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Expressions for integrals of non-elementary integrals
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Functions like $frac1log x$ have no elementary integrals in terms of standard functions; they are instead represented using special function notations, such as $mathrmli(x)$.
That's all and well, but how does one obtain an expression for the integral of these nonelementary integrals?
I would expect that there would be "no result found in terms of standard mathematical functions", but WolframAlpha actually gives an expression for the logarithmic integral:
$$int mathrmli(x)dx = x , mathrmli(x) - mathrmEi(2 log x) + C.$$
(Or $x , mathrmli(x) - mathrmli(x^2) + C$, for an appropriate choice of domain.)
How does WolframAlpha obtain such a result? Is there a special technique that I am unaware of?
calculus integration indefinite-integrals
asked Aug 3 at 21:56
Bladewood
19513
add a comment |Â
up vote
6
down vote
favorite
Functions like $frac1log x$ have no elementary integrals in terms of standard functions; they are instead represented using special function notations, such as $mathrmli(x)$.
That's all and well, but how does one obtain an expression for the integral of these nonelementary integrals?
I would expect that there would be "no result found in terms of standard mathematical functions", but WolframAlpha actually gives an expression for the logarithmic integral:
$$int mathrmli(x)dx = x , mathrmli(x) - mathrmEi(2 log x) + C.$$
(Or $x , mathrmli(x) - mathrmli(x^2) + C$, for an appropriate choice of domain.)
How does WolframAlpha obtain such a result? Is there a special technique that I am unaware of?
calculus integration indefinite-integrals
asked Aug 3 at 21:56
Bladewood
19513
2
That particular example is just an application of integration by parts and some manipulation.
– Simply Beautiful Art
Aug 3 at 21:59
1
Similarly it may be deduced by applying Cauchy's repeated integral formula.
– Simply Beautiful Art
Aug 3 at 22:00
So then it was just a lucky hit, rearranging the integral of $mathrmli$ into a form that just happens to contain $frac1log x$ itself. What should I do with this question, then?
– Bladewood
Aug 3 at 23:14
You could self-answer your question :-)
– Simply Beautiful Art
Aug 3 at 23:20
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Functions like $frac1log x$ have no elementary integrals in terms of standard functions; they are instead represented using special function notations, such as $mathrmli(x)$.
That's all and well, but how does one obtain an expression for the integral of these nonelementary integrals?
I would expect that there would be "no result found in terms of standard mathematical functions", but WolframAlpha actually gives an expression for the logarithmic integral:
$$int mathrmli(x)dx = x , mathrmli(x) - mathrmEi(2 log x) + C.$$
(Or $x , mathrmli(x) - mathrmli(x^2) + C$, for an appropriate choice of domain.)
How does WolframAlpha obtain such a result? Is there a special technique that I am unaware of?
calculus integration indefinite-integrals
asked Aug 3 at 21:56
Bladewood
19513
Functions like $frac1log x$ have no elementary integrals in terms of standard functions; they are instead represented using special function notations, such as $mathrmli(x)$.
That's all and well, but how does one obtain an expression for the integral of these nonelementary integrals?
I would expect that there would be "no result found in terms of standard mathematical functions", but WolframAlpha actually gives an expression for the logarithmic integral:
$$int mathrmli(x)dx = x , mathrmli(x) - mathrmEi(2 log x) + C.$$
(Or $x , mathrmli(x) - mathrmli(x^2) + C$, for an appropriate choice of domain.)
How does WolframAlpha obtain such a result? Is there a special technique that I am unaware of?
calculus integration indefinite-integrals
asked Aug 3 at 21:56
Bladewood
19513
asked Aug 3 at 21:56
Bladewood
19513
asked Aug 3 at 21:56
Bladewood
19513
asked Aug 3 at 21:56
Bladewood
19513
19513
2
That particular example is just an application of integration by parts and some manipulation.
– Simply Beautiful Art
Aug 3 at 21:59
1
Similarly it may be deduced by applying Cauchy's repeated integral formula.
– Simply Beautiful Art
Aug 3 at 22:00
So then it was just a lucky hit, rearranging the integral of $mathrmli$ into a form that just happens to contain $frac1log x$ itself. What should I do with this question, then?
– Bladewood
Aug 3 at 23:14
You could self-answer your question :-)
– Simply Beautiful Art
Aug 3 at 23:20
add a comment |Â
2
That particular example is just an application of integration by parts and some manipulation.
– Simply Beautiful Art
Aug 3 at 21:59
1
Similarly it may be deduced by applying Cauchy's repeated integral formula.
– Simply Beautiful Art
Aug 3 at 22:00
So then it was just a lucky hit, rearranging the integral of $mathrmli$ into a form that just happens to contain $frac1log x$ itself. What should I do with this question, then?
– Bladewood
Aug 3 at 23:14
You could self-answer your question :-)
– Simply Beautiful Art
Aug 3 at 23:20
2
2
That particular example is just an application of integration by parts and some manipulation.
– Simply Beautiful Art
Aug 3 at 21:59
That particular example is just an application of integration by parts and some manipulation.
– Simply Beautiful Art
Aug 3 at 21:59
1
1
Similarly it may be deduced by applying Cauchy's repeated integral formula.
– Simply Beautiful Art
Aug 3 at 22:00
Similarly it may be deduced by applying Cauchy's repeated integral formula.
– Simply Beautiful Art
Aug 3 at 22:00
So then it was just a lucky hit, rearranging the integral of $mathrmli$ into a form that just happens to contain $frac1log x$ itself. What should I do with this question, then?
– Bladewood
Aug 3 at 23:14
So then it was just a lucky hit, rearranging the integral of $mathrmli$ into a form that just happens to contain $frac1log x$ itself. What should I do with this question, then?
– Bladewood
Aug 3 at 23:14
You could self-answer your question :-)
– Simply Beautiful Art
Aug 3 at 23:20
You could self-answer your question :-)
– Simply Beautiful Art
Aug 3 at 23:20
add a comment |Â
2 Answers
2
active
oldest
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up vote
2
down vote
In general, there is still often no elementary solution to these kinds of integrals.
In this case, however, there is fortunately a relatively simple way to evaluate the integral of $mathrmli(x)$, involving integration by parts to first break down the nonelementary integral inside and to hope for a way to rearrange it into a simpler form.
So, using integration by parts, we find
$$int mathrmli(x) , dx = x , mathrmli(x) - int fracxlog x , dx.$$
We can evaluate this new integral using the substitution $u = log x$, which leads to $x = e^u$ and $dx = e^u , du$.
beginalign
int fracxlog x , dx & = int frace^uu e^u , du \
& = int frace^2uu , du.
endalign
This is itself a nonelementary integral, but a common special function exists that can describe it: the exponential integral, $$mathrmEi(x) equiv - int_-z^infty frace^-tt , dt.$$
In this case, though, we're only worrying about the indefinite integral, so we'll just use
$$mathrmEi(x) equiv int frace^xx , dx.$$
So, using this, our integral becomes with $v = 2u$
beginalign
int frace^2uu , du & = int frace^vfracv2 , fracdv2 \
& = int frace^vv , dv \
& = mathrmEi(v) \
& = mathrmEi(2 log x).
endalign
Substituting this back into our original integral, we obtain
$$int mathrmli(x) , dx = x , mathrmli(x) - mathrmEi(2 log x) + C.$$
answered Aug 4 at 0:02
Bladewood
19513
add a comment |Â
up vote
1
down vote
Not being a pure mathematician it is not at first obvious to me that integration by parts should automatically work for a function like $textli(x)=int_0^x frac1log t ,dt$ for which there is no infinite series expansion.
It's seems slightly clearer to me in this case to take the proposed solution and approach this problem by differentiating $x ,textli(x)$ and $textli(x^2)$ instead (even though I am differentiating a definite integral which is not the normal case):
$$fracd ,(x ,textli(x))d x=textli(x)+fracxlog (x)$$
Then rearranging and integrating gives
$$int textli(x) , dx=x ,textli(x)-int fracxlog (x) dx$$
and then since
$$fracd, (textli(x^2))d x=frac2xlog (x^2)=fracxlog (x)$$
$$int fracxlog (x), dx= textli(x^2)$$
Giving
$$int textli(x) , dx=x ,textli(x)-textli(x^2)$$
Then finally finding $textli(x^2)$ in terms of the exponential integral as in @Bladewood's answer.
[I am not happy with this - is it more rigorous to prove that this only works when the lower limit in the $textli(x)$ function integral is constant?]
answered Aug 4 at 0:30
James Arathoon
1,155420
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
In general, there is still often no elementary solution to these kinds of integrals.
In this case, however, there is fortunately a relatively simple way to evaluate the integral of $mathrmli(x)$, involving integration by parts to first break down the nonelementary integral inside and to hope for a way to rearrange it into a simpler form.
So, using integration by parts, we find
$$int mathrmli(x) , dx = x , mathrmli(x) - int fracxlog x , dx.$$
We can evaluate this new integral using the substitution $u = log x$, which leads to $x = e^u$ and $dx = e^u , du$.
beginalign
int fracxlog x , dx & = int frace^uu e^u , du \
& = int frace^2uu , du.
endalign
This is itself a nonelementary integral, but a common special function exists that can describe it: the exponential integral, $$mathrmEi(x) equiv - int_-z^infty frace^-tt , dt.$$
In this case, though, we're only worrying about the indefinite integral, so we'll just use
$$mathrmEi(x) equiv int frace^xx , dx.$$
So, using this, our integral becomes with $v = 2u$
beginalign
int frace^2uu , du & = int frace^vfracv2 , fracdv2 \
& = int frace^vv , dv \
& = mathrmEi(v) \
& = mathrmEi(2 log x).
endalign
Substituting this back into our original integral, we obtain
$$int mathrmli(x) , dx = x , mathrmli(x) - mathrmEi(2 log x) + C.$$
answered Aug 4 at 0:02
Bladewood
19513
add a comment |Â
up vote
2
down vote
In general, there is still often no elementary solution to these kinds of integrals.
In this case, however, there is fortunately a relatively simple way to evaluate the integral of $mathrmli(x)$, involving integration by parts to first break down the nonelementary integral inside and to hope for a way to rearrange it into a simpler form.
So, using integration by parts, we find
$$int mathrmli(x) , dx = x , mathrmli(x) - int fracxlog x , dx.$$
We can evaluate this new integral using the substitution $u = log x$, which leads to $x = e^u$ and $dx = e^u , du$.
beginalign
int fracxlog x , dx & = int frace^uu e^u , du \
& = int frace^2uu , du.
endalign
This is itself a nonelementary integral, but a common special function exists that can describe it: the exponential integral, $$mathrmEi(x) equiv - int_-z^infty frace^-tt , dt.$$
In this case, though, we're only worrying about the indefinite integral, so we'll just use
$$mathrmEi(x) equiv int frace^xx , dx.$$
So, using this, our integral becomes with $v = 2u$
beginalign
int frace^2uu , du & = int frace^vfracv2 , fracdv2 \
& = int frace^vv , dv \
& = mathrmEi(v) \
& = mathrmEi(2 log x).
endalign
Substituting this back into our original integral, we obtain
$$int mathrmli(x) , dx = x , mathrmli(x) - mathrmEi(2 log x) + C.$$
answered Aug 4 at 0:02
Bladewood
19513
add a comment |Â
up vote
2
down vote
up vote
2
down vote
In general, there is still often no elementary solution to these kinds of integrals.
In this case, however, there is fortunately a relatively simple way to evaluate the integral of $mathrmli(x)$, involving integration by parts to first break down the nonelementary integral inside and to hope for a way to rearrange it into a simpler form.
So, using integration by parts, we find
$$int mathrmli(x) , dx = x , mathrmli(x) - int fracxlog x , dx.$$
We can evaluate this new integral using the substitution $u = log x$, which leads to $x = e^u$ and $dx = e^u , du$.
beginalign
int fracxlog x , dx & = int frace^uu e^u , du \
& = int frace^2uu , du.
endalign
This is itself a nonelementary integral, but a common special function exists that can describe it: the exponential integral, $$mathrmEi(x) equiv - int_-z^infty frace^-tt , dt.$$
In this case, though, we're only worrying about the indefinite integral, so we'll just use
$$mathrmEi(x) equiv int frace^xx , dx.$$
So, using this, our integral becomes with $v = 2u$
beginalign
int frace^2uu , du & = int frace^vfracv2 , fracdv2 \
& = int frace^vv , dv \
& = mathrmEi(v) \
& = mathrmEi(2 log x).
endalign
Substituting this back into our original integral, we obtain
$$int mathrmli(x) , dx = x , mathrmli(x) - mathrmEi(2 log x) + C.$$
answered Aug 4 at 0:02
Bladewood
19513
In general, there is still often no elementary solution to these kinds of integrals.
In this case, however, there is fortunately a relatively simple way to evaluate the integral of $mathrmli(x)$, involving integration by parts to first break down the nonelementary integral inside and to hope for a way to rearrange it into a simpler form.
So, using integration by parts, we find
$$int mathrmli(x) , dx = x , mathrmli(x) - int fracxlog x , dx.$$
We can evaluate this new integral using the substitution $u = log x$, which leads to $x = e^u$ and $dx = e^u , du$.
beginalign
int fracxlog x , dx & = int frace^uu e^u , du \
& = int frace^2uu , du.
endalign
This is itself a nonelementary integral, but a common special function exists that can describe it: the exponential integral, $$mathrmEi(x) equiv - int_-z^infty frace^-tt , dt.$$
In this case, though, we're only worrying about the indefinite integral, so we'll just use
$$mathrmEi(x) equiv int frace^xx , dx.$$
So, using this, our integral becomes with $v = 2u$
beginalign
int frace^2uu , du & = int frace^vfracv2 , fracdv2 \
& = int frace^vv , dv \
& = mathrmEi(v) \
& = mathrmEi(2 log x).
endalign
Substituting this back into our original integral, we obtain
$$int mathrmli(x) , dx = x , mathrmli(x) - mathrmEi(2 log x) + C.$$
answered Aug 4 at 0:02
Bladewood
19513
answered Aug 4 at 0:02
Bladewood
19513
answered Aug 4 at 0:02
Bladewood
19513
answered Aug 4 at 0:02
Bladewood
19513
19513
add a comment |Â
add a comment |Â
up vote
1
down vote
Not being a pure mathematician it is not at first obvious to me that integration by parts should automatically work for a function like $textli(x)=int_0^x frac1log t ,dt$ for which there is no infinite series expansion.
It's seems slightly clearer to me in this case to take the proposed solution and approach this problem by differentiating $x ,textli(x)$ and $textli(x^2)$ instead (even though I am differentiating a definite integral which is not the normal case):
$$fracd ,(x ,textli(x))d x=textli(x)+fracxlog (x)$$
Then rearranging and integrating gives
$$int textli(x) , dx=x ,textli(x)-int fracxlog (x) dx$$
and then since
$$fracd, (textli(x^2))d x=frac2xlog (x^2)=fracxlog (x)$$
$$int fracxlog (x), dx= textli(x^2)$$
Giving
$$int textli(x) , dx=x ,textli(x)-textli(x^2)$$
Then finally finding $textli(x^2)$ in terms of the exponential integral as in @Bladewood's answer.
[I am not happy with this - is it more rigorous to prove that this only works when the lower limit in the $textli(x)$ function integral is constant?]
answered Aug 4 at 0:30
James Arathoon
1,155420
add a comment |Â
up vote
1
down vote
Not being a pure mathematician it is not at first obvious to me that integration by parts should automatically work for a function like $textli(x)=int_0^x frac1log t ,dt$ for which there is no infinite series expansion.
It's seems slightly clearer to me in this case to take the proposed solution and approach this problem by differentiating $x ,textli(x)$ and $textli(x^2)$ instead (even though I am differentiating a definite integral which is not the normal case):
$$fracd ,(x ,textli(x))d x=textli(x)+fracxlog (x)$$
Then rearranging and integrating gives
$$int textli(x) , dx=x ,textli(x)-int fracxlog (x) dx$$
and then since
$$fracd, (textli(x^2))d x=frac2xlog (x^2)=fracxlog (x)$$
$$int fracxlog (x), dx= textli(x^2)$$
Giving
$$int textli(x) , dx=x ,textli(x)-textli(x^2)$$
Then finally finding $textli(x^2)$ in terms of the exponential integral as in @Bladewood's answer.
[I am not happy with this - is it more rigorous to prove that this only works when the lower limit in the $textli(x)$ function integral is constant?]
answered Aug 4 at 0:30
James Arathoon
1,155420
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Not being a pure mathematician it is not at first obvious to me that integration by parts should automatically work for a function like $textli(x)=int_0^x frac1log t ,dt$ for which there is no infinite series expansion.
It's seems slightly clearer to me in this case to take the proposed solution and approach this problem by differentiating $x ,textli(x)$ and $textli(x^2)$ instead (even though I am differentiating a definite integral which is not the normal case):
$$fracd ,(x ,textli(x))d x=textli(x)+fracxlog (x)$$
Then rearranging and integrating gives
$$int textli(x) , dx=x ,textli(x)-int fracxlog (x) dx$$
and then since
$$fracd, (textli(x^2))d x=frac2xlog (x^2)=fracxlog (x)$$
$$int fracxlog (x), dx= textli(x^2)$$
Giving
$$int textli(x) , dx=x ,textli(x)-textli(x^2)$$
Then finally finding $textli(x^2)$ in terms of the exponential integral as in @Bladewood's answer.
[I am not happy with this - is it more rigorous to prove that this only works when the lower limit in the $textli(x)$ function integral is constant?]
answered Aug 4 at 0:30
James Arathoon
1,155420
Not being a pure mathematician it is not at first obvious to me that integration by parts should automatically work for a function like $textli(x)=int_0^x frac1log t ,dt$ for which there is no infinite series expansion.
It's seems slightly clearer to me in this case to take the proposed solution and approach this problem by differentiating $x ,textli(x)$ and $textli(x^2)$ instead (even though I am differentiating a definite integral which is not the normal case):
$$fracd ,(x ,textli(x))d x=textli(x)+fracxlog (x)$$
Then rearranging and integrating gives
$$int textli(x) , dx=x ,textli(x)-int fracxlog (x) dx$$
and then since
$$fracd, (textli(x^2))d x=frac2xlog (x^2)=fracxlog (x)$$
$$int fracxlog (x), dx= textli(x^2)$$
Giving
$$int textli(x) , dx=x ,textli(x)-textli(x^2)$$
Then finally finding $textli(x^2)$ in terms of the exponential integral as in @Bladewood's answer.
[I am not happy with this - is it more rigorous to prove that this only works when the lower limit in the $textli(x)$ function integral is constant?]
answered Aug 4 at 0:30
James Arathoon
1,155420
edited 2 days ago
answered Aug 4 at 0:30
James Arathoon
1,155420
answered Aug 4 at 0:30
James Arathoon
1,155420
answered Aug 4 at 0:30
James Arathoon
1,155420
1,155420
add a comment |Â
add a comment |Â
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2
That particular example is just an application of integration by parts and some manipulation.
– Simply Beautiful Art
Aug 3 at 21:59
1
Similarly it may be deduced by applying Cauchy's repeated integral formula.
– Simply Beautiful Art
Aug 3 at 22:00
So then it was just a lucky hit, rearranging the integral of $mathrmli$ into a form that just happens to contain $frac1log x$ itself. What should I do with this question, then?
– Bladewood
Aug 3 at 23:14
You could self-answer your question :-)
– Simply Beautiful Art
Aug 3 at 23:20