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Does $intfracxcos(x)dx$ have an elementary solution?
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I need to solve the following integral: $displaystyleintfracxcos(x),dx$. My procedure is the following:
beginalign*intfracxcos(x),dx &= int xsec(x),dx\
&=xln(tan(x)+sec(x))-intln(tan(x)+sec(x)),dx.
endalign*
But, I'm stuck at this step, after using integration by parts I have $ln(tan(x)+sec(x))$ inside the new integral and then I do not know how to solve it, I was trying using by parts again but it gets more complicated.
Any advice on how to continue? I looked for related questions to this problem here in math.stackexchange but did not find anything useful.
integration indefinite-integrals
asked Jul 19 at 18:07
J. Doe
122
 |Â
show 7 more comments
up vote
1
down vote
favorite
I need to solve the following integral: $displaystyleintfracxcos(x),dx$. My procedure is the following:
beginalign*intfracxcos(x),dx &= int xsec(x),dx\
&=xln(tan(x)+sec(x))-intln(tan(x)+sec(x)),dx.
endalign*
But, I'm stuck at this step, after using integration by parts I have $ln(tan(x)+sec(x))$ inside the new integral and then I do not know how to solve it, I was trying using by parts again but it gets more complicated.
Any advice on how to continue? I looked for related questions to this problem here in math.stackexchange but did not find anything useful.
integration indefinite-integrals
asked Jul 19 at 18:07
J. Doe
122
3
I very much doubt that $frac xcos x$ has an elementary primitive.
– José Carlos Santos
Jul 19 at 18:09
wolframalpha.com/input/?i=integral+of+x%2Fcos(x) , it is looks like José is correct
– Holo
Jul 19 at 18:09
Can I trust on WolframAlpha results? I have heard it does not provide good results at the majority of problems.
– J. Doe
Jul 19 at 18:14
What is the source of the problem,?
– clark
Jul 19 at 18:15
6
@J.Doe: I am always very shocked to read negative comments about Wolfram Alpha. This is a wonderful piece of software including many man-years of extraordinary work by highly talented mathematicians. It is extremely rigorous and powerful, and have never heard of a single mistake it made (but many interpretation errors by the users). And in any case, it is more reliable than the average human. "it does not provide good results at the majority of problems" is shameful disinformation.
– Yves Daoust
Jul 19 at 18:23
 |Â
show 7 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need to solve the following integral: $displaystyleintfracxcos(x),dx$. My procedure is the following:
beginalign*intfracxcos(x),dx &= int xsec(x),dx\
&=xln(tan(x)+sec(x))-intln(tan(x)+sec(x)),dx.
endalign*
But, I'm stuck at this step, after using integration by parts I have $ln(tan(x)+sec(x))$ inside the new integral and then I do not know how to solve it, I was trying using by parts again but it gets more complicated.
Any advice on how to continue? I looked for related questions to this problem here in math.stackexchange but did not find anything useful.
integration indefinite-integrals
asked Jul 19 at 18:07
J. Doe
122
I need to solve the following integral: $displaystyleintfracxcos(x),dx$. My procedure is the following:
beginalign*intfracxcos(x),dx &= int xsec(x),dx\
&=xln(tan(x)+sec(x))-intln(tan(x)+sec(x)),dx.
endalign*
But, I'm stuck at this step, after using integration by parts I have $ln(tan(x)+sec(x))$ inside the new integral and then I do not know how to solve it, I was trying using by parts again but it gets more complicated.
Any advice on how to continue? I looked for related questions to this problem here in math.stackexchange but did not find anything useful.
integration indefinite-integrals
asked Jul 19 at 18:07
J. Doe
122
edited Jul 20 at 22:34
asked Jul 19 at 18:07
J. Doe
122
asked Jul 19 at 18:07
J. Doe
122
asked Jul 19 at 18:07
J. Doe
122
122
3
I very much doubt that $frac xcos x$ has an elementary primitive.
– José Carlos Santos
Jul 19 at 18:09
wolframalpha.com/input/?i=integral+of+x%2Fcos(x) , it is looks like José is correct
– Holo
Jul 19 at 18:09
Can I trust on WolframAlpha results? I have heard it does not provide good results at the majority of problems.
– J. Doe
Jul 19 at 18:14
What is the source of the problem,?
– clark
Jul 19 at 18:15
6
@J.Doe: I am always very shocked to read negative comments about Wolfram Alpha. This is a wonderful piece of software including many man-years of extraordinary work by highly talented mathematicians. It is extremely rigorous and powerful, and have never heard of a single mistake it made (but many interpretation errors by the users). And in any case, it is more reliable than the average human. "it does not provide good results at the majority of problems" is shameful disinformation.
– Yves Daoust
Jul 19 at 18:23
 |Â
show 7 more comments
3
I very much doubt that $frac xcos x$ has an elementary primitive.
– José Carlos Santos
Jul 19 at 18:09
wolframalpha.com/input/?i=integral+of+x%2Fcos(x) , it is looks like José is correct
– Holo
Jul 19 at 18:09
Can I trust on WolframAlpha results? I have heard it does not provide good results at the majority of problems.
– J. Doe
Jul 19 at 18:14
What is the source of the problem,?
– clark
Jul 19 at 18:15
6
@J.Doe: I am always very shocked to read negative comments about Wolfram Alpha. This is a wonderful piece of software including many man-years of extraordinary work by highly talented mathematicians. It is extremely rigorous and powerful, and have never heard of a single mistake it made (but many interpretation errors by the users). And in any case, it is more reliable than the average human. "it does not provide good results at the majority of problems" is shameful disinformation.
– Yves Daoust
Jul 19 at 18:23
3
3
I very much doubt that $frac xcos x$ has an elementary primitive.
– José Carlos Santos
Jul 19 at 18:09
I very much doubt that $frac xcos x$ has an elementary primitive.
– José Carlos Santos
Jul 19 at 18:09
wolframalpha.com/input/?i=integral+of+x%2Fcos(x) , it is looks like José is correct
– Holo
Jul 19 at 18:09
wolframalpha.com/input/?i=integral+of+x%2Fcos(x) , it is looks like José is correct
– Holo
Jul 19 at 18:09
Can I trust on WolframAlpha results? I have heard it does not provide good results at the majority of problems.
– J. Doe
Jul 19 at 18:14
Can I trust on WolframAlpha results? I have heard it does not provide good results at the majority of problems.
– J. Doe
Jul 19 at 18:14
What is the source of the problem,?
– clark
Jul 19 at 18:15
What is the source of the problem,?
– clark
Jul 19 at 18:15
6
6
@J.Doe: I am always very shocked to read negative comments about Wolfram Alpha. This is a wonderful piece of software including many man-years of extraordinary work by highly talented mathematicians. It is extremely rigorous and powerful, and have never heard of a single mistake it made (but many interpretation errors by the users). And in any case, it is more reliable than the average human. "it does not provide good results at the majority of problems" is shameful disinformation.
– Yves Daoust
Jul 19 at 18:23
@J.Doe: I am always very shocked to read negative comments about Wolfram Alpha. This is a wonderful piece of software including many man-years of extraordinary work by highly talented mathematicians. It is extremely rigorous and powerful, and have never heard of a single mistake it made (but many interpretation errors by the users). And in any case, it is more reliable than the average human. "it does not provide good results at the majority of problems" is shameful disinformation.
– Yves Daoust
Jul 19 at 18:23
 |Â
show 7 more comments
3 Answers
3
active
oldest
votes
up vote
1
down vote
Your integration by parts can be written
$$beginalign*intfracxcos(x),dx &= int xsec(x),dx\
&=x ln tanleft(fracpi4+fracx2right) -intln tanleft(fracpi4+fracx2right),dx.
endalign*$$
and the infinite series for $ln tanleft(fracpi4+fracx2right)$ is
$$ln tanleft(fracpi4+fracx2right)=2sum_k=1^infty frac2^2k-1 beta(2k-1)(2k-1)pi^2k-1x^2k-1 tag1$$
where $beta(n)$ is the Dirichlet Beta Function.
Then you also can establish the Cosine Series for $ln tanleft(fracpi4+fracx2right)$
$$ln tanleft(fracpi4+fracx2right)=-2 sum _k=1^infty fraccos left(2 (2 k-1) left(fracpi 4+fracx2right)right)2 k-1 tag2$$
from the well known Cosine Series for $ln sin x$ and $ln cos x$.
Therefore using the approach above you should be able to calculate, for example, the integral $$int_0^pi/3fracxcos(x),dx=frac13 pi log left(tan left(fracpi 4+fracpi 6right)right)-frac2 G3$$
where $G$ is Catalans Constant and without simplification
$$frac2 G3=2sum_k=1^infty frac2^2k-1 beta(2k-1)(2k-1),2k, pi^2k-1left(fracpi3right)^2k$$
The pattern appears to be that directly integrating any of the functions $ln sin x$, $ln cos x$, $ln tan x$, $ln tan (pi/4+x/2)$, $ln sinh x$, $ln cosh x$, $ln tanh x$ and $ln tanh (pi/4+x/2)$ does not lead to a new elementary primitive. That associated definite integral closed forms can be evaluated using the above techniques, without introducing complex numbers, is worth remembering.
[Note the infinite series for $ln tan x$ is different to (1) given above for $ln tan (pi/4+x/2)$, since the phase shifted $ln tan fracx2$ function appears when integrating the $csc$ function, not the $sec$ function as above. In the the equivalent notation
$ln tan x= log x + sum_k=1^infty frac2^2k ,eta(2k)k ,pi^2kx^2k$, where $eta(k)$ is the Dirichlet Eta Function]
answered Jul 25 at 23:35
James Arathoon
1,159420
add a comment |Â
up vote
0
down vote
Wolfram Alpha returns
$$x (log(1 - i e^i x) - log(1 + i e^i x)) + i (textLi_2(-i e^i x) - textLi_2(i e^i x))$$
which you can trust with closed eyes.
Alpha uses to return solutions valid in $mathbb C$, which sometimes makes the expression look complicated. In the case on hand, maybe the logarithmic terms can be rewritten with reals only. But maybe not the dilogarithmic ones.
And you can be sure that no elementary solution is possible.
answered Jul 19 at 18:35
Yves Daoust
111k665204
i will only be sure when you proof this statement.
– Dr. Sonnhard Graubner
Jul 19 at 18:46
@Dr.SonnhardGraubner: my mastery of Liouville's theorem and the Risch algorithm is insufficient. But Alpha knows them well. Presumably, Mathematica could give more details.
– Yves Daoust
Jul 19 at 18:55
@Dr.SonnhardGraubner: also note that if an elementary solution exists, the dilogarithm has an elementary expression as well.
– Yves Daoust
Jul 19 at 19:18
add a comment |Â
up vote
0
down vote
Perhaps the expanding of $e^x$ be useful, well
beginalign
intdfracxcos xdx
&= intdfrac2xe^-ix1+e^-2ixdx \
&= int 2xe^-ixsum_ngeq0e^-2inxdx \
&= sum_ngeq0int 2xe^-i(2n+1)xdx \
&= sum_ngeq02e^-i(2n+1)xleft(dfracix2n+1+dfrac1(2n+1)^2right) \
endalign
answered Jul 19 at 18:39
Nosrati
19.6k41544
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Your integration by parts can be written
$$beginalign*intfracxcos(x),dx &= int xsec(x),dx\
&=x ln tanleft(fracpi4+fracx2right) -intln tanleft(fracpi4+fracx2right),dx.
endalign*$$
and the infinite series for $ln tanleft(fracpi4+fracx2right)$ is
$$ln tanleft(fracpi4+fracx2right)=2sum_k=1^infty frac2^2k-1 beta(2k-1)(2k-1)pi^2k-1x^2k-1 tag1$$
where $beta(n)$ is the Dirichlet Beta Function.
Then you also can establish the Cosine Series for $ln tanleft(fracpi4+fracx2right)$
$$ln tanleft(fracpi4+fracx2right)=-2 sum _k=1^infty fraccos left(2 (2 k-1) left(fracpi 4+fracx2right)right)2 k-1 tag2$$
from the well known Cosine Series for $ln sin x$ and $ln cos x$.
Therefore using the approach above you should be able to calculate, for example, the integral $$int_0^pi/3fracxcos(x),dx=frac13 pi log left(tan left(fracpi 4+fracpi 6right)right)-frac2 G3$$
where $G$ is Catalans Constant and without simplification
$$frac2 G3=2sum_k=1^infty frac2^2k-1 beta(2k-1)(2k-1),2k, pi^2k-1left(fracpi3right)^2k$$
The pattern appears to be that directly integrating any of the functions $ln sin x$, $ln cos x$, $ln tan x$, $ln tan (pi/4+x/2)$, $ln sinh x$, $ln cosh x$, $ln tanh x$ and $ln tanh (pi/4+x/2)$ does not lead to a new elementary primitive. That associated definite integral closed forms can be evaluated using the above techniques, without introducing complex numbers, is worth remembering.
[Note the infinite series for $ln tan x$ is different to (1) given above for $ln tan (pi/4+x/2)$, since the phase shifted $ln tan fracx2$ function appears when integrating the $csc$ function, not the $sec$ function as above. In the the equivalent notation
$ln tan x= log x + sum_k=1^infty frac2^2k ,eta(2k)k ,pi^2kx^2k$, where $eta(k)$ is the Dirichlet Eta Function]
answered Jul 25 at 23:35
James Arathoon
1,159420
add a comment |Â
up vote
1
down vote
Your integration by parts can be written
$$beginalign*intfracxcos(x),dx &= int xsec(x),dx\
&=x ln tanleft(fracpi4+fracx2right) -intln tanleft(fracpi4+fracx2right),dx.
endalign*$$
and the infinite series for $ln tanleft(fracpi4+fracx2right)$ is
$$ln tanleft(fracpi4+fracx2right)=2sum_k=1^infty frac2^2k-1 beta(2k-1)(2k-1)pi^2k-1x^2k-1 tag1$$
where $beta(n)$ is the Dirichlet Beta Function.
Then you also can establish the Cosine Series for $ln tanleft(fracpi4+fracx2right)$
$$ln tanleft(fracpi4+fracx2right)=-2 sum _k=1^infty fraccos left(2 (2 k-1) left(fracpi 4+fracx2right)right)2 k-1 tag2$$
from the well known Cosine Series for $ln sin x$ and $ln cos x$.
Therefore using the approach above you should be able to calculate, for example, the integral $$int_0^pi/3fracxcos(x),dx=frac13 pi log left(tan left(fracpi 4+fracpi 6right)right)-frac2 G3$$
where $G$ is Catalans Constant and without simplification
$$frac2 G3=2sum_k=1^infty frac2^2k-1 beta(2k-1)(2k-1),2k, pi^2k-1left(fracpi3right)^2k$$
The pattern appears to be that directly integrating any of the functions $ln sin x$, $ln cos x$, $ln tan x$, $ln tan (pi/4+x/2)$, $ln sinh x$, $ln cosh x$, $ln tanh x$ and $ln tanh (pi/4+x/2)$ does not lead to a new elementary primitive. That associated definite integral closed forms can be evaluated using the above techniques, without introducing complex numbers, is worth remembering.
[Note the infinite series for $ln tan x$ is different to (1) given above for $ln tan (pi/4+x/2)$, since the phase shifted $ln tan fracx2$ function appears when integrating the $csc$ function, not the $sec$ function as above. In the the equivalent notation
$ln tan x= log x + sum_k=1^infty frac2^2k ,eta(2k)k ,pi^2kx^2k$, where $eta(k)$ is the Dirichlet Eta Function]
answered Jul 25 at 23:35
James Arathoon
1,159420
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your integration by parts can be written
$$beginalign*intfracxcos(x),dx &= int xsec(x),dx\
&=x ln tanleft(fracpi4+fracx2right) -intln tanleft(fracpi4+fracx2right),dx.
endalign*$$
and the infinite series for $ln tanleft(fracpi4+fracx2right)$ is
$$ln tanleft(fracpi4+fracx2right)=2sum_k=1^infty frac2^2k-1 beta(2k-1)(2k-1)pi^2k-1x^2k-1 tag1$$
where $beta(n)$ is the Dirichlet Beta Function.
Then you also can establish the Cosine Series for $ln tanleft(fracpi4+fracx2right)$
$$ln tanleft(fracpi4+fracx2right)=-2 sum _k=1^infty fraccos left(2 (2 k-1) left(fracpi 4+fracx2right)right)2 k-1 tag2$$
from the well known Cosine Series for $ln sin x$ and $ln cos x$.
Therefore using the approach above you should be able to calculate, for example, the integral $$int_0^pi/3fracxcos(x),dx=frac13 pi log left(tan left(fracpi 4+fracpi 6right)right)-frac2 G3$$
where $G$ is Catalans Constant and without simplification
$$frac2 G3=2sum_k=1^infty frac2^2k-1 beta(2k-1)(2k-1),2k, pi^2k-1left(fracpi3right)^2k$$
The pattern appears to be that directly integrating any of the functions $ln sin x$, $ln cos x$, $ln tan x$, $ln tan (pi/4+x/2)$, $ln sinh x$, $ln cosh x$, $ln tanh x$ and $ln tanh (pi/4+x/2)$ does not lead to a new elementary primitive. That associated definite integral closed forms can be evaluated using the above techniques, without introducing complex numbers, is worth remembering.
[Note the infinite series for $ln tan x$ is different to (1) given above for $ln tan (pi/4+x/2)$, since the phase shifted $ln tan fracx2$ function appears when integrating the $csc$ function, not the $sec$ function as above. In the the equivalent notation
$ln tan x= log x + sum_k=1^infty frac2^2k ,eta(2k)k ,pi^2kx^2k$, where $eta(k)$ is the Dirichlet Eta Function]
answered Jul 25 at 23:35
James Arathoon
1,159420
Your integration by parts can be written
$$beginalign*intfracxcos(x),dx &= int xsec(x),dx\
&=x ln tanleft(fracpi4+fracx2right) -intln tanleft(fracpi4+fracx2right),dx.
endalign*$$
and the infinite series for $ln tanleft(fracpi4+fracx2right)$ is
$$ln tanleft(fracpi4+fracx2right)=2sum_k=1^infty frac2^2k-1 beta(2k-1)(2k-1)pi^2k-1x^2k-1 tag1$$
where $beta(n)$ is the Dirichlet Beta Function.
Then you also can establish the Cosine Series for $ln tanleft(fracpi4+fracx2right)$
$$ln tanleft(fracpi4+fracx2right)=-2 sum _k=1^infty fraccos left(2 (2 k-1) left(fracpi 4+fracx2right)right)2 k-1 tag2$$
from the well known Cosine Series for $ln sin x$ and $ln cos x$.
Therefore using the approach above you should be able to calculate, for example, the integral $$int_0^pi/3fracxcos(x),dx=frac13 pi log left(tan left(fracpi 4+fracpi 6right)right)-frac2 G3$$
where $G$ is Catalans Constant and without simplification
$$frac2 G3=2sum_k=1^infty frac2^2k-1 beta(2k-1)(2k-1),2k, pi^2k-1left(fracpi3right)^2k$$
The pattern appears to be that directly integrating any of the functions $ln sin x$, $ln cos x$, $ln tan x$, $ln tan (pi/4+x/2)$, $ln sinh x$, $ln cosh x$, $ln tanh x$ and $ln tanh (pi/4+x/2)$ does not lead to a new elementary primitive. That associated definite integral closed forms can be evaluated using the above techniques, without introducing complex numbers, is worth remembering.
[Note the infinite series for $ln tan x$ is different to (1) given above for $ln tan (pi/4+x/2)$, since the phase shifted $ln tan fracx2$ function appears when integrating the $csc$ function, not the $sec$ function as above. In the the equivalent notation
$ln tan x= log x + sum_k=1^infty frac2^2k ,eta(2k)k ,pi^2kx^2k$, where $eta(k)$ is the Dirichlet Eta Function]
answered Jul 25 at 23:35
James Arathoon
1,159420
edited Jul 26 at 6:03
answered Jul 25 at 23:35
James Arathoon
1,159420
answered Jul 25 at 23:35
James Arathoon
1,159420
answered Jul 25 at 23:35
James Arathoon
1,159420
1,159420
add a comment |Â
add a comment |Â
up vote
0
down vote
Wolfram Alpha returns
$$x (log(1 - i e^i x) - log(1 + i e^i x)) + i (textLi_2(-i e^i x) - textLi_2(i e^i x))$$
which you can trust with closed eyes.
Alpha uses to return solutions valid in $mathbb C$, which sometimes makes the expression look complicated. In the case on hand, maybe the logarithmic terms can be rewritten with reals only. But maybe not the dilogarithmic ones.
And you can be sure that no elementary solution is possible.
answered Jul 19 at 18:35
Yves Daoust
111k665204
i will only be sure when you proof this statement.
– Dr. Sonnhard Graubner
Jul 19 at 18:46
@Dr.SonnhardGraubner: my mastery of Liouville's theorem and the Risch algorithm is insufficient. But Alpha knows them well. Presumably, Mathematica could give more details.
– Yves Daoust
Jul 19 at 18:55
@Dr.SonnhardGraubner: also note that if an elementary solution exists, the dilogarithm has an elementary expression as well.
– Yves Daoust
Jul 19 at 19:18
add a comment |Â
up vote
0
down vote
Wolfram Alpha returns
$$x (log(1 - i e^i x) - log(1 + i e^i x)) + i (textLi_2(-i e^i x) - textLi_2(i e^i x))$$
which you can trust with closed eyes.
Alpha uses to return solutions valid in $mathbb C$, which sometimes makes the expression look complicated. In the case on hand, maybe the logarithmic terms can be rewritten with reals only. But maybe not the dilogarithmic ones.
And you can be sure that no elementary solution is possible.
answered Jul 19 at 18:35
Yves Daoust
111k665204
i will only be sure when you proof this statement.
– Dr. Sonnhard Graubner
Jul 19 at 18:46
@Dr.SonnhardGraubner: my mastery of Liouville's theorem and the Risch algorithm is insufficient. But Alpha knows them well. Presumably, Mathematica could give more details.
– Yves Daoust
Jul 19 at 18:55
@Dr.SonnhardGraubner: also note that if an elementary solution exists, the dilogarithm has an elementary expression as well.
– Yves Daoust
Jul 19 at 19:18
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Wolfram Alpha returns
$$x (log(1 - i e^i x) - log(1 + i e^i x)) + i (textLi_2(-i e^i x) - textLi_2(i e^i x))$$
which you can trust with closed eyes.
Alpha uses to return solutions valid in $mathbb C$, which sometimes makes the expression look complicated. In the case on hand, maybe the logarithmic terms can be rewritten with reals only. But maybe not the dilogarithmic ones.
And you can be sure that no elementary solution is possible.
answered Jul 19 at 18:35
Yves Daoust
111k665204
Wolfram Alpha returns
$$x (log(1 - i e^i x) - log(1 + i e^i x)) + i (textLi_2(-i e^i x) - textLi_2(i e^i x))$$
which you can trust with closed eyes.
Alpha uses to return solutions valid in $mathbb C$, which sometimes makes the expression look complicated. In the case on hand, maybe the logarithmic terms can be rewritten with reals only. But maybe not the dilogarithmic ones.
And you can be sure that no elementary solution is possible.
answered Jul 19 at 18:35
Yves Daoust
111k665204
answered Jul 19 at 18:35
Yves Daoust
111k665204
answered Jul 19 at 18:35
Yves Daoust
111k665204
answered Jul 19 at 18:35
Yves Daoust
111k665204
111k665204
i will only be sure when you proof this statement.
– Dr. Sonnhard Graubner
Jul 19 at 18:46
@Dr.SonnhardGraubner: my mastery of Liouville's theorem and the Risch algorithm is insufficient. But Alpha knows them well. Presumably, Mathematica could give more details.
– Yves Daoust
Jul 19 at 18:55
@Dr.SonnhardGraubner: also note that if an elementary solution exists, the dilogarithm has an elementary expression as well.
– Yves Daoust
Jul 19 at 19:18
add a comment |Â
i will only be sure when you proof this statement.
– Dr. Sonnhard Graubner
Jul 19 at 18:46
@Dr.SonnhardGraubner: my mastery of Liouville's theorem and the Risch algorithm is insufficient. But Alpha knows them well. Presumably, Mathematica could give more details.
– Yves Daoust
Jul 19 at 18:55
@Dr.SonnhardGraubner: also note that if an elementary solution exists, the dilogarithm has an elementary expression as well.
– Yves Daoust
Jul 19 at 19:18
i will only be sure when you proof this statement.
– Dr. Sonnhard Graubner
Jul 19 at 18:46
i will only be sure when you proof this statement.
– Dr. Sonnhard Graubner
Jul 19 at 18:46
@Dr.SonnhardGraubner: my mastery of Liouville's theorem and the Risch algorithm is insufficient. But Alpha knows them well. Presumably, Mathematica could give more details.
– Yves Daoust
Jul 19 at 18:55
@Dr.SonnhardGraubner: my mastery of Liouville's theorem and the Risch algorithm is insufficient. But Alpha knows them well. Presumably, Mathematica could give more details.
– Yves Daoust
Jul 19 at 18:55
@Dr.SonnhardGraubner: also note that if an elementary solution exists, the dilogarithm has an elementary expression as well.
– Yves Daoust
Jul 19 at 19:18
@Dr.SonnhardGraubner: also note that if an elementary solution exists, the dilogarithm has an elementary expression as well.
– Yves Daoust
Jul 19 at 19:18
add a comment |Â
up vote
0
down vote
Perhaps the expanding of $e^x$ be useful, well
beginalign
intdfracxcos xdx
&= intdfrac2xe^-ix1+e^-2ixdx \
&= int 2xe^-ixsum_ngeq0e^-2inxdx \
&= sum_ngeq0int 2xe^-i(2n+1)xdx \
&= sum_ngeq02e^-i(2n+1)xleft(dfracix2n+1+dfrac1(2n+1)^2right) \
endalign
answered Jul 19 at 18:39
Nosrati
19.6k41544
add a comment |Â
up vote
0
down vote
Perhaps the expanding of $e^x$ be useful, well
beginalign
intdfracxcos xdx
&= intdfrac2xe^-ix1+e^-2ixdx \
&= int 2xe^-ixsum_ngeq0e^-2inxdx \
&= sum_ngeq0int 2xe^-i(2n+1)xdx \
&= sum_ngeq02e^-i(2n+1)xleft(dfracix2n+1+dfrac1(2n+1)^2right) \
endalign
answered Jul 19 at 18:39
Nosrati
19.6k41544
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Perhaps the expanding of $e^x$ be useful, well
beginalign
intdfracxcos xdx
&= intdfrac2xe^-ix1+e^-2ixdx \
&= int 2xe^-ixsum_ngeq0e^-2inxdx \
&= sum_ngeq0int 2xe^-i(2n+1)xdx \
&= sum_ngeq02e^-i(2n+1)xleft(dfracix2n+1+dfrac1(2n+1)^2right) \
endalign
answered Jul 19 at 18:39
Nosrati
19.6k41544
Perhaps the expanding of $e^x$ be useful, well
beginalign
intdfracxcos xdx
&= intdfrac2xe^-ix1+e^-2ixdx \
&= int 2xe^-ixsum_ngeq0e^-2inxdx \
&= sum_ngeq0int 2xe^-i(2n+1)xdx \
&= sum_ngeq02e^-i(2n+1)xleft(dfracix2n+1+dfrac1(2n+1)^2right) \
endalign
answered Jul 19 at 18:39
Nosrati
19.6k41544
edited Jul 19 at 18:48
answered Jul 19 at 18:39
Nosrati
19.6k41544
answered Jul 19 at 18:39
Nosrati
19.6k41544
answered Jul 19 at 18:39
Nosrati
19.6k41544
19.6k41544
add a comment |Â
add a comment |Â
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3
I very much doubt that $frac xcos x$ has an elementary primitive.
– José Carlos Santos
Jul 19 at 18:09
wolframalpha.com/input/?i=integral+of+x%2Fcos(x) , it is looks like José is correct
– Holo
Jul 19 at 18:09
Can I trust on WolframAlpha results? I have heard it does not provide good results at the majority of problems.
– J. Doe
Jul 19 at 18:14
What is the source of the problem,?
– clark
Jul 19 at 18:15
6
@J.Doe: I am always very shocked to read negative comments about Wolfram Alpha. This is a wonderful piece of software including many man-years of extraordinary work by highly talented mathematicians. It is extremely rigorous and powerful, and have never heard of a single mistake it made (but many interpretation errors by the users). And in any case, it is more reliable than the average human. "it does not provide good results at the majority of problems" is shameful disinformation.
– Yves Daoust
Jul 19 at 18:23