 19:42
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 20 '18 at 19:42
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 20 '18 at 19:42
hamam_Abdallahhamam_Abdallah
38k21634
answered Dec 20 '18 at 19:42
hamam_Abdallahhamam_Abdallah
38k21634
38k21634
add a comment |
add a comment |
