Mixing
Calculate $int_0^1 frac{ln(1-x+x^2)}{x-x^2}dx$
$begingroup$
I am trying to calculate:
$$int_0^1 frac{ln(1-x+x^2)}{x-x^2}dx$$
I am not looking for an answer but simply a nudge in the right direction. A strategy, just something that would get me started.
So, after doing the Taylor Expansion on the $ln(1-x+x^2)$ ig to the following: Let $x=x-x^2$ then $ln(1-x)$ then,
begin{align*}
=&-x-frac{x^2}{2}-frac{x^3}{3}-frac{x^4}{4}-...\
=&-(x-x^2)-frac{(x-x^2)^2}{2}-...\
=&-x(1-x)+frac{x^2}{2}(1-x)^2-frac{x^3}{3}(1-x)^3\
text{thus the pattern is:}\
=&frac{x^n(1-x)^n}{n}
end{align*}
Am I right?
Then our Integral would be: $$sum_{n=0}^{infty} frac{1}{n+1} int_0^1 x^n(1-x)^n$$
Am I on the right track? Suggestions, tips, comments?
$underline{NEW EDIT:}$
SO after integrating the function I got the following after a couple of iterations:
begin{align*}
frac{n(n-1)...1}{(n+1)(n+2)...(2n)}int_0^1 x^{2n} dx
end{align*}
This shows a pattern:
begin{align*}
=&frac{(n!)^2}{(2n)!} (frac{1}{2n+1})\
=& frac{(n!)^2}{(2n+1)!}
end{align*}
So my question is, what to do from here. I have done all this but still have no clue how to actually solve the integral. Can somebody shed some light on this!
Thanks
calculus real-analysis integration definite-integrals
edited Jun 1 '14 at 12:16
bobbym
2,31121123
asked Apr 16 '14 at 16:53
user2233524user2233524
663
$endgroup$
|
show 4 more comments
$begingroup$
I am trying to calculate:
$$int_0^1 frac{ln(1-x+x^2)}{x-x^2}dx$$
I am not looking for an answer but simply a nudge in the right direction. A strategy, just something that would get me started.
So, after doing the Taylor Expansion on the $ln(1-x+x^2)$ ig to the following: Let $x=x-x^2$ then $ln(1-x)$ then,
begin{align*}
=&-x-frac{x^2}{2}-frac{x^3}{3}-frac{x^4}{4}-...\
=&-(x-x^2)-frac{(x-x^2)^2}{2}-...\
=&-x(1-x)+frac{x^2}{2}(1-x)^2-frac{x^3}{3}(1-x)^3\
text{thus the pattern is:}\
=&frac{x^n(1-x)^n}{n}
end{align*}
Am I right?
Then our Integral would be: $$sum_{n=0}^{infty} frac{1}{n+1} int_0^1 x^n(1-x)^n$$
Am I on the right track? Suggestions, tips, comments?
$underline{NEW EDIT:}$
SO after integrating the function I got the following after a couple of iterations:
begin{align*}
frac{n(n-1)...1}{(n+1)(n+2)...(2n)}int_0^1 x^{2n} dx
end{align*}
This shows a pattern:
begin{align*}
=&frac{(n!)^2}{(2n)!} (frac{1}{2n+1})\
=& frac{(n!)^2}{(2n+1)!}
end{align*}
So my question is, what to do from here. I have done all this but still have no clue how to actually solve the integral. Can somebody shed some light on this!
Thanks
calculus real-analysis integration definite-integrals
edited Jun 1 '14 at 12:16
bobbym
2,31121123
asked Apr 16 '14 at 16:53
user2233524user2233524
663
$endgroup$
$begingroup$
Can you start with en.wikipedia.org/wiki/Integration_by_parts#LIATE_rule ?
$endgroup$
– lab bhattacharjee
Apr 16 '14 at 16:56
4
$begingroup$
Expand the log into a Taylor series about $x-x^2$, perform the integration, and then recognize the resulting series as an expansion of a (sort of) well-known function evaluated at a particular value.
$endgroup$
– Ron Gordon
Apr 16 '14 at 16:59
$begingroup$
@Ron Gordon Mmmmh. Very interesting. I will work on it and will post what I come up with. Thanks
$endgroup$
– user2233524
Apr 16 '14 at 20:28
$begingroup$
Ok guys I added some stuff. Still need some help. Thanks
$endgroup$
– user2233524
Apr 18 '14 at 19:30
$begingroup$
@user2233524 $int_0^1 frac{ln(1-x+x^2)}{x-x^2}dx$ is less than zero, whereas $sum_{n=0}^{infty} frac{1}{n+1} int_0^1 x^n(1-x)^n$ is clearly positive, so you must have made a mistake somewhere. They seem to have equal magnitude though, so it's probably just a simple sign error.
$endgroup$
– David H
Apr 18 '14 at 21:47
|
show 4 more comments
$begingroup$
I am trying to calculate:
$$int_0^1 frac{ln(1-x+x^2)}{x-x^2}dx$$
I am not looking for an answer but simply a nudge in the right direction. A strategy, just something that would get me started.
So, after doing the Taylor Expansion on the $ln(1-x+x^2)$ ig to the following: Let $x=x-x^2$ then $ln(1-x)$ then,
begin{align*}
=&-x-frac{x^2}{2}-frac{x^3}{3}-frac{x^4}{4}-...\
=&-(x-x^2)-frac{(x-x^2)^2}{2}-...\
=&-x(1-x)+frac{x^2}{2}(1-x)^2-frac{x^3}{3}(1-x)^3\
text{thus the pattern is:}\
=&frac{x^n(1-x)^n}{n}
end{align*}
Am I right?
Then our Integral would be: $$sum_{n=0}^{infty} frac{1}{n+1} int_0^1 x^n(1-x)^n$$
Am I on the right track? Suggestions, tips, comments?
$underline{NEW EDIT:}$
SO after integrating the function I got the following after a couple of iterations:
begin{align*}
frac{n(n-1)...1}{(n+1)(n+2)...(2n)}int_0^1 x^{2n} dx
end{align*}
This shows a pattern:
begin{align*}
=&frac{(n!)^2}{(2n)!} (frac{1}{2n+1})\
=& frac{(n!)^2}{(2n+1)!}
end{align*}
So my question is, what to do from here. I have done all this but still have no clue how to actually solve the integral. Can somebody shed some light on this!
Thanks
calculus real-analysis integration definite-integrals
edited Jun 1 '14 at 12:16
bobbym
2,31121123
asked Apr 16 '14 at 16:53
user2233524user2233524
663
$endgroup$
I am trying to calculate:
$$int_0^1 frac{ln(1-x+x^2)}{x-x^2}dx$$
I am not looking for an answer but simply a nudge in the right direction. A strategy, just something that would get me started.
So, after doing the Taylor Expansion on the $ln(1-x+x^2)$ ig to the following: Let $x=x-x^2$ then $ln(1-x)$ then,
begin{align*}
=&-x-frac{x^2}{2}-frac{x^3}{3}-frac{x^4}{4}-...\
=&-(x-x^2)-frac{(x-x^2)^2}{2}-...\
=&-x(1-x)+frac{x^2}{2}(1-x)^2-frac{x^3}{3}(1-x)^3\
text{thus the pattern is:}\
=&frac{x^n(1-x)^n}{n}
end{align*}
Am I right?
Then our Integral would be: $$sum_{n=0}^{infty} frac{1}{n+1} int_0^1 x^n(1-x)^n$$
Am I on the right track? Suggestions, tips, comments?
$underline{NEW EDIT:}$
SO after integrating the function I got the following after a couple of iterations:
begin{align*}
frac{n(n-1)...1}{(n+1)(n+2)...(2n)}int_0^1 x^{2n} dx
end{align*}
This shows a pattern:
begin{align*}
=&frac{(n!)^2}{(2n)!} (frac{1}{2n+1})\
=& frac{(n!)^2}{(2n+1)!}
end{align*}
So my question is, what to do from here. I have done all this but still have no clue how to actually solve the integral. Can somebody shed some light on this!
Thanks
calculus real-analysis integration definite-integrals
calculus real-analysis integration definite-integrals
edited Jun 1 '14 at 12:16
bobbym
2,31121123
asked Apr 16 '14 at 16:53
user2233524user2233524
663
edited Jun 1 '14 at 12:16
bobbym
2,31121123
asked Apr 16 '14 at 16:53
user2233524user2233524
663
edited Jun 1 '14 at 12:16
bobbym
2,31121123
edited Jun 1 '14 at 12:16
bobbym
2,31121123
edited Jun 1 '14 at 12:16
bobbym
2,31121123
2,31121123
asked Apr 16 '14 at 16:53
user2233524user2233524
663
asked Apr 16 '14 at 16:53
user2233524user2233524
663
asked Apr 16 '14 at 16:53
user2233524user2233524
663
663
$begingroup$
Can you start with en.wikipedia.org/wiki/Integration_by_parts#LIATE_rule ?
$endgroup$
– lab bhattacharjee
Apr 16 '14 at 16:56
4
$begingroup$
Expand the log into a Taylor series about $x-x^2$, perform the integration, and then recognize the resulting series as an expansion of a (sort of) well-known function evaluated at a particular value.
$endgroup$
– Ron Gordon
Apr 16 '14 at 16:59
$begingroup$
@Ron Gordon Mmmmh. Very interesting. I will work on it and will post what I come up with. Thanks
$endgroup$
– user2233524
Apr 16 '14 at 20:28
$begingroup$
Ok guys I added some stuff. Still need some help. Thanks
$endgroup$
– user2233524
Apr 18 '14 at 19:30
$begingroup$
@user2233524 $int_0^1 frac{ln(1-x+x^2)}{x-x^2}dx$ is less than zero, whereas $sum_{n=0}^{infty} frac{1}{n+1} int_0^1 x^n(1-x)^n$ is clearly positive, so you must have made a mistake somewhere. They seem to have equal magnitude though, so it's probably just a simple sign error.
$endgroup$
– David H
Apr 18 '14 at 21:47
|
show 4 more comments
$begingroup$
Can you start with en.wikipedia.org/wiki/Integration_by_parts#LIATE_rule ?
$endgroup$
– lab bhattacharjee
Apr 16 '14 at 16:56
4
$begingroup$
Expand the log into a Taylor series about $x-x^2$, perform the integration, and then recognize the resulting series as an expansion of a (sort of) well-known function evaluated at a particular value.
$endgroup$
– Ron Gordon
Apr 16 '14 at 16:59
$begingroup$
@Ron Gordon Mmmmh. Very interesting. I will work on it and will post what I come up with. Thanks
$endgroup$
– user2233524
Apr 16 '14 at 20:28
$begingroup$
Ok guys I added some stuff. Still need some help. Thanks
$endgroup$
– user2233524
Apr 18 '14 at 19:30
$begingroup$
@user2233524 $int_0^1 frac{ln(1-x+x^2)}{x-x^2}dx$ is less than zero, whereas $sum_{n=0}^{infty} frac{1}{n+1} int_0^1 x^n(1-x)^n$ is clearly positive, so you must have made a mistake somewhere. They seem to have equal magnitude though, so it's probably just a simple sign error.
$endgroup$
– David H
Apr 18 '14 at 21:47
$begingroup$
Can you start with en.wikipedia.org/wiki/Integration_by_parts#LIATE_rule ?
$endgroup$
– lab bhattacharjee
Apr 16 '14 at 16:56
$begingroup$
Can you start with en.wikipedia.org/wiki/Integration_by_parts#LIATE_rule ?
$endgroup$
– lab bhattacharjee
Apr 16 '14 at 16:56
4
4
$begingroup$
Expand the log into a Taylor series about $x-x^2$, perform the integration, and then recognize the resulting series as an expansion of a (sort of) well-known function evaluated at a particular value.
$endgroup$
– Ron Gordon
Apr 16 '14 at 16:59
$begingroup$
Expand the log into a Taylor series about $x-x^2$, perform the integration, and then recognize the resulting series as an expansion of a (sort of) well-known function evaluated at a particular value.
$endgroup$
– Ron Gordon
Apr 16 '14 at 16:59
$begingroup$
@Ron Gordon Mmmmh. Very interesting. I will work on it and will post what I come up with. Thanks
$endgroup$
– user2233524
Apr 16 '14 at 20:28
$begingroup$
@Ron Gordon Mmmmh. Very interesting. I will work on it and will post what I come up with. Thanks
$endgroup$
– user2233524
Apr 16 '14 at 20:28
$begingroup$
Ok guys I added some stuff. Still need some help. Thanks
$endgroup$
– user2233524
Apr 18 '14 at 19:30
$begingroup$
Ok guys I added some stuff. Still need some help. Thanks
$endgroup$
– user2233524
Apr 18 '14 at 19:30
$begingroup$
@user2233524 $int_0^1 frac{ln(1-x+x^2)}{x-x^2}dx$ is less than zero, whereas $sum_{n=0}^{infty} frac{1}{n+1} int_0^1 x^n(1-x)^n$ is clearly positive, so you must have made a mistake somewhere. They seem to have equal magnitude though, so it's probably just a simple sign error.
$endgroup$
– David H
Apr 18 '14 at 21:47
$begingroup$
@user2233524 $int_0^1 frac{ln(1-x+x^2)}{x-x^2}dx$ is less than zero, whereas $sum_{n=0}^{infty} frac{1}{n+1} int_0^1 x^n(1-x)^n$ is clearly positive, so you must have made a mistake somewhere. They seem to have equal magnitude though, so it's probably just a simple sign error.
$endgroup$
– David H
Apr 18 '14 at 21:47
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Just to simplify the things, make the change of variables $s=2x-1$. The integral then reduces to
$$I=2int_{-1}^1frac{lnfrac{3+s^2}{4}}{1-s^2}ds.tag{1}$$
The antiderivative of any expression of the type $displaystylefrac{ln P(x)}{Q(x)}$ is computable in terms of dilogarithms, essentially due to
$$displaystyle intfrac{ln(a-x)}{x+b}dx=mathrm{Li}_2left(frac{a-x}{a+b}right)+ln(a-x)lnfrac{x+b}{a+b}.tag{2}$$
Hence the answer can be certainly expressed in terms of dilogarithm values.
Let us spell this out more explicitly. It is convenient to integrate once by parts and rewrite (1) as
begin{align}I=&-int_{-1}^1frac{2s}{3+s^2}lnfrac{1+s}{1-s}ds=
4Reint_{-1}^1frac{ln(1-s)}{s+isqrt3}ds.
end{align}
Applying (2), this reduces to
$$I=-4Re,mathrm{Li}_2left(e^{ipi/3}right)=-frac{pi^2}{9},$$
where at the last step we have used that for $zin(0,1)$ one has
$$Re,mathrm{Li}_2left(e^{2ipi z}right)=pi^2left(z^2-z+frac16right).$$
answered Apr 19 '14 at 19:08
Start wearing purpleStart wearing purple
47.3k12135192
$endgroup$
$begingroup$
I don't understand the substitution you made in the integral.
$endgroup$
– user2233524
Apr 19 '14 at 20:25
$begingroup$
@user2233524 As $x^2-x= left(x-frac12right)^2-frac14$, I set $x-frac12=frac{s}{2}$ $Longleftrightarrow$ $s=2x-1$.
$endgroup$
– Start wearing purple
Apr 19 '14 at 20:29
$begingroup$
It is still unclear to me how you get the equalities after "It is convenient to integrate once by parts and rewrite (1) as." Where does the imaginary part come form?
$endgroup$
– user103828
Apr 20 '14 at 8:21
1
$begingroup$
@user103828 The first equality comes from $left(lnfrac{1+s}{1-s}right)'=frac{2}{1-s^2}$ and integration by parts. The second equality comes from the fact that $left(ln(3+s^2)right)'=frac{2s}{3+s^2}$ is odd function (so that we can replace $lnfrac{1+s}{1-s}$ by $-2ln(1-s)$) and the decomposition $frac{2s}{3+s^2}=frac{1}{s+isqrt3} + frac{1}{s-isqrt3} = 2Re frac{1}{s+isqrt3}$.
$endgroup$
– Start wearing purple
Apr 20 '14 at 8:40
$begingroup$
@O.L. +1, do you have proof/reference for (2). Seems like a useful thing to know. thanks
$endgroup$
– Jeff Faraci
Apr 21 '14 at 16:19
add a comment |
$begingroup$
Differentiation under the integral sign can be applied to this.
Consider
begin{align}
I(a)&=int_0^1 , frac{ln{(1-a,(x-x^2))}}{x-x^2}, dx tag 1\
frac{partial}{partial a} I(a)&=int_0^1, -frac{1}{1-a,(x-x^2)}, dx \
&= -frac{4 , sqrt{-y^{2} + 4 , y} arctanleft(frac{sqrt{-y^{2} + 4 , y}}{y - 4}right)}{y^{2} - 4 , y}tag 2
end{align}
Integrating $(2)$ gives us (I used Sage for that)
begin{align}
I(a)=sqrt{a} sqrt{-a + 4} arcsinleft(frac{1}{2} , a - 1right) + 2 , arcsinleft(frac{1}{2} , a - 1right)^{2} - 4 , arctanleft(frac{{left(a - 2right)} sqrt{a} sqrt{-a + 4}}{a^{2} - 4 , a}right) arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right) + 4 , arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right)^{2} + 4 , arcsinleft(frac{1}{2} , a - 1right) arctanleft(frac{sqrt{-a^{2} + 4 , a}}{a - 4}right) - sqrt{-a^{2} + 4 , a} arcsinleft(frac{1}{2} , a - 1right) + C \
tag 3
end{align}
Setting $a=0$ in $(2)$ and $(3)$, $C=-frac{pi^2}{2}$, hence:
begin{align}
I(a)=sqrt{a} sqrt{-a + 4} arcsinleft(frac{1}{2} , a - 1right) + 2 , arcsinleft(frac{1}{2} , a - 1right)^{2} - 4 , arctanleft(frac{{left(a - 2right)} sqrt{a} sqrt{-a + 4}}{a^{2} - 4 , a}right) arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right) + 4 , arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right)^{2} + 4 , arcsinleft(frac{1}{2} , a - 1right) arctanleft(frac{sqrt{-a^{2} + 4 , a}}{a - 4}right) - sqrt{-a^{2} + 4 , a} arcsinleft(frac{1}{2} , a - 1right)-frac{pi^2}{2}
end{align}
Therefore, the required answer is
$$
I(1)=-frac{pi^2}{9}approx -1.09662271123215
$$
=== Update ===
There is a simpler general form as I suspected:
Instead of $(1)$, consider $I(a)$ to be
begin{align}
I(a)&=int_0^1 , frac{ln{(1+a,(x-x^2))}}{x-x^2}, dx
end{align}
Applying the differentiation under the integral sign now yields a nice simpler form:
begin{align}
I(a)&=lnleft( frac{a+2 -sqrt{a^{2} + 4 , a}}{2}right)^{2}
end{align}
answered Apr 21 '14 at 5:20
gargar
4,6581123
$endgroup$
$begingroup$
Do you have an analytical proof of (2)? A numerical answer isn't much of a proof. The method of differentiation under the integral sign is very nice here though, +1
$endgroup$
– Jeff Faraci
Apr 21 '14 at 16:16
1
$begingroup$
I do not know how $(2)$ is evaluated by hand, there may be an alternative simpler answer since sometimes CAS evaluates to more complicated expressions. The differentiation under the integral sign makes possible to evaluate some integrals which otherwise cannot be evaluated even by a CAS.
$endgroup$
– gar
Apr 21 '14 at 17:18
add a comment |
$begingroup$
Here is a relatively simple way that only relies on knowing the following Maclarin series expansion for the square of the inverse sine function (for several different proofs of this, see here)
$$(sin^{-1} x)^2 = frac{1}{2} sum_{n = 1}^infty frac{(2x)^{2n}}{n^2 binom{2n}{n}}, qquad |x| leqslant 1.$$
Note that if we set $x = 1/2$ one obtains:
$$sum_{n = 1}^infty frac{1}{n binom{2n}{n}} = frac{pi^2}{18}. qquad (*)$$
Now
begin{align}
int_0^1 frac{ln (1 - x + x^2)}{x - x^2} , dx &= int_0^1 frac{ln [1 - (x - x^2)]}{x - x^2} , dx\
&= -int_0^1 sum_{n = 1}^infty frac{(x - x^2)^n}{n} frac{dx}{x - x^2} tag1\
&= -sum_{n = 1}^infty frac{1}{n} int_0^1 (x - x^2)^{n - 1} , dx tag2\
&= -sum_{n = 1}^infty frac{1}{n} int_0^1 x^{n - 1} (1 - x)^{n - 1} , dx\
&= -sum_{n = 1}^infty frac{1}{n} operatorname{B} (n,n) tag3\
&= -sum_{n = 1}^infty frac{1}{n} frac{Gamma (n) Gamma (n)}{Gamma (2n)} tag4\
&= -sum_{n = 1}^infty frac{1}{n} frac{(n - 1)! (n - 1)!}{(2n - 1)!}\
&= -2sum_{n = 1}^infty frac{1}{n} frac{(n!)^2}{(2n)!}\
&= -2 sum_{n = 1}^infty frac{1}{n binom{2n}{n}}\
&= -frac{pi^2}{9} tag5
end{align}
Explanation
(1): Maclaurin series expansion for $ln (1 - z)$.
(2): The dominated convergence theorem allows the summation and integration signs to be interchanged.
(3): Integral representation for the beta function.
(4): Using the property $operatorname{B} (x,y) = frac{Gamma (x) Gamma (y)}{Gamma (x + y)}$.
(5): Using the result given above in ($*$).
answered Jan 27 at 10:11
omegadotomegadot
6,4592829
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
$begingroup$
Just to simplify the things, make the change of variables $s=2x-1$. The integral then reduces to
$$I=2int_{-1}^1frac{lnfrac{3+s^2}{4}}{1-s^2}ds.tag{1}$$
The antiderivative of any expression of the type $displaystylefrac{ln P(x)}{Q(x)}$ is computable in terms of dilogarithms, essentially due to
$$displaystyle intfrac{ln(a-x)}{x+b}dx=mathrm{Li}_2left(frac{a-x}{a+b}right)+ln(a-x)lnfrac{x+b}{a+b}.tag{2}$$
Hence the answer can be certainly expressed in terms of dilogarithm values.
Let us spell this out more explicitly. It is convenient to integrate once by parts and rewrite (1) as
begin{align}I=&-int_{-1}^1frac{2s}{3+s^2}lnfrac{1+s}{1-s}ds=
4Reint_{-1}^1frac{ln(1-s)}{s+isqrt3}ds.
end{align}
Applying (2), this reduces to
$$I=-4Re,mathrm{Li}_2left(e^{ipi/3}right)=-frac{pi^2}{9},$$
where at the last step we have used that for $zin(0,1)$ one has
$$Re,mathrm{Li}_2left(e^{2ipi z}right)=pi^2left(z^2-z+frac16right).$$
answered Apr 19 '14 at 19:08
Start wearing purpleStart wearing purple
47.3k12135192
$endgroup$
$begingroup$
I don't understand the substitution you made in the integral.
$endgroup$
– user2233524
Apr 19 '14 at 20:25
$begingroup$
@user2233524 As $x^2-x= left(x-frac12right)^2-frac14$, I set $x-frac12=frac{s}{2}$ $Longleftrightarrow$ $s=2x-1$.
$endgroup$
– Start wearing purple
Apr 19 '14 at 20:29
$begingroup$
It is still unclear to me how you get the equalities after "It is convenient to integrate once by parts and rewrite (1) as." Where does the imaginary part come form?
$endgroup$
– user103828
Apr 20 '14 at 8:21
1
$begingroup$
@user103828 The first equality comes from $left(lnfrac{1+s}{1-s}right)'=frac{2}{1-s^2}$ and integration by parts. The second equality comes from the fact that $left(ln(3+s^2)right)'=frac{2s}{3+s^2}$ is odd function (so that we can replace $lnfrac{1+s}{1-s}$ by $-2ln(1-s)$) and the decomposition $frac{2s}{3+s^2}=frac{1}{s+isqrt3} + frac{1}{s-isqrt3} = 2Re frac{1}{s+isqrt3}$.
$endgroup$
– Start wearing purple
Apr 20 '14 at 8:40
$begingroup$
@O.L. +1, do you have proof/reference for (2). Seems like a useful thing to know. thanks
$endgroup$
– Jeff Faraci
Apr 21 '14 at 16:19
add a comment |
$begingroup$
Just to simplify the things, make the change of variables $s=2x-1$. The integral then reduces to
$$I=2int_{-1}^1frac{lnfrac{3+s^2}{4}}{1-s^2}ds.tag{1}$$
The antiderivative of any expression of the type $displaystylefrac{ln P(x)}{Q(x)}$ is computable in terms of dilogarithms, essentially due to
$$displaystyle intfrac{ln(a-x)}{x+b}dx=mathrm{Li}_2left(frac{a-x}{a+b}right)+ln(a-x)lnfrac{x+b}{a+b}.tag{2}$$
Hence the answer can be certainly expressed in terms of dilogarithm values.
Let us spell this out more explicitly. It is convenient to integrate once by parts and rewrite (1) as
begin{align}I=&-int_{-1}^1frac{2s}{3+s^2}lnfrac{1+s}{1-s}ds=
4Reint_{-1}^1frac{ln(1-s)}{s+isqrt3}ds.
end{align}
Applying (2), this reduces to
$$I=-4Re,mathrm{Li}_2left(e^{ipi/3}right)=-frac{pi^2}{9},$$
where at the last step we have used that for $zin(0,1)$ one has
$$Re,mathrm{Li}_2left(e^{2ipi z}right)=pi^2left(z^2-z+frac16right).$$
answered Apr 19 '14 at 19:08
Start wearing purpleStart wearing purple
47.3k12135192
$endgroup$
$begingroup$
I don't understand the substitution you made in the integral.
$endgroup$
– user2233524
Apr 19 '14 at 20:25
$begingroup$
@user2233524 As $x^2-x= left(x-frac12right)^2-frac14$, I set $x-frac12=frac{s}{2}$ $Longleftrightarrow$ $s=2x-1$.
$endgroup$
– Start wearing purple
Apr 19 '14 at 20:29
$begingroup$
It is still unclear to me how you get the equalities after "It is convenient to integrate once by parts and rewrite (1) as." Where does the imaginary part come form?
$endgroup$
– user103828
Apr 20 '14 at 8:21
1
$begingroup$
@user103828 The first equality comes from $left(lnfrac{1+s}{1-s}right)'=frac{2}{1-s^2}$ and integration by parts. The second equality comes from the fact that $left(ln(3+s^2)right)'=frac{2s}{3+s^2}$ is odd function (so that we can replace $lnfrac{1+s}{1-s}$ by $-2ln(1-s)$) and the decomposition $frac{2s}{3+s^2}=frac{1}{s+isqrt3} + frac{1}{s-isqrt3} = 2Re frac{1}{s+isqrt3}$.
$endgroup$
– Start wearing purple
Apr 20 '14 at 8:40
$begingroup$
@O.L. +1, do you have proof/reference for (2). Seems like a useful thing to know. thanks
$endgroup$
– Jeff Faraci
Apr 21 '14 at 16:19
add a comment |
$begingroup$
Just to simplify the things, make the change of variables $s=2x-1$. The integral then reduces to
$$I=2int_{-1}^1frac{lnfrac{3+s^2}{4}}{1-s^2}ds.tag{1}$$
The antiderivative of any expression of the type $displaystylefrac{ln P(x)}{Q(x)}$ is computable in terms of dilogarithms, essentially due to
$$displaystyle intfrac{ln(a-x)}{x+b}dx=mathrm{Li}_2left(frac{a-x}{a+b}right)+ln(a-x)lnfrac{x+b}{a+b}.tag{2}$$
Hence the answer can be certainly expressed in terms of dilogarithm values.
Let us spell this out more explicitly. It is convenient to integrate once by parts and rewrite (1) as
begin{align}I=&-int_{-1}^1frac{2s}{3+s^2}lnfrac{1+s}{1-s}ds=
4Reint_{-1}^1frac{ln(1-s)}{s+isqrt3}ds.
end{align}
Applying (2), this reduces to
$$I=-4Re,mathrm{Li}_2left(e^{ipi/3}right)=-frac{pi^2}{9},$$
where at the last step we have used that for $zin(0,1)$ one has
$$Re,mathrm{Li}_2left(e^{2ipi z}right)=pi^2left(z^2-z+frac16right).$$
answered Apr 19 '14 at 19:08
Start wearing purpleStart wearing purple
47.3k12135192
$endgroup$
Just to simplify the things, make the change of variables $s=2x-1$. The integral then reduces to
$$I=2int_{-1}^1frac{lnfrac{3+s^2}{4}}{1-s^2}ds.tag{1}$$
The antiderivative of any expression of the type $displaystylefrac{ln P(x)}{Q(x)}$ is computable in terms of dilogarithms, essentially due to
$$displaystyle intfrac{ln(a-x)}{x+b}dx=mathrm{Li}_2left(frac{a-x}{a+b}right)+ln(a-x)lnfrac{x+b}{a+b}.tag{2}$$
Hence the answer can be certainly expressed in terms of dilogarithm values.
Let us spell this out more explicitly. It is convenient to integrate once by parts and rewrite (1) as
begin{align}I=&-int_{-1}^1frac{2s}{3+s^2}lnfrac{1+s}{1-s}ds=
4Reint_{-1}^1frac{ln(1-s)}{s+isqrt3}ds.
end{align}
Applying (2), this reduces to
$$I=-4Re,mathrm{Li}_2left(e^{ipi/3}right)=-frac{pi^2}{9},$$
where at the last step we have used that for $zin(0,1)$ one has
$$Re,mathrm{Li}_2left(e^{2ipi z}right)=pi^2left(z^2-z+frac16right).$$
answered Apr 19 '14 at 19:08
Start wearing purpleStart wearing purple
47.3k12135192
answered Apr 19 '14 at 19:08
Start wearing purpleStart wearing purple
47.3k12135192
answered Apr 19 '14 at 19:08
Start wearing purpleStart wearing purple
47.3k12135192
answered Apr 19 '14 at 19:08
Start wearing purpleStart wearing purple
47.3k12135192
47.3k12135192
$begingroup$
I don't understand the substitution you made in the integral.
$endgroup$
– user2233524
Apr 19 '14 at 20:25
$begingroup$
@user2233524 As $x^2-x= left(x-frac12right)^2-frac14$, I set $x-frac12=frac{s}{2}$ $Longleftrightarrow$ $s=2x-1$.
$endgroup$
– Start wearing purple
Apr 19 '14 at 20:29
$begingroup$
It is still unclear to me how you get the equalities after "It is convenient to integrate once by parts and rewrite (1) as." Where does the imaginary part come form?
$endgroup$
– user103828
Apr 20 '14 at 8:21
1
$begingroup$
@user103828 The first equality comes from $left(lnfrac{1+s}{1-s}right)'=frac{2}{1-s^2}$ and integration by parts. The second equality comes from the fact that $left(ln(3+s^2)right)'=frac{2s}{3+s^2}$ is odd function (so that we can replace $lnfrac{1+s}{1-s}$ by $-2ln(1-s)$) and the decomposition $frac{2s}{3+s^2}=frac{1}{s+isqrt3} + frac{1}{s-isqrt3} = 2Re frac{1}{s+isqrt3}$.
$endgroup$
– Start wearing purple
Apr 20 '14 at 8:40
$begingroup$
@O.L. +1, do you have proof/reference for (2). Seems like a useful thing to know. thanks
$endgroup$
– Jeff Faraci
Apr 21 '14 at 16:19
add a comment |
$begingroup$
I don't understand the substitution you made in the integral.
$endgroup$
– user2233524
Apr 19 '14 at 20:25
$begingroup$
@user2233524 As $x^2-x= left(x-frac12right)^2-frac14$, I set $x-frac12=frac{s}{2}$ $Longleftrightarrow$ $s=2x-1$.
$endgroup$
– Start wearing purple
Apr 19 '14 at 20:29
$begingroup$
It is still unclear to me how you get the equalities after "It is convenient to integrate once by parts and rewrite (1) as." Where does the imaginary part come form?
$endgroup$
– user103828
Apr 20 '14 at 8:21
1
$begingroup$
@user103828 The first equality comes from $left(lnfrac{1+s}{1-s}right)'=frac{2}{1-s^2}$ and integration by parts. The second equality comes from the fact that $left(ln(3+s^2)right)'=frac{2s}{3+s^2}$ is odd function (so that we can replace $lnfrac{1+s}{1-s}$ by $-2ln(1-s)$) and the decomposition $frac{2s}{3+s^2}=frac{1}{s+isqrt3} + frac{1}{s-isqrt3} = 2Re frac{1}{s+isqrt3}$.
$endgroup$
– Start wearing purple
Apr 20 '14 at 8:40
$begingroup$
@O.L. +1, do you have proof/reference for (2). Seems like a useful thing to know. thanks
$endgroup$
– Jeff Faraci
Apr 21 '14 at 16:19
$begingroup$
I don't understand the substitution you made in the integral.
$endgroup$
– user2233524
Apr 19 '14 at 20:25
$begingroup$
I don't understand the substitution you made in the integral.
$endgroup$
– user2233524
Apr 19 '14 at 20:25
$begingroup$
@user2233524 As $x^2-x= left(x-frac12right)^2-frac14$, I set $x-frac12=frac{s}{2}$ $Longleftrightarrow$ $s=2x-1$.
$endgroup$
– Start wearing purple
Apr 19 '14 at 20:29
$begingroup$
@user2233524 As $x^2-x= left(x-frac12right)^2-frac14$, I set $x-frac12=frac{s}{2}$ $Longleftrightarrow$ $s=2x-1$.
$endgroup$
– Start wearing purple
Apr 19 '14 at 20:29
$begingroup$
It is still unclear to me how you get the equalities after "It is convenient to integrate once by parts and rewrite (1) as." Where does the imaginary part come form?
$endgroup$
– user103828
Apr 20 '14 at 8:21
$begingroup$
It is still unclear to me how you get the equalities after "It is convenient to integrate once by parts and rewrite (1) as." Where does the imaginary part come form?
$endgroup$
– user103828
Apr 20 '14 at 8:21
1
1
$begingroup$
@user103828 The first equality comes from $left(lnfrac{1+s}{1-s}right)'=frac{2}{1-s^2}$ and integration by parts. The second equality comes from the fact that $left(ln(3+s^2)right)'=frac{2s}{3+s^2}$ is odd function (so that we can replace $lnfrac{1+s}{1-s}$ by $-2ln(1-s)$) and the decomposition $frac{2s}{3+s^2}=frac{1}{s+isqrt3} + frac{1}{s-isqrt3} = 2Re frac{1}{s+isqrt3}$.
$endgroup$
– Start wearing purple
Apr 20 '14 at 8:40
$begingroup$
@user103828 The first equality comes from $left(lnfrac{1+s}{1-s}right)'=frac{2}{1-s^2}$ and integration by parts. The second equality comes from the fact that $left(ln(3+s^2)right)'=frac{2s}{3+s^2}$ is odd function (so that we can replace $lnfrac{1+s}{1-s}$ by $-2ln(1-s)$) and the decomposition $frac{2s}{3+s^2}=frac{1}{s+isqrt3} + frac{1}{s-isqrt3} = 2Re frac{1}{s+isqrt3}$.
$endgroup$
– Start wearing purple
Apr 20 '14 at 8:40
$begingroup$
@O.L. +1, do you have proof/reference for (2). Seems like a useful thing to know. thanks
$endgroup$
– Jeff Faraci
Apr 21 '14 at 16:19
$begingroup$
@O.L. +1, do you have proof/reference for (2). Seems like a useful thing to know. thanks
$endgroup$
– Jeff Faraci
Apr 21 '14 at 16:19
add a comment |
$begingroup$
Differentiation under the integral sign can be applied to this.
Consider
begin{align}
I(a)&=int_0^1 , frac{ln{(1-a,(x-x^2))}}{x-x^2}, dx tag 1\
frac{partial}{partial a} I(a)&=int_0^1, -frac{1}{1-a,(x-x^2)}, dx \
&= -frac{4 , sqrt{-y^{2} + 4 , y} arctanleft(frac{sqrt{-y^{2} + 4 , y}}{y - 4}right)}{y^{2} - 4 , y}tag 2
end{align}
Integrating $(2)$ gives us (I used Sage for that)
begin{align}
I(a)=sqrt{a} sqrt{-a + 4} arcsinleft(frac{1}{2} , a - 1right) + 2 , arcsinleft(frac{1}{2} , a - 1right)^{2} - 4 , arctanleft(frac{{left(a - 2right)} sqrt{a} sqrt{-a + 4}}{a^{2} - 4 , a}right) arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right) + 4 , arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right)^{2} + 4 , arcsinleft(frac{1}{2} , a - 1right) arctanleft(frac{sqrt{-a^{2} + 4 , a}}{a - 4}right) - sqrt{-a^{2} + 4 , a} arcsinleft(frac{1}{2} , a - 1right) + C \
tag 3
end{align}
Setting $a=0$ in $(2)$ and $(3)$, $C=-frac{pi^2}{2}$, hence:
begin{align}
I(a)=sqrt{a} sqrt{-a + 4} arcsinleft(frac{1}{2} , a - 1right) + 2 , arcsinleft(frac{1}{2} , a - 1right)^{2} - 4 , arctanleft(frac{{left(a - 2right)} sqrt{a} sqrt{-a + 4}}{a^{2} - 4 , a}right) arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right) + 4 , arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right)^{2} + 4 , arcsinleft(frac{1}{2} , a - 1right) arctanleft(frac{sqrt{-a^{2} + 4 , a}}{a - 4}right) - sqrt{-a^{2} + 4 , a} arcsinleft(frac{1}{2} , a - 1right)-frac{pi^2}{2}
end{align}
Therefore, the required answer is
$$
I(1)=-frac{pi^2}{9}approx -1.09662271123215
$$
=== Update ===
There is a simpler general form as I suspected:
Instead of $(1)$, consider $I(a)$ to be
begin{align}
I(a)&=int_0^1 , frac{ln{(1+a,(x-x^2))}}{x-x^2}, dx
end{align}
Applying the differentiation under the integral sign now yields a nice simpler form:
begin{align}
I(a)&=lnleft( frac{a+2 -sqrt{a^{2} + 4 , a}}{2}right)^{2}
end{align}
answered Apr 21 '14 at 5:20
gargar
4,6581123
$endgroup$
$begingroup$
Do you have an analytical proof of (2)? A numerical answer isn't much of a proof. The method of differentiation under the integral sign is very nice here though, +1
$endgroup$
– Jeff Faraci
Apr 21 '14 at 16:16
1
$begingroup$
I do not know how $(2)$ is evaluated by hand, there may be an alternative simpler answer since sometimes CAS evaluates to more complicated expressions. The differentiation under the integral sign makes possible to evaluate some integrals which otherwise cannot be evaluated even by a CAS.
$endgroup$
– gar
Apr 21 '14 at 17:18
add a comment |
$begingroup$
Differentiation under the integral sign can be applied to this.
Consider
begin{align}
I(a)&=int_0^1 , frac{ln{(1-a,(x-x^2))}}{x-x^2}, dx tag 1\
frac{partial}{partial a} I(a)&=int_0^1, -frac{1}{1-a,(x-x^2)}, dx \
&= -frac{4 , sqrt{-y^{2} + 4 , y} arctanleft(frac{sqrt{-y^{2} + 4 , y}}{y - 4}right)}{y^{2} - 4 , y}tag 2
end{align}
Integrating $(2)$ gives us (I used Sage for that)
begin{align}
I(a)=sqrt{a} sqrt{-a + 4} arcsinleft(frac{1}{2} , a - 1right) + 2 , arcsinleft(frac{1}{2} , a - 1right)^{2} - 4 , arctanleft(frac{{left(a - 2right)} sqrt{a} sqrt{-a + 4}}{a^{2} - 4 , a}right) arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right) + 4 , arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right)^{2} + 4 , arcsinleft(frac{1}{2} , a - 1right) arctanleft(frac{sqrt{-a^{2} + 4 , a}}{a - 4}right) - sqrt{-a^{2} + 4 , a} arcsinleft(frac{1}{2} , a - 1right) + C \
tag 3
end{align}
Setting $a=0$ in $(2)$ and $(3)$, $C=-frac{pi^2}{2}$, hence:
begin{align}
I(a)=sqrt{a} sqrt{-a + 4} arcsinleft(frac{1}{2} , a - 1right) + 2 , arcsinleft(frac{1}{2} , a - 1right)^{2} - 4 , arctanleft(frac{{left(a - 2right)} sqrt{a} sqrt{-a + 4}}{a^{2} - 4 , a}right) arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right) + 4 , arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right)^{2} + 4 , arcsinleft(frac{1}{2} , a - 1right) arctanleft(frac{sqrt{-a^{2} + 4 , a}}{a - 4}right) - sqrt{-a^{2} + 4 , a} arcsinleft(frac{1}{2} , a - 1right)-frac{pi^2}{2}
end{align}
Therefore, the required answer is
$$
I(1)=-frac{pi^2}{9}approx -1.09662271123215
$$
=== Update ===
There is a simpler general form as I suspected:
Instead of $(1)$, consider $I(a)$ to be
begin{align}
I(a)&=int_0^1 , frac{ln{(1+a,(x-x^2))}}{x-x^2}, dx
end{align}
Applying the differentiation under the integral sign now yields a nice simpler form:
begin{align}
I(a)&=lnleft( frac{a+2 -sqrt{a^{2} + 4 , a}}{2}right)^{2}
end{align}
answered Apr 21 '14 at 5:20
gargar
4,6581123
$endgroup$
$begingroup$
Do you have an analytical proof of (2)? A numerical answer isn't much of a proof. The method of differentiation under the integral sign is very nice here though, +1
$endgroup$
– Jeff Faraci
Apr 21 '14 at 16:16
1
$begingroup$
I do not know how $(2)$ is evaluated by hand, there may be an alternative simpler answer since sometimes CAS evaluates to more complicated expressions. The differentiation under the integral sign makes possible to evaluate some integrals which otherwise cannot be evaluated even by a CAS.
$endgroup$
– gar
Apr 21 '14 at 17:18
add a comment |
$begingroup$
Differentiation under the integral sign can be applied to this.
Consider
begin{align}
I(a)&=int_0^1 , frac{ln{(1-a,(x-x^2))}}{x-x^2}, dx tag 1\
frac{partial}{partial a} I(a)&=int_0^1, -frac{1}{1-a,(x-x^2)}, dx \
&= -frac{4 , sqrt{-y^{2} + 4 , y} arctanleft(frac{sqrt{-y^{2} + 4 , y}}{y - 4}right)}{y^{2} - 4 , y}tag 2
end{align}
Integrating $(2)$ gives us (I used Sage for that)
begin{align}
I(a)=sqrt{a} sqrt{-a + 4} arcsinleft(frac{1}{2} , a - 1right) + 2 , arcsinleft(frac{1}{2} , a - 1right)^{2} - 4 , arctanleft(frac{{left(a - 2right)} sqrt{a} sqrt{-a + 4}}{a^{2} - 4 , a}right) arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right) + 4 , arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right)^{2} + 4 , arcsinleft(frac{1}{2} , a - 1right) arctanleft(frac{sqrt{-a^{2} + 4 , a}}{a - 4}right) - sqrt{-a^{2} + 4 , a} arcsinleft(frac{1}{2} , a - 1right) + C \
tag 3
end{align}
Setting $a=0$ in $(2)$ and $(3)$, $C=-frac{pi^2}{2}$, hence:
begin{align}
I(a)=sqrt{a} sqrt{-a + 4} arcsinleft(frac{1}{2} , a - 1right) + 2 , arcsinleft(frac{1}{2} , a - 1right)^{2} - 4 , arctanleft(frac{{left(a - 2right)} sqrt{a} sqrt{-a + 4}}{a^{2} - 4 , a}right) arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right) + 4 , arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right)^{2} + 4 , arcsinleft(frac{1}{2} , a - 1right) arctanleft(frac{sqrt{-a^{2} + 4 , a}}{a - 4}right) - sqrt{-a^{2} + 4 , a} arcsinleft(frac{1}{2} , a - 1right)-frac{pi^2}{2}
end{align}
Therefore, the required answer is
$$
I(1)=-frac{pi^2}{9}approx -1.09662271123215
$$
=== Update ===
There is a simpler general form as I suspected:
Instead of $(1)$, consider $I(a)$ to be
begin{align}
I(a)&=int_0^1 , frac{ln{(1+a,(x-x^2))}}{x-x^2}, dx
end{align}
Applying the differentiation under the integral sign now yields a nice simpler form:
begin{align}
I(a)&=lnleft( frac{a+2 -sqrt{a^{2} + 4 , a}}{2}right)^{2}
end{align}
answered Apr 21 '14 at 5:20
gargar
4,6581123
$endgroup$
Differentiation under the integral sign can be applied to this.
Consider
begin{align}
I(a)&=int_0^1 , frac{ln{(1-a,(x-x^2))}}{x-x^2}, dx tag 1\
frac{partial}{partial a} I(a)&=int_0^1, -frac{1}{1-a,(x-x^2)}, dx \
&= -frac{4 , sqrt{-y^{2} + 4 , y} arctanleft(frac{sqrt{-y^{2} + 4 , y}}{y - 4}right)}{y^{2} - 4 , y}tag 2
end{align}
Integrating $(2)$ gives us (I used Sage for that)
begin{align}
I(a)=sqrt{a} sqrt{-a + 4} arcsinleft(frac{1}{2} , a - 1right) + 2 , arcsinleft(frac{1}{2} , a - 1right)^{2} - 4 , arctanleft(frac{{left(a - 2right)} sqrt{a} sqrt{-a + 4}}{a^{2} - 4 , a}right) arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right) + 4 , arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right)^{2} + 4 , arcsinleft(frac{1}{2} , a - 1right) arctanleft(frac{sqrt{-a^{2} + 4 , a}}{a - 4}right) - sqrt{-a^{2} + 4 , a} arcsinleft(frac{1}{2} , a - 1right) + C \
tag 3
end{align}
Setting $a=0$ in $(2)$ and $(3)$, $C=-frac{pi^2}{2}$, hence:
begin{align}
I(a)=sqrt{a} sqrt{-a + 4} arcsinleft(frac{1}{2} , a - 1right) + 2 , arcsinleft(frac{1}{2} , a - 1right)^{2} - 4 , arctanleft(frac{{left(a - 2right)} sqrt{a} sqrt{-a + 4}}{a^{2} - 4 , a}right) arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right) + 4 , arctanleft(frac{sqrt{-a + 4}}{sqrt{a}}right)^{2} + 4 , arcsinleft(frac{1}{2} , a - 1right) arctanleft(frac{sqrt{-a^{2} + 4 , a}}{a - 4}right) - sqrt{-a^{2} + 4 , a} arcsinleft(frac{1}{2} , a - 1right)-frac{pi^2}{2}
end{align}
Therefore, the required answer is
$$
I(1)=-frac{pi^2}{9}approx -1.09662271123215
$$
=== Update ===
There is a simpler general form as I suspected:
Instead of $(1)$, consider $I(a)$ to be
begin{align}
I(a)&=int_0^1 , frac{ln{(1+a,(x-x^2))}}{x-x^2}, dx
end{align}
Applying the differentiation under the integral sign now yields a nice simpler form:
begin{align}
I(a)&=lnleft( frac{a+2 -sqrt{a^{2} + 4 , a}}{2}right)^{2}
end{align}
answered Apr 21 '14 at 5:20
gargar
4,6581123
edited Apr 22 '14 at 10:28
answered Apr 21 '14 at 5:20
gargar
4,6581123
answered Apr 21 '14 at 5:20
gargar
4,6581123
answered Apr 21 '14 at 5:20
gargar
4,6581123
4,6581123
$begingroup$
Do you have an analytical proof of (2)? A numerical answer isn't much of a proof. The method of differentiation under the integral sign is very nice here though, +1
$endgroup$
– Jeff Faraci
Apr 21 '14 at 16:16
1
$begingroup$
I do not know how $(2)$ is evaluated by hand, there may be an alternative simpler answer since sometimes CAS evaluates to more complicated expressions. The differentiation under the integral sign makes possible to evaluate some integrals which otherwise cannot be evaluated even by a CAS.
$endgroup$
– gar
Apr 21 '14 at 17:18
add a comment |
$begingroup$
Do you have an analytical proof of (2)? A numerical answer isn't much of a proof. The method of differentiation under the integral sign is very nice here though, +1
$endgroup$
– Jeff Faraci
Apr 21 '14 at 16:16
1
$begingroup$
I do not know how $(2)$ is evaluated by hand, there may be an alternative simpler answer since sometimes CAS evaluates to more complicated expressions. The differentiation under the integral sign makes possible to evaluate some integrals which otherwise cannot be evaluated even by a CAS.
$endgroup$
– gar
Apr 21 '14 at 17:18
$begingroup$
Do you have an analytical proof of (2)? A numerical answer isn't much of a proof. The method of differentiation under the integral sign is very nice here though, +1
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– Jeff Faraci
Apr 21 '14 at 16:16
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Do you have an analytical proof of (2)? A numerical answer isn't much of a proof. The method of differentiation under the integral sign is very nice here though, +1
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– Jeff Faraci
Apr 21 '14 at 16:16
1
1
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I do not know how $(2)$ is evaluated by hand, there may be an alternative simpler answer since sometimes CAS evaluates to more complicated expressions. The differentiation under the integral sign makes possible to evaluate some integrals which otherwise cannot be evaluated even by a CAS.
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– gar
Apr 21 '14 at 17:18
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I do not know how $(2)$ is evaluated by hand, there may be an alternative simpler answer since sometimes CAS evaluates to more complicated expressions. The differentiation under the integral sign makes possible to evaluate some integrals which otherwise cannot be evaluated even by a CAS.
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– gar
Apr 21 '14 at 17:18
add a comment |
$begingroup$
Here is a relatively simple way that only relies on knowing the following Maclarin series expansion for the square of the inverse sine function (for several different proofs of this, see here)
$$(sin^{-1} x)^2 = frac{1}{2} sum_{n = 1}^infty frac{(2x)^{2n}}{n^2 binom{2n}{n}}, qquad |x| leqslant 1.$$
Note that if we set $x = 1/2$ one obtains:
$$sum_{n = 1}^infty frac{1}{n binom{2n}{n}} = frac{pi^2}{18}. qquad (*)$$
Now
begin{align}
int_0^1 frac{ln (1 - x + x^2)}{x - x^2} , dx &= int_0^1 frac{ln [1 - (x - x^2)]}{x - x^2} , dx\
&= -int_0^1 sum_{n = 1}^infty frac{(x - x^2)^n}{n} frac{dx}{x - x^2} tag1\
&= -sum_{n = 1}^infty frac{1}{n} int_0^1 (x - x^2)^{n - 1} , dx tag2\
&= -sum_{n = 1}^infty frac{1}{n} int_0^1 x^{n - 1} (1 - x)^{n - 1} , dx\
&= -sum_{n = 1}^infty frac{1}{n} operatorname{B} (n,n) tag3\
&= -sum_{n = 1}^infty frac{1}{n} frac{Gamma (n) Gamma (n)}{Gamma (2n)} tag4\
&= -sum_{n = 1}^infty frac{1}{n} frac{(n - 1)! (n - 1)!}{(2n - 1)!}\
&= -2sum_{n = 1}^infty frac{1}{n} frac{(n!)^2}{(2n)!}\
&= -2 sum_{n = 1}^infty frac{1}{n binom{2n}{n}}\
&= -frac{pi^2}{9} tag5
end{align}
Explanation
(1): Maclaurin series expansion for $ln (1 - z)$.
(2): The dominated convergence theorem allows the summation and integration signs to be interchanged.
(3): Integral representation for the beta function.
(4): Using the property $operatorname{B} (x,y) = frac{Gamma (x) Gamma (y)}{Gamma (x + y)}$.
(5): Using the result given above in ($*$).
answered Jan 27 at 10:11
omegadotomegadot
6,4592829
$endgroup$
add a comment |
$begingroup$
Here is a relatively simple way that only relies on knowing the following Maclarin series expansion for the square of the inverse sine function (for several different proofs of this, see here)
$$(sin^{-1} x)^2 = frac{1}{2} sum_{n = 1}^infty frac{(2x)^{2n}}{n^2 binom{2n}{n}}, qquad |x| leqslant 1.$$
Note that if we set $x = 1/2$ one obtains:
$$sum_{n = 1}^infty frac{1}{n binom{2n}{n}} = frac{pi^2}{18}. qquad (*)$$
Now
begin{align}
int_0^1 frac{ln (1 - x + x^2)}{x - x^2} , dx &= int_0^1 frac{ln [1 - (x - x^2)]}{x - x^2} , dx\
&= -int_0^1 sum_{n = 1}^infty frac{(x - x^2)^n}{n} frac{dx}{x - x^2} tag1\
&= -sum_{n = 1}^infty frac{1}{n} int_0^1 (x - x^2)^{n - 1} , dx tag2\
&= -sum_{n = 1}^infty frac{1}{n} int_0^1 x^{n - 1} (1 - x)^{n - 1} , dx\
&= -sum_{n = 1}^infty frac{1}{n} operatorname{B} (n,n) tag3\
&= -sum_{n = 1}^infty frac{1}{n} frac{Gamma (n) Gamma (n)}{Gamma (2n)} tag4\
&= -sum_{n = 1}^infty frac{1}{n} frac{(n - 1)! (n - 1)!}{(2n - 1)!}\
&= -2sum_{n = 1}^infty frac{1}{n} frac{(n!)^2}{(2n)!}\
&= -2 sum_{n = 1}^infty frac{1}{n binom{2n}{n}}\
&= -frac{pi^2}{9} tag5
end{align}
Explanation
(1): Maclaurin series expansion for $ln (1 - z)$.
(2): The dominated convergence theorem allows the summation and integration signs to be interchanged.
(3): Integral representation for the beta function.
(4): Using the property $operatorname{B} (x,y) = frac{Gamma (x) Gamma (y)}{Gamma (x + y)}$.
(5): Using the result given above in ($*$).
answered Jan 27 at 10:11
omegadotomegadot
6,4592829
$endgroup$
add a comment |
$begingroup$
Here is a relatively simple way that only relies on knowing the following Maclarin series expansion for the square of the inverse sine function (for several different proofs of this, see here)
$$(sin^{-1} x)^2 = frac{1}{2} sum_{n = 1}^infty frac{(2x)^{2n}}{n^2 binom{2n}{n}}, qquad |x| leqslant 1.$$
Note that if we set $x = 1/2$ one obtains:
$$sum_{n = 1}^infty frac{1}{n binom{2n}{n}} = frac{pi^2}{18}. qquad (*)$$
Now
begin{align}
int_0^1 frac{ln (1 - x + x^2)}{x - x^2} , dx &= int_0^1 frac{ln [1 - (x - x^2)]}{x - x^2} , dx\
&= -int_0^1 sum_{n = 1}^infty frac{(x - x^2)^n}{n} frac{dx}{x - x^2} tag1\
&= -sum_{n = 1}^infty frac{1}{n} int_0^1 (x - x^2)^{n - 1} , dx tag2\
&= -sum_{n = 1}^infty frac{1}{n} int_0^1 x^{n - 1} (1 - x)^{n - 1} , dx\
&= -sum_{n = 1}^infty frac{1}{n} operatorname{B} (n,n) tag3\
&= -sum_{n = 1}^infty frac{1}{n} frac{Gamma (n) Gamma (n)}{Gamma (2n)} tag4\
&= -sum_{n = 1}^infty frac{1}{n} frac{(n - 1)! (n - 1)!}{(2n - 1)!}\
&= -2sum_{n = 1}^infty frac{1}{n} frac{(n!)^2}{(2n)!}\
&= -2 sum_{n = 1}^infty frac{1}{n binom{2n}{n}}\
&= -frac{pi^2}{9} tag5
end{align}
Explanation
(1): Maclaurin series expansion for $ln (1 - z)$.
(2): The dominated convergence theorem allows the summation and integration signs to be interchanged.
(3): Integral representation for the beta function.
(4): Using the property $operatorname{B} (x,y) = frac{Gamma (x) Gamma (y)}{Gamma (x + y)}$.
(5): Using the result given above in ($*$).
answered Jan 27 at 10:11
omegadotomegadot
6,4592829
$endgroup$
Here is a relatively simple way that only relies on knowing the following Maclarin series expansion for the square of the inverse sine function (for several different proofs of this, see here)
$$(sin^{-1} x)^2 = frac{1}{2} sum_{n = 1}^infty frac{(2x)^{2n}}{n^2 binom{2n}{n}}, qquad |x| leqslant 1.$$
Note that if we set $x = 1/2$ one obtains:
$$sum_{n = 1}^infty frac{1}{n binom{2n}{n}} = frac{pi^2}{18}. qquad (*)$$
Now
begin{align}
int_0^1 frac{ln (1 - x + x^2)}{x - x^2} , dx &= int_0^1 frac{ln [1 - (x - x^2)]}{x - x^2} , dx\
&= -int_0^1 sum_{n = 1}^infty frac{(x - x^2)^n}{n} frac{dx}{x - x^2} tag1\
&= -sum_{n = 1}^infty frac{1}{n} int_0^1 (x - x^2)^{n - 1} , dx tag2\
&= -sum_{n = 1}^infty frac{1}{n} int_0^1 x^{n - 1} (1 - x)^{n - 1} , dx\
&= -sum_{n = 1}^infty frac{1}{n} operatorname{B} (n,n) tag3\
&= -sum_{n = 1}^infty frac{1}{n} frac{Gamma (n) Gamma (n)}{Gamma (2n)} tag4\
&= -sum_{n = 1}^infty frac{1}{n} frac{(n - 1)! (n - 1)!}{(2n - 1)!}\
&= -2sum_{n = 1}^infty frac{1}{n} frac{(n!)^2}{(2n)!}\
&= -2 sum_{n = 1}^infty frac{1}{n binom{2n}{n}}\
&= -frac{pi^2}{9} tag5
end{align}
Explanation
(1): Maclaurin series expansion for $ln (1 - z)$.
(2): The dominated convergence theorem allows the summation and integration signs to be interchanged.
(3): Integral representation for the beta function.
(4): Using the property $operatorname{B} (x,y) = frac{Gamma (x) Gamma (y)}{Gamma (x + y)}$.
(5): Using the result given above in ($*$).
answered Jan 27 at 10:11
omegadotomegadot
6,4592829
answered Jan 27 at 10:11
omegadotomegadot
6,4592829
answered Jan 27 at 10:11
omegadotomegadot
6,4592829
answered Jan 27 at 10:11
omegadotomegadot
6,4592829
6,4592829
add a comment |
add a comment |
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Can you start with en.wikipedia.org/wiki/Integration_by_parts#LIATE_rule ?
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– lab bhattacharjee
Apr 16 '14 at 16:56
4
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Expand the log into a Taylor series about $x-x^2$, perform the integration, and then recognize the resulting series as an expansion of a (sort of) well-known function evaluated at a particular value.
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– Ron Gordon
Apr 16 '14 at 16:59
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@Ron Gordon Mmmmh. Very interesting. I will work on it and will post what I come up with. Thanks
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– user2233524
Apr 16 '14 at 20:28
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Ok guys I added some stuff. Still need some help. Thanks
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– user2233524
Apr 18 '14 at 19:30
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@user2233524 $int_0^1 frac{ln(1-x+x^2)}{x-x^2}dx$ is less than zero, whereas $sum_{n=0}^{infty} frac{1}{n+1} int_0^1 x^n(1-x)^n$ is clearly positive, so you must have made a mistake somewhere. They seem to have equal magnitude though, so it's probably just a simple sign error.
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– David H
Apr 18 '14 at 21:47