Evaluating a seemingly simple integral
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
I'm trying to evaluate the following integral, which arised while attempting to find the sum of a series :
$$int_0^1 fracln(x)x-1 ln(1+sqrtx)textdx$$
I've tried unsuccessfully some substitutions, integration by parts, feynman integration... I'm not familiar with more advanced integration techniques like residues theorem etc. ,so maybe that's the way to go.
Any hint, solution or partial solution would be nice !
integration
asked Aug 18 at 5:37
Harmonic Sun
2126
add a comment |Â
up vote
5
down vote
favorite
I'm trying to evaluate the following integral, which arised while attempting to find the sum of a series :
$$int_0^1 fracln(x)x-1 ln(1+sqrtx)textdx$$
I've tried unsuccessfully some substitutions, integration by parts, feynman integration... I'm not familiar with more advanced integration techniques like residues theorem etc. ,so maybe that's the way to go.
Any hint, solution or partial solution would be nice !
integration
asked Aug 18 at 5:37
Harmonic Sun
2126
Not relevant, but this integral is not "simple" at the first glimpse…
– xbh
Aug 18 at 5:43
2
Mathematica gives the value $$frac14(pi^2 log 4 - 9 zeta(3)),$$ which suggests that one might be able to rewrite the integrand in a clever way, but I don't have a solution at this time.
– heropup
Aug 18 at 5:52
@heropup. What is funny is that the antiderivative seems to exist in terms of polylogarithms. Would you check that Mathematica gives it ? Thanks.
– Claude Leibovici
Aug 18 at 6:07
When I saw the zeta(3) in the value you just gave me, I immediatly thought that there's ought to be some trilogarithm hidden in there... Thank you by the way. I don't have a mathematica lisence though, and wolfram is unable to proceed it
– Harmonic Sun
Aug 18 at 6:08
@ClaudeLeibovici Yes, it does, but it is quite lengthy and its evaluation at the endpoints requires some delicacy as it appears to involve taking limits. I think an approach specific to the definite integral is more likely to be fruitful.
– heropup
Aug 18 at 6:11
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I'm trying to evaluate the following integral, which arised while attempting to find the sum of a series :
$$int_0^1 fracln(x)x-1 ln(1+sqrtx)textdx$$
I've tried unsuccessfully some substitutions, integration by parts, feynman integration... I'm not familiar with more advanced integration techniques like residues theorem etc. ,so maybe that's the way to go.
Any hint, solution or partial solution would be nice !
integration
asked Aug 18 at 5:37
Harmonic Sun
2126
I'm trying to evaluate the following integral, which arised while attempting to find the sum of a series :
$$int_0^1 fracln(x)x-1 ln(1+sqrtx)textdx$$
I've tried unsuccessfully some substitutions, integration by parts, feynman integration... I'm not familiar with more advanced integration techniques like residues theorem etc. ,so maybe that's the way to go.
Any hint, solution or partial solution would be nice !
integration
asked Aug 18 at 5:37
Harmonic Sun
2126
asked Aug 18 at 5:37
Harmonic Sun
2126
asked Aug 18 at 5:37
Harmonic Sun
2126
asked Aug 18 at 5:37
Harmonic Sun
2126
2126
Not relevant, but this integral is not "simple" at the first glimpse…
– xbh
Aug 18 at 5:43
2
Mathematica gives the value $$frac14(pi^2 log 4 - 9 zeta(3)),$$ which suggests that one might be able to rewrite the integrand in a clever way, but I don't have a solution at this time.
– heropup
Aug 18 at 5:52
@heropup. What is funny is that the antiderivative seems to exist in terms of polylogarithms. Would you check that Mathematica gives it ? Thanks.
– Claude Leibovici
Aug 18 at 6:07
When I saw the zeta(3) in the value you just gave me, I immediatly thought that there's ought to be some trilogarithm hidden in there... Thank you by the way. I don't have a mathematica lisence though, and wolfram is unable to proceed it
– Harmonic Sun
Aug 18 at 6:08
@ClaudeLeibovici Yes, it does, but it is quite lengthy and its evaluation at the endpoints requires some delicacy as it appears to involve taking limits. I think an approach specific to the definite integral is more likely to be fruitful.
– heropup
Aug 18 at 6:11
add a comment |Â
Not relevant, but this integral is not "simple" at the first glimpse…
– xbh
Aug 18 at 5:43
2
Mathematica gives the value $$frac14(pi^2 log 4 - 9 zeta(3)),$$ which suggests that one might be able to rewrite the integrand in a clever way, but I don't have a solution at this time.
– heropup
Aug 18 at 5:52
@heropup. What is funny is that the antiderivative seems to exist in terms of polylogarithms. Would you check that Mathematica gives it ? Thanks.
– Claude Leibovici
Aug 18 at 6:07
When I saw the zeta(3) in the value you just gave me, I immediatly thought that there's ought to be some trilogarithm hidden in there... Thank you by the way. I don't have a mathematica lisence though, and wolfram is unable to proceed it
– Harmonic Sun
Aug 18 at 6:08
@ClaudeLeibovici Yes, it does, but it is quite lengthy and its evaluation at the endpoints requires some delicacy as it appears to involve taking limits. I think an approach specific to the definite integral is more likely to be fruitful.
– heropup
Aug 18 at 6:11
Not relevant, but this integral is not "simple" at the first glimpse…
– xbh
Aug 18 at 5:43
Not relevant, but this integral is not "simple" at the first glimpse…
– xbh
Aug 18 at 5:43
2
2
Mathematica gives the value $$frac14(pi^2 log 4 - 9 zeta(3)),$$ which suggests that one might be able to rewrite the integrand in a clever way, but I don't have a solution at this time.
– heropup
Aug 18 at 5:52
Mathematica gives the value $$frac14(pi^2 log 4 - 9 zeta(3)),$$ which suggests that one might be able to rewrite the integrand in a clever way, but I don't have a solution at this time.
– heropup
Aug 18 at 5:52
@heropup. What is funny is that the antiderivative seems to exist in terms of polylogarithms. Would you check that Mathematica gives it ? Thanks.
– Claude Leibovici
Aug 18 at 6:07
@heropup. What is funny is that the antiderivative seems to exist in terms of polylogarithms. Would you check that Mathematica gives it ? Thanks.
– Claude Leibovici
Aug 18 at 6:07
When I saw the zeta(3) in the value you just gave me, I immediatly thought that there's ought to be some trilogarithm hidden in there... Thank you by the way. I don't have a mathematica lisence though, and wolfram is unable to proceed it
– Harmonic Sun
Aug 18 at 6:08
When I saw the zeta(3) in the value you just gave me, I immediatly thought that there's ought to be some trilogarithm hidden in there... Thank you by the way. I don't have a mathematica lisence though, and wolfram is unable to proceed it
– Harmonic Sun
Aug 18 at 6:08
@ClaudeLeibovici Yes, it does, but it is quite lengthy and its evaluation at the endpoints requires some delicacy as it appears to involve taking limits. I think an approach specific to the definite integral is more likely to be fruitful.
– heropup
Aug 18 at 6:11
@ClaudeLeibovici Yes, it does, but it is quite lengthy and its evaluation at the endpoints requires some delicacy as it appears to involve taking limits. I think an approach specific to the definite integral is more likely to be fruitful.
– heropup
Aug 18 at 6:11
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
You can reduce your integral to two integrals which have already been solved on this site:
Let $x=t^2$, perform a partial fraction decomposition and integrate by parts in the second integral to find
beginalign
I &equiv intlimits_0^1 frac-ln(x) ln(1+sqrtx)1-x , mathrmdx \
&= 4 int limits_0^1 [-ln(t) ln(1+t)] fract(1-t)(1+t) , mathrmdt \
&= 2 int limits_0^1 frac-ln(t) ln(1+t)1-t , mathrmd t - 2 int limits_0^1 frac-ln(t) ln(1+t)1+t , mathrmd t \
&= 2 int limits_0^1 frac-ln(t) ln(1+t)1-t , mathrmd t - int limits_0^1 fracln^2 (1+t)t , mathrmd t \
&equiv 2 I_1 - I_2 , .
endalign
In this question $I_2 = fraczeta(3)4$ is proved (FDP's answer does not use contour integration or the polylogarithm). $I_1 = fracpi^24 ln(2) - zeta(3)$ is derived here (the answer relies on contour integration though and it would be nice to have a simpler proof). Thus we obtain the result suggested in the comments:
$$ I = 2 left(fracpi^24 ln(2) - zeta(3)right) - fraczeta(3)4 = fracpi^22 ln(2) - frac94 zeta(3) , . $$
answered Aug 18 at 9:40
ComplexYetTrivial
3,007626
Thank you very much !
– Harmonic Sun
Aug 18 at 22:18
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
You can reduce your integral to two integrals which have already been solved on this site:
Let $x=t^2$, perform a partial fraction decomposition and integrate by parts in the second integral to find
beginalign
I &equiv intlimits_0^1 frac-ln(x) ln(1+sqrtx)1-x , mathrmdx \
&= 4 int limits_0^1 [-ln(t) ln(1+t)] fract(1-t)(1+t) , mathrmdt \
&= 2 int limits_0^1 frac-ln(t) ln(1+t)1-t , mathrmd t - 2 int limits_0^1 frac-ln(t) ln(1+t)1+t , mathrmd t \
&= 2 int limits_0^1 frac-ln(t) ln(1+t)1-t , mathrmd t - int limits_0^1 fracln^2 (1+t)t , mathrmd t \
&equiv 2 I_1 - I_2 , .
endalign
In this question $I_2 = fraczeta(3)4$ is proved (FDP's answer does not use contour integration or the polylogarithm). $I_1 = fracpi^24 ln(2) - zeta(3)$ is derived here (the answer relies on contour integration though and it would be nice to have a simpler proof). Thus we obtain the result suggested in the comments:
$$ I = 2 left(fracpi^24 ln(2) - zeta(3)right) - fraczeta(3)4 = fracpi^22 ln(2) - frac94 zeta(3) , . $$
answered Aug 18 at 9:40
ComplexYetTrivial
3,007626
Thank you very much !
– Harmonic Sun
Aug 18 at 22:18
add a comment |Â
up vote
4
down vote
accepted
You can reduce your integral to two integrals which have already been solved on this site:
Let $x=t^2$, perform a partial fraction decomposition and integrate by parts in the second integral to find
beginalign
I &equiv intlimits_0^1 frac-ln(x) ln(1+sqrtx)1-x , mathrmdx \
&= 4 int limits_0^1 [-ln(t) ln(1+t)] fract(1-t)(1+t) , mathrmdt \
&= 2 int limits_0^1 frac-ln(t) ln(1+t)1-t , mathrmd t - 2 int limits_0^1 frac-ln(t) ln(1+t)1+t , mathrmd t \
&= 2 int limits_0^1 frac-ln(t) ln(1+t)1-t , mathrmd t - int limits_0^1 fracln^2 (1+t)t , mathrmd t \
&equiv 2 I_1 - I_2 , .
endalign
In this question $I_2 = fraczeta(3)4$ is proved (FDP's answer does not use contour integration or the polylogarithm). $I_1 = fracpi^24 ln(2) - zeta(3)$ is derived here (the answer relies on contour integration though and it would be nice to have a simpler proof). Thus we obtain the result suggested in the comments:
$$ I = 2 left(fracpi^24 ln(2) - zeta(3)right) - fraczeta(3)4 = fracpi^22 ln(2) - frac94 zeta(3) , . $$
answered Aug 18 at 9:40
ComplexYetTrivial
3,007626
Thank you very much !
– Harmonic Sun
Aug 18 at 22:18
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
You can reduce your integral to two integrals which have already been solved on this site:
Let $x=t^2$, perform a partial fraction decomposition and integrate by parts in the second integral to find
beginalign
I &equiv intlimits_0^1 frac-ln(x) ln(1+sqrtx)1-x , mathrmdx \
&= 4 int limits_0^1 [-ln(t) ln(1+t)] fract(1-t)(1+t) , mathrmdt \
&= 2 int limits_0^1 frac-ln(t) ln(1+t)1-t , mathrmd t - 2 int limits_0^1 frac-ln(t) ln(1+t)1+t , mathrmd t \
&= 2 int limits_0^1 frac-ln(t) ln(1+t)1-t , mathrmd t - int limits_0^1 fracln^2 (1+t)t , mathrmd t \
&equiv 2 I_1 - I_2 , .
endalign
In this question $I_2 = fraczeta(3)4$ is proved (FDP's answer does not use contour integration or the polylogarithm). $I_1 = fracpi^24 ln(2) - zeta(3)$ is derived here (the answer relies on contour integration though and it would be nice to have a simpler proof). Thus we obtain the result suggested in the comments:
$$ I = 2 left(fracpi^24 ln(2) - zeta(3)right) - fraczeta(3)4 = fracpi^22 ln(2) - frac94 zeta(3) , . $$
answered Aug 18 at 9:40
ComplexYetTrivial
3,007626
You can reduce your integral to two integrals which have already been solved on this site:
Let $x=t^2$, perform a partial fraction decomposition and integrate by parts in the second integral to find
beginalign
I &equiv intlimits_0^1 frac-ln(x) ln(1+sqrtx)1-x , mathrmdx \
&= 4 int limits_0^1 [-ln(t) ln(1+t)] fract(1-t)(1+t) , mathrmdt \
&= 2 int limits_0^1 frac-ln(t) ln(1+t)1-t , mathrmd t - 2 int limits_0^1 frac-ln(t) ln(1+t)1+t , mathrmd t \
&= 2 int limits_0^1 frac-ln(t) ln(1+t)1-t , mathrmd t - int limits_0^1 fracln^2 (1+t)t , mathrmd t \
&equiv 2 I_1 - I_2 , .
endalign
In this question $I_2 = fraczeta(3)4$ is proved (FDP's answer does not use contour integration or the polylogarithm). $I_1 = fracpi^24 ln(2) - zeta(3)$ is derived here (the answer relies on contour integration though and it would be nice to have a simpler proof). Thus we obtain the result suggested in the comments:
$$ I = 2 left(fracpi^24 ln(2) - zeta(3)right) - fraczeta(3)4 = fracpi^22 ln(2) - frac94 zeta(3) , . $$
answered Aug 18 at 9:40
ComplexYetTrivial
3,007626
answered Aug 18 at 9:40
ComplexYetTrivial
3,007626
answered Aug 18 at 9:40
ComplexYetTrivial
3,007626
answered Aug 18 at 9:40
ComplexYetTrivial
3,007626
3,007626
Thank you very much !
– Harmonic Sun
Aug 18 at 22:18
add a comment |Â
Thank you very much !
– Harmonic Sun
Aug 18 at 22:18
Thank you very much !
– Harmonic Sun
Aug 18 at 22:18
Thank you very much !
– Harmonic Sun
Aug 18 at 22:18
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2886447%2fevaluating-a-seemingly-simple-integral%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Not relevant, but this integral is not "simple" at the first glimpse…
– xbh
Aug 18 at 5:43
2
Mathematica gives the value $$frac14(pi^2 log 4 - 9 zeta(3)),$$ which suggests that one might be able to rewrite the integrand in a clever way, but I don't have a solution at this time.
– heropup
Aug 18 at 5:52
@heropup. What is funny is that the antiderivative seems to exist in terms of polylogarithms. Would you check that Mathematica gives it ? Thanks.
– Claude Leibovici
Aug 18 at 6:07
When I saw the zeta(3) in the value you just gave me, I immediatly thought that there's ought to be some trilogarithm hidden in there... Thank you by the way. I don't have a mathematica lisence though, and wolfram is unable to proceed it
– Harmonic Sun
Aug 18 at 6:08
@ClaudeLeibovici Yes, it does, but it is quite lengthy and its evaluation at the endpoints requires some delicacy as it appears to involve taking limits. I think an approach specific to the definite integral is more likely to be fruitful.
– heropup
Aug 18 at 6:11