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A double sum or a definite integral.
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I am trying to evaluate the following double sum
begineqnarray*
sum_n=1^infty sum_m=1^infty frac(-1)^n+mn(3n+m).
endeqnarray*
Using the integral trick
begineqnarray*
frac13n+m =int_0^1 y^3n+m-1 dy,
endeqnarray*
the sum can be transformed into integral
begineqnarray*
int_0^1 frac ln(1+y^3)1+y dy.
endeqnarray*
Now "half" of this is easy (IBP & rearrange)
begineqnarray*
int_0^1 frac ln(1+y)1+y dy = frac12 (ln 2)^2.
endeqnarray*
So we are left with
begineqnarray*
int_0^1 frac ln(1-y+y^2)1+y dy = int_0^1 frac (1-2y)ln(1+y)1-y+y^2 dy.
endeqnarray*
Now, apart from the obvious IBP done above, this integral has me stumped.
An exact evaluation would be nice, barring that, an expression in terms of dilogrithmic values or something similar would be helpful. Any comments or answers, gratefully received.
If this integral has been seen before a reference & advise on how I would search & find something similar in the future would be gratefully appreciated.
integration sequences-and-series definite-integrals summation
edited 55 mins ago
Frank W.
2,1461315
asked 6 hours ago
Donald Splutterwit
21.5k21244
add a comment |Â
up vote
5
down vote
favorite
I am trying to evaluate the following double sum
begineqnarray*
sum_n=1^infty sum_m=1^infty frac(-1)^n+mn(3n+m).
endeqnarray*
Using the integral trick
begineqnarray*
frac13n+m =int_0^1 y^3n+m-1 dy,
endeqnarray*
the sum can be transformed into integral
begineqnarray*
int_0^1 frac ln(1+y^3)1+y dy.
endeqnarray*
Now "half" of this is easy (IBP & rearrange)
begineqnarray*
int_0^1 frac ln(1+y)1+y dy = frac12 (ln 2)^2.
endeqnarray*
So we are left with
begineqnarray*
int_0^1 frac ln(1-y+y^2)1+y dy = int_0^1 frac (1-2y)ln(1+y)1-y+y^2 dy.
endeqnarray*
Now, apart from the obvious IBP done above, this integral has me stumped.
An exact evaluation would be nice, barring that, an expression in terms of dilogrithmic values or something similar would be helpful. Any comments or answers, gratefully received.
If this integral has been seen before a reference & advise on how I would search & find something similar in the future would be gratefully appreciated.
integration sequences-and-series definite-integrals summation
edited 55 mins ago
Frank W.
2,1461315
asked 6 hours ago
Donald Splutterwit
21.5k21244
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I am trying to evaluate the following double sum
begineqnarray*
sum_n=1^infty sum_m=1^infty frac(-1)^n+mn(3n+m).
endeqnarray*
Using the integral trick
begineqnarray*
frac13n+m =int_0^1 y^3n+m-1 dy,
endeqnarray*
the sum can be transformed into integral
begineqnarray*
int_0^1 frac ln(1+y^3)1+y dy.
endeqnarray*
Now "half" of this is easy (IBP & rearrange)
begineqnarray*
int_0^1 frac ln(1+y)1+y dy = frac12 (ln 2)^2.
endeqnarray*
So we are left with
begineqnarray*
int_0^1 frac ln(1-y+y^2)1+y dy = int_0^1 frac (1-2y)ln(1+y)1-y+y^2 dy.
endeqnarray*
Now, apart from the obvious IBP done above, this integral has me stumped.
An exact evaluation would be nice, barring that, an expression in terms of dilogrithmic values or something similar would be helpful. Any comments or answers, gratefully received.
If this integral has been seen before a reference & advise on how I would search & find something similar in the future would be gratefully appreciated.
integration sequences-and-series definite-integrals summation
edited 55 mins ago
Frank W.
2,1461315
asked 6 hours ago
Donald Splutterwit
21.5k21244
I am trying to evaluate the following double sum
begineqnarray*
sum_n=1^infty sum_m=1^infty frac(-1)^n+mn(3n+m).
endeqnarray*
Using the integral trick
begineqnarray*
frac13n+m =int_0^1 y^3n+m-1 dy,
endeqnarray*
the sum can be transformed into integral
begineqnarray*
int_0^1 frac ln(1+y^3)1+y dy.
endeqnarray*
Now "half" of this is easy (IBP & rearrange)
begineqnarray*
int_0^1 frac ln(1+y)1+y dy = frac12 (ln 2)^2.
endeqnarray*
So we are left with
begineqnarray*
int_0^1 frac ln(1-y+y^2)1+y dy = int_0^1 frac (1-2y)ln(1+y)1-y+y^2 dy.
endeqnarray*
Now, apart from the obvious IBP done above, this integral has me stumped.
An exact evaluation would be nice, barring that, an expression in terms of dilogrithmic values or something similar would be helpful. Any comments or answers, gratefully received.
If this integral has been seen before a reference & advise on how I would search & find something similar in the future would be gratefully appreciated.
integration sequences-and-series definite-integrals summation
integration sequences-and-series definite-integrals summation
edited 55 mins ago
Frank W.
2,1461315
asked 6 hours ago
Donald Splutterwit
21.5k21244
edited 55 mins ago
Frank W.
2,1461315
asked 6 hours ago
Donald Splutterwit
21.5k21244
edited 55 mins ago
Frank W.
2,1461315
edited 55 mins ago
Frank W.
2,1461315
edited 55 mins ago
Frank W.
2,1461315
2,1461315
asked 6 hours ago
Donald Splutterwit
21.5k21244
asked 6 hours ago
Donald Splutterwit
21.5k21244
asked 6 hours ago
Donald Splutterwit
21.5k21244
21.5k21244
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
If it helps: Maple 7 computes the expression
$$ - mathrmln(2),mathrmln(3) - mathrmln(2),mathrm
ln(displaystyle frac 13 ,I) - mathrmln(2),mathrm
ln(displaystyle frac -13 ,I)
\- mathrmdilog( -
displaystyle frac 13 + I,sqrt3 + displaystyle
frac I,sqrt33 + I,sqrt3 )
\ - mathrmdilog(displaystyle frac 1 - 3 + I,
sqrt3 + displaystyle frac I,sqrt3 - 3 + I,sqrt
3 )
\ + mathrmdilog( - displaystyle frac 1 - 3 + I,
sqrt3 + displaystyle frac I,sqrt3 - 3 + I,sqrt
3 ) \
+ mathrmdilog(displaystyle frac 13 + I,sqrt3
+ displaystyle frac I,sqrt33 + I,sqrt3 ) $$
This will evaluate to $-0.12693500084879648110964091818$ and is compatible with Wolfram Alpha's $-0.126935$ (no symbolic answer here).
Please note that Maple's dilog is related to the standard polylogarithms by
$$operatornamedilog(x) = operatornameLi_2(1-x)$$
answered 5 hours ago
gammatester
15.9k21529
Thanks you for this ... I will try to neaten up these dilogarithms.
– Donald Splutterwit
5 hours ago
1
With Maple 2018 I get $$-frac left( ln left( 3 right) right) ^24+frac 5,pi^2 36+ln left( 2 right) ln left( 3 right) -rm dilog left( i/sqrt 3 right) -rm dilog left( -i/sqrt 3 right) $$
– Robert Israel
5 hours ago
@RobertIsrael Isn't there a functional equation for dilogarithms where you can simplify the last two terms into a real term? Or am I mistaken?
– Frank W.
5 hours ago
@FrankW. I don't know of one. Using the definition of dilog just gets me back to an integral similar to the one we started with.
– Robert Israel
2 hours ago
@RobertIsrael Okay, maybe I'm thinking of something else.
– Frank W.
1 hour ago
add a comment |Â
up vote
2
down vote
This integral won't evaluate to anything pretty, but we can write it in terms of the dilogarithm $operatorname Li_2(x)$. Factoring a $-2$ out of the numerator and completing the square in the denominator gives us:
$$int_0^1 frac (1-2y)ln(1+y)1-y+y^2 dy = -2int_0^1 frac (y-frac12)ln(1+y)(y-frac12)^2 +frac34 dy.$$
Then apply the substitution $x=y-frac12$.
$$int_-frac 12^frac 12 fracx ln(x+ frac32)x^2+ frac 34dx = int_-frac 12^frac 12 overbraceln(x+ frac32)^u cdot underbrace frac xx^2+ frac 34 dx _dv$$ Integration by parts show the original integral is equivalent to $displaystyle int_-frac 12^frac12 fracln(x^2+frac34)x+frac 32dx$. Now we can introduce yet another substitution, say $z=x+frac 32$. Now we get:
$$int_1^2 frac ln bigl((z- frac 32)^2 + frac 34big)zdz$$
We can factor the sum of squares by using the fact that $a^2 +b^2 = (a+bi)(a-bi)$.
$$int_1^2 frac lnbig((z- frac 32 + frac sqrt 32i)(z- frac 32 - frac sqrt 32i)big)zdz$$
Using the log rules, you can rewrite this as the sum of two integrals:
$$int_1^2 frac ln(z- frac 32 + frac sqrt 32i)zdz + int_1^2 frac ln (z- frac 32 - frac sqrt 32i)zdz tag1$$
Now lets look for a general solution to $displaystyle int_1^2 frac ln (x+a)xdx $, where $a$ is any constant.
$$int_1^2 frac ln (x+a)xdx = int_1^2 frac lnbig(a(frac xa +1)big)xdx = ln(a) int_1^2 frac dxx+ int_1^2 frac ln(frac xa +1)x dx.$$ Apply the substitution $u = - frac xa$. This changes it to:
$$ ln(a)ln(2)+ int frac ln(1-u)u du = Big[ln(a)ln(x)+ operatorname Li_2 (- frac xa)+ CBig]_1^2$$
$$ int_1^2 frac ln (x+a)xdx =ln(a)ln(2)+ operatorname Li_2 (- frac 2a)- operatorname Li_2 (-frac 1a) tag2$$
Now, putting (2) into (1) yields:
$$ ln(- frac 32 + frac sqrt 32i)ln(2)+ operatorname Li_2 (- frac 2- frac 32 + frac sqrt 32i)- operatorname Li_2 (-frac 1- frac 32 + frac sqrt 32i) + ln(- frac 32 - frac sqrt 32i)ln(2)+ operatorname Li_2 ( frac 2 frac 32 + frac sqrt 32i)- operatorname Li_2 (frac 1 frac 32 + frac sqrt 32i) $$
There's probably a way to simplify this, but I bet it's very tedious. The imaginary part should end up being 0.
answered 2 hours ago
guy600
655
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up vote
1
down vote
$$intlimits_0^1mathrm dx,frac log(1+x^3)1+xcolorblue=frac 12operatornameLi_2left(-frac 13right)+frac 14log^23+frac 12log^22-frac pi^236$$Confirmed by Wolfram Alpha.
Call the integral $mathfrakI$ and make the transformation $xmapstotfrac 1-x1+x$. What's left is$$mathfrakI=log^22+intlimits_0^1mathrm dx,frac log(1+3x^2)1+x-3intlimits_0^1mathrm dx,frac log(1+x)1+x$$
The last integral is trivially equal to$$intlimits_0^1mathrm dx,frac log(1+x)1+xcolorred=frac 12log^22$$The middle integral is difficult and I am still not aware of a full solution to the integral by hand. Replace the three with a parameter say, $a$, and differentiate with respect to $a$$$mathfrakI(a)=intlimits_0^1mathrm dxspacefrac log(1+ax^2)1+x$$Thus$$beginalign*mathfrakI'(a) & =intlimits_0^1mathrm dxspaceleft[frac 1(1+a)(1+x)+frac x(1+a)(1+ax^2)-frac 1(1+a)(1+ax^2)right]\ & =frac 11+alog 2+frac log(1+a)2a(1+a)-frac 1sqrt a(1+a)arctansqrt aendalign*$$Integrate both sides with respect to $a$ and from zero to three to get (I used Wolfram Alpha on this step)$$intlimits_0^1mathrm dxspacefrac log(1+3x^2)1+xcolorbrown=log^22+frac 12operatornameLi_2left(-frac 13right)+frac 14log^23-frac pi^236$$Now add everything together and you should get the result I stated at the beginning of my answer.
answered 1 hour ago
Frank W.
2,1461315
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If it helps: Maple 7 computes the expression
$$ - mathrmln(2),mathrmln(3) - mathrmln(2),mathrm
ln(displaystyle frac 13 ,I) - mathrmln(2),mathrm
ln(displaystyle frac -13 ,I)
\- mathrmdilog( -
displaystyle frac 13 + I,sqrt3 + displaystyle
frac I,sqrt33 + I,sqrt3 )
\ - mathrmdilog(displaystyle frac 1 - 3 + I,
sqrt3 + displaystyle frac I,sqrt3 - 3 + I,sqrt
3 )
\ + mathrmdilog( - displaystyle frac 1 - 3 + I,
sqrt3 + displaystyle frac I,sqrt3 - 3 + I,sqrt
3 ) \
+ mathrmdilog(displaystyle frac 13 + I,sqrt3
+ displaystyle frac I,sqrt33 + I,sqrt3 ) $$
This will evaluate to $-0.12693500084879648110964091818$ and is compatible with Wolfram Alpha's $-0.126935$ (no symbolic answer here).
Please note that Maple's dilog is related to the standard polylogarithms by
$$operatornamedilog(x) = operatornameLi_2(1-x)$$
answered 5 hours ago
gammatester
15.9k21529
Thanks you for this ... I will try to neaten up these dilogarithms.
– Donald Splutterwit
5 hours ago
1
With Maple 2018 I get $$-frac left( ln left( 3 right) right) ^24+frac 5,pi^2 36+ln left( 2 right) ln left( 3 right) -rm dilog left( i/sqrt 3 right) -rm dilog left( -i/sqrt 3 right) $$
– Robert Israel
5 hours ago
@RobertIsrael Isn't there a functional equation for dilogarithms where you can simplify the last two terms into a real term? Or am I mistaken?
– Frank W.
5 hours ago
@FrankW. I don't know of one. Using the definition of dilog just gets me back to an integral similar to the one we started with.
– Robert Israel
2 hours ago
@RobertIsrael Okay, maybe I'm thinking of something else.
– Frank W.
1 hour ago
add a comment |Â
up vote
2
down vote
If it helps: Maple 7 computes the expression
$$ - mathrmln(2),mathrmln(3) - mathrmln(2),mathrm
ln(displaystyle frac 13 ,I) - mathrmln(2),mathrm
ln(displaystyle frac -13 ,I)
\- mathrmdilog( -
displaystyle frac 13 + I,sqrt3 + displaystyle
frac I,sqrt33 + I,sqrt3 )
\ - mathrmdilog(displaystyle frac 1 - 3 + I,
sqrt3 + displaystyle frac I,sqrt3 - 3 + I,sqrt
3 )
\ + mathrmdilog( - displaystyle frac 1 - 3 + I,
sqrt3 + displaystyle frac I,sqrt3 - 3 + I,sqrt
3 ) \
+ mathrmdilog(displaystyle frac 13 + I,sqrt3
+ displaystyle frac I,sqrt33 + I,sqrt3 ) $$
This will evaluate to $-0.12693500084879648110964091818$ and is compatible with Wolfram Alpha's $-0.126935$ (no symbolic answer here).
Please note that Maple's dilog is related to the standard polylogarithms by
$$operatornamedilog(x) = operatornameLi_2(1-x)$$
answered 5 hours ago
gammatester
15.9k21529
Thanks you for this ... I will try to neaten up these dilogarithms.
– Donald Splutterwit
5 hours ago
1
With Maple 2018 I get $$-frac left( ln left( 3 right) right) ^24+frac 5,pi^2 36+ln left( 2 right) ln left( 3 right) -rm dilog left( i/sqrt 3 right) -rm dilog left( -i/sqrt 3 right) $$
– Robert Israel
5 hours ago
@RobertIsrael Isn't there a functional equation for dilogarithms where you can simplify the last two terms into a real term? Or am I mistaken?
– Frank W.
5 hours ago
@FrankW. I don't know of one. Using the definition of dilog just gets me back to an integral similar to the one we started with.
– Robert Israel
2 hours ago
@RobertIsrael Okay, maybe I'm thinking of something else.
– Frank W.
1 hour ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If it helps: Maple 7 computes the expression
$$ - mathrmln(2),mathrmln(3) - mathrmln(2),mathrm
ln(displaystyle frac 13 ,I) - mathrmln(2),mathrm
ln(displaystyle frac -13 ,I)
\- mathrmdilog( -
displaystyle frac 13 + I,sqrt3 + displaystyle
frac I,sqrt33 + I,sqrt3 )
\ - mathrmdilog(displaystyle frac 1 - 3 + I,
sqrt3 + displaystyle frac I,sqrt3 - 3 + I,sqrt
3 )
\ + mathrmdilog( - displaystyle frac 1 - 3 + I,
sqrt3 + displaystyle frac I,sqrt3 - 3 + I,sqrt
3 ) \
+ mathrmdilog(displaystyle frac 13 + I,sqrt3
+ displaystyle frac I,sqrt33 + I,sqrt3 ) $$
This will evaluate to $-0.12693500084879648110964091818$ and is compatible with Wolfram Alpha's $-0.126935$ (no symbolic answer here).
Please note that Maple's dilog is related to the standard polylogarithms by
$$operatornamedilog(x) = operatornameLi_2(1-x)$$
answered 5 hours ago
gammatester
15.9k21529
If it helps: Maple 7 computes the expression
$$ - mathrmln(2),mathrmln(3) - mathrmln(2),mathrm
ln(displaystyle frac 13 ,I) - mathrmln(2),mathrm
ln(displaystyle frac -13 ,I)
\- mathrmdilog( -
displaystyle frac 13 + I,sqrt3 + displaystyle
frac I,sqrt33 + I,sqrt3 )
\ - mathrmdilog(displaystyle frac 1 - 3 + I,
sqrt3 + displaystyle frac I,sqrt3 - 3 + I,sqrt
3 )
\ + mathrmdilog( - displaystyle frac 1 - 3 + I,
sqrt3 + displaystyle frac I,sqrt3 - 3 + I,sqrt
3 ) \
+ mathrmdilog(displaystyle frac 13 + I,sqrt3
+ displaystyle frac I,sqrt33 + I,sqrt3 ) $$
This will evaluate to $-0.12693500084879648110964091818$ and is compatible with Wolfram Alpha's $-0.126935$ (no symbolic answer here).
Please note that Maple's dilog is related to the standard polylogarithms by
$$operatornamedilog(x) = operatornameLi_2(1-x)$$
answered 5 hours ago
gammatester
15.9k21529
edited 5 hours ago
answered 5 hours ago
gammatester
15.9k21529
answered 5 hours ago
gammatester
15.9k21529
answered 5 hours ago
gammatester
15.9k21529
15.9k21529
Thanks you for this ... I will try to neaten up these dilogarithms.
– Donald Splutterwit
5 hours ago
1
With Maple 2018 I get $$-frac left( ln left( 3 right) right) ^24+frac 5,pi^2 36+ln left( 2 right) ln left( 3 right) -rm dilog left( i/sqrt 3 right) -rm dilog left( -i/sqrt 3 right) $$
– Robert Israel
5 hours ago
@RobertIsrael Isn't there a functional equation for dilogarithms where you can simplify the last two terms into a real term? Or am I mistaken?
– Frank W.
5 hours ago
@FrankW. I don't know of one. Using the definition of dilog just gets me back to an integral similar to the one we started with.
– Robert Israel
2 hours ago
@RobertIsrael Okay, maybe I'm thinking of something else.
– Frank W.
1 hour ago
add a comment |Â
Thanks you for this ... I will try to neaten up these dilogarithms.
– Donald Splutterwit
5 hours ago
1
With Maple 2018 I get $$-frac left( ln left( 3 right) right) ^24+frac 5,pi^2 36+ln left( 2 right) ln left( 3 right) -rm dilog left( i/sqrt 3 right) -rm dilog left( -i/sqrt 3 right) $$
– Robert Israel
5 hours ago
@RobertIsrael Isn't there a functional equation for dilogarithms where you can simplify the last two terms into a real term? Or am I mistaken?
– Frank W.
5 hours ago
@FrankW. I don't know of one. Using the definition of dilog just gets me back to an integral similar to the one we started with.
– Robert Israel
2 hours ago
@RobertIsrael Okay, maybe I'm thinking of something else.
– Frank W.
1 hour ago
Thanks you for this ... I will try to neaten up these dilogarithms.
– Donald Splutterwit
5 hours ago
Thanks you for this ... I will try to neaten up these dilogarithms.
– Donald Splutterwit
5 hours ago
1
1
With Maple 2018 I get $$-frac left( ln left( 3 right) right) ^24+frac 5,pi^2 36+ln left( 2 right) ln left( 3 right) -rm dilog left( i/sqrt 3 right) -rm dilog left( -i/sqrt 3 right) $$
– Robert Israel
5 hours ago
With Maple 2018 I get $$-frac left( ln left( 3 right) right) ^24+frac 5,pi^2 36+ln left( 2 right) ln left( 3 right) -rm dilog left( i/sqrt 3 right) -rm dilog left( -i/sqrt 3 right) $$
– Robert Israel
5 hours ago
@RobertIsrael Isn't there a functional equation for dilogarithms where you can simplify the last two terms into a real term? Or am I mistaken?
– Frank W.
5 hours ago
@RobertIsrael Isn't there a functional equation for dilogarithms where you can simplify the last two terms into a real term? Or am I mistaken?
– Frank W.
5 hours ago
@FrankW. I don't know of one. Using the definition of dilog just gets me back to an integral similar to the one we started with.
– Robert Israel
2 hours ago
@FrankW. I don't know of one. Using the definition of dilog just gets me back to an integral similar to the one we started with.
– Robert Israel
2 hours ago
@RobertIsrael Okay, maybe I'm thinking of something else.
– Frank W.
1 hour ago
@RobertIsrael Okay, maybe I'm thinking of something else.
– Frank W.
1 hour ago
add a comment |Â
up vote
2
down vote
This integral won't evaluate to anything pretty, but we can write it in terms of the dilogarithm $operatorname Li_2(x)$. Factoring a $-2$ out of the numerator and completing the square in the denominator gives us:
$$int_0^1 frac (1-2y)ln(1+y)1-y+y^2 dy = -2int_0^1 frac (y-frac12)ln(1+y)(y-frac12)^2 +frac34 dy.$$
Then apply the substitution $x=y-frac12$.
$$int_-frac 12^frac 12 fracx ln(x+ frac32)x^2+ frac 34dx = int_-frac 12^frac 12 overbraceln(x+ frac32)^u cdot underbrace frac xx^2+ frac 34 dx _dv$$ Integration by parts show the original integral is equivalent to $displaystyle int_-frac 12^frac12 fracln(x^2+frac34)x+frac 32dx$. Now we can introduce yet another substitution, say $z=x+frac 32$. Now we get:
$$int_1^2 frac ln bigl((z- frac 32)^2 + frac 34big)zdz$$
We can factor the sum of squares by using the fact that $a^2 +b^2 = (a+bi)(a-bi)$.
$$int_1^2 frac lnbig((z- frac 32 + frac sqrt 32i)(z- frac 32 - frac sqrt 32i)big)zdz$$
Using the log rules, you can rewrite this as the sum of two integrals:
$$int_1^2 frac ln(z- frac 32 + frac sqrt 32i)zdz + int_1^2 frac ln (z- frac 32 - frac sqrt 32i)zdz tag1$$
Now lets look for a general solution to $displaystyle int_1^2 frac ln (x+a)xdx $, where $a$ is any constant.
$$int_1^2 frac ln (x+a)xdx = int_1^2 frac lnbig(a(frac xa +1)big)xdx = ln(a) int_1^2 frac dxx+ int_1^2 frac ln(frac xa +1)x dx.$$ Apply the substitution $u = - frac xa$. This changes it to:
$$ ln(a)ln(2)+ int frac ln(1-u)u du = Big[ln(a)ln(x)+ operatorname Li_2 (- frac xa)+ CBig]_1^2$$
$$ int_1^2 frac ln (x+a)xdx =ln(a)ln(2)+ operatorname Li_2 (- frac 2a)- operatorname Li_2 (-frac 1a) tag2$$
Now, putting (2) into (1) yields:
$$ ln(- frac 32 + frac sqrt 32i)ln(2)+ operatorname Li_2 (- frac 2- frac 32 + frac sqrt 32i)- operatorname Li_2 (-frac 1- frac 32 + frac sqrt 32i) + ln(- frac 32 - frac sqrt 32i)ln(2)+ operatorname Li_2 ( frac 2 frac 32 + frac sqrt 32i)- operatorname Li_2 (frac 1 frac 32 + frac sqrt 32i) $$
There's probably a way to simplify this, but I bet it's very tedious. The imaginary part should end up being 0.
answered 2 hours ago
guy600
655
add a comment |Â
up vote
2
down vote
This integral won't evaluate to anything pretty, but we can write it in terms of the dilogarithm $operatorname Li_2(x)$. Factoring a $-2$ out of the numerator and completing the square in the denominator gives us:
$$int_0^1 frac (1-2y)ln(1+y)1-y+y^2 dy = -2int_0^1 frac (y-frac12)ln(1+y)(y-frac12)^2 +frac34 dy.$$
Then apply the substitution $x=y-frac12$.
$$int_-frac 12^frac 12 fracx ln(x+ frac32)x^2+ frac 34dx = int_-frac 12^frac 12 overbraceln(x+ frac32)^u cdot underbrace frac xx^2+ frac 34 dx _dv$$ Integration by parts show the original integral is equivalent to $displaystyle int_-frac 12^frac12 fracln(x^2+frac34)x+frac 32dx$. Now we can introduce yet another substitution, say $z=x+frac 32$. Now we get:
$$int_1^2 frac ln bigl((z- frac 32)^2 + frac 34big)zdz$$
We can factor the sum of squares by using the fact that $a^2 +b^2 = (a+bi)(a-bi)$.
$$int_1^2 frac lnbig((z- frac 32 + frac sqrt 32i)(z- frac 32 - frac sqrt 32i)big)zdz$$
Using the log rules, you can rewrite this as the sum of two integrals:
$$int_1^2 frac ln(z- frac 32 + frac sqrt 32i)zdz + int_1^2 frac ln (z- frac 32 - frac sqrt 32i)zdz tag1$$
Now lets look for a general solution to $displaystyle int_1^2 frac ln (x+a)xdx $, where $a$ is any constant.
$$int_1^2 frac ln (x+a)xdx = int_1^2 frac lnbig(a(frac xa +1)big)xdx = ln(a) int_1^2 frac dxx+ int_1^2 frac ln(frac xa +1)x dx.$$ Apply the substitution $u = - frac xa$. This changes it to:
$$ ln(a)ln(2)+ int frac ln(1-u)u du = Big[ln(a)ln(x)+ operatorname Li_2 (- frac xa)+ CBig]_1^2$$
$$ int_1^2 frac ln (x+a)xdx =ln(a)ln(2)+ operatorname Li_2 (- frac 2a)- operatorname Li_2 (-frac 1a) tag2$$
Now, putting (2) into (1) yields:
$$ ln(- frac 32 + frac sqrt 32i)ln(2)+ operatorname Li_2 (- frac 2- frac 32 + frac sqrt 32i)- operatorname Li_2 (-frac 1- frac 32 + frac sqrt 32i) + ln(- frac 32 - frac sqrt 32i)ln(2)+ operatorname Li_2 ( frac 2 frac 32 + frac sqrt 32i)- operatorname Li_2 (frac 1 frac 32 + frac sqrt 32i) $$
There's probably a way to simplify this, but I bet it's very tedious. The imaginary part should end up being 0.
answered 2 hours ago
guy600
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This integral won't evaluate to anything pretty, but we can write it in terms of the dilogarithm $operatorname Li_2(x)$. Factoring a $-2$ out of the numerator and completing the square in the denominator gives us:
$$int_0^1 frac (1-2y)ln(1+y)1-y+y^2 dy = -2int_0^1 frac (y-frac12)ln(1+y)(y-frac12)^2 +frac34 dy.$$
Then apply the substitution $x=y-frac12$.
$$int_-frac 12^frac 12 fracx ln(x+ frac32)x^2+ frac 34dx = int_-frac 12^frac 12 overbraceln(x+ frac32)^u cdot underbrace frac xx^2+ frac 34 dx _dv$$ Integration by parts show the original integral is equivalent to $displaystyle int_-frac 12^frac12 fracln(x^2+frac34)x+frac 32dx$. Now we can introduce yet another substitution, say $z=x+frac 32$. Now we get:
$$int_1^2 frac ln bigl((z- frac 32)^2 + frac 34big)zdz$$
We can factor the sum of squares by using the fact that $a^2 +b^2 = (a+bi)(a-bi)$.
$$int_1^2 frac lnbig((z- frac 32 + frac sqrt 32i)(z- frac 32 - frac sqrt 32i)big)zdz$$
Using the log rules, you can rewrite this as the sum of two integrals:
$$int_1^2 frac ln(z- frac 32 + frac sqrt 32i)zdz + int_1^2 frac ln (z- frac 32 - frac sqrt 32i)zdz tag1$$
Now lets look for a general solution to $displaystyle int_1^2 frac ln (x+a)xdx $, where $a$ is any constant.
$$int_1^2 frac ln (x+a)xdx = int_1^2 frac lnbig(a(frac xa +1)big)xdx = ln(a) int_1^2 frac dxx+ int_1^2 frac ln(frac xa +1)x dx.$$ Apply the substitution $u = - frac xa$. This changes it to:
$$ ln(a)ln(2)+ int frac ln(1-u)u du = Big[ln(a)ln(x)+ operatorname Li_2 (- frac xa)+ CBig]_1^2$$
$$ int_1^2 frac ln (x+a)xdx =ln(a)ln(2)+ operatorname Li_2 (- frac 2a)- operatorname Li_2 (-frac 1a) tag2$$
Now, putting (2) into (1) yields:
$$ ln(- frac 32 + frac sqrt 32i)ln(2)+ operatorname Li_2 (- frac 2- frac 32 + frac sqrt 32i)- operatorname Li_2 (-frac 1- frac 32 + frac sqrt 32i) + ln(- frac 32 - frac sqrt 32i)ln(2)+ operatorname Li_2 ( frac 2 frac 32 + frac sqrt 32i)- operatorname Li_2 (frac 1 frac 32 + frac sqrt 32i) $$
There's probably a way to simplify this, but I bet it's very tedious. The imaginary part should end up being 0.
answered 2 hours ago
guy600
655
This integral won't evaluate to anything pretty, but we can write it in terms of the dilogarithm $operatorname Li_2(x)$. Factoring a $-2$ out of the numerator and completing the square in the denominator gives us:
$$int_0^1 frac (1-2y)ln(1+y)1-y+y^2 dy = -2int_0^1 frac (y-frac12)ln(1+y)(y-frac12)^2 +frac34 dy.$$
Then apply the substitution $x=y-frac12$.
$$int_-frac 12^frac 12 fracx ln(x+ frac32)x^2+ frac 34dx = int_-frac 12^frac 12 overbraceln(x+ frac32)^u cdot underbrace frac xx^2+ frac 34 dx _dv$$ Integration by parts show the original integral is equivalent to $displaystyle int_-frac 12^frac12 fracln(x^2+frac34)x+frac 32dx$. Now we can introduce yet another substitution, say $z=x+frac 32$. Now we get:
$$int_1^2 frac ln bigl((z- frac 32)^2 + frac 34big)zdz$$
We can factor the sum of squares by using the fact that $a^2 +b^2 = (a+bi)(a-bi)$.
$$int_1^2 frac lnbig((z- frac 32 + frac sqrt 32i)(z- frac 32 - frac sqrt 32i)big)zdz$$
Using the log rules, you can rewrite this as the sum of two integrals:
$$int_1^2 frac ln(z- frac 32 + frac sqrt 32i)zdz + int_1^2 frac ln (z- frac 32 - frac sqrt 32i)zdz tag1$$
Now lets look for a general solution to $displaystyle int_1^2 frac ln (x+a)xdx $, where $a$ is any constant.
$$int_1^2 frac ln (x+a)xdx = int_1^2 frac lnbig(a(frac xa +1)big)xdx = ln(a) int_1^2 frac dxx+ int_1^2 frac ln(frac xa +1)x dx.$$ Apply the substitution $u = - frac xa$. This changes it to:
$$ ln(a)ln(2)+ int frac ln(1-u)u du = Big[ln(a)ln(x)+ operatorname Li_2 (- frac xa)+ CBig]_1^2$$
$$ int_1^2 frac ln (x+a)xdx =ln(a)ln(2)+ operatorname Li_2 (- frac 2a)- operatorname Li_2 (-frac 1a) tag2$$
Now, putting (2) into (1) yields:
$$ ln(- frac 32 + frac sqrt 32i)ln(2)+ operatorname Li_2 (- frac 2- frac 32 + frac sqrt 32i)- operatorname Li_2 (-frac 1- frac 32 + frac sqrt 32i) + ln(- frac 32 - frac sqrt 32i)ln(2)+ operatorname Li_2 ( frac 2 frac 32 + frac sqrt 32i)- operatorname Li_2 (frac 1 frac 32 + frac sqrt 32i) $$
There's probably a way to simplify this, but I bet it's very tedious. The imaginary part should end up being 0.
answered 2 hours ago
guy600
655
answered 2 hours ago
guy600
655
answered 2 hours ago
guy600
655
answered 2 hours ago
guy600
655
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$$intlimits_0^1mathrm dx,frac log(1+x^3)1+xcolorblue=frac 12operatornameLi_2left(-frac 13right)+frac 14log^23+frac 12log^22-frac pi^236$$Confirmed by Wolfram Alpha.
Call the integral $mathfrakI$ and make the transformation $xmapstotfrac 1-x1+x$. What's left is$$mathfrakI=log^22+intlimits_0^1mathrm dx,frac log(1+3x^2)1+x-3intlimits_0^1mathrm dx,frac log(1+x)1+x$$
The last integral is trivially equal to$$intlimits_0^1mathrm dx,frac log(1+x)1+xcolorred=frac 12log^22$$The middle integral is difficult and I am still not aware of a full solution to the integral by hand. Replace the three with a parameter say, $a$, and differentiate with respect to $a$$$mathfrakI(a)=intlimits_0^1mathrm dxspacefrac log(1+ax^2)1+x$$Thus$$beginalign*mathfrakI'(a) & =intlimits_0^1mathrm dxspaceleft[frac 1(1+a)(1+x)+frac x(1+a)(1+ax^2)-frac 1(1+a)(1+ax^2)right]\ & =frac 11+alog 2+frac log(1+a)2a(1+a)-frac 1sqrt a(1+a)arctansqrt aendalign*$$Integrate both sides with respect to $a$ and from zero to three to get (I used Wolfram Alpha on this step)$$intlimits_0^1mathrm dxspacefrac log(1+3x^2)1+xcolorbrown=log^22+frac 12operatornameLi_2left(-frac 13right)+frac 14log^23-frac pi^236$$Now add everything together and you should get the result I stated at the beginning of my answer.
answered 1 hour ago
Frank W.
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$$intlimits_0^1mathrm dx,frac log(1+x^3)1+xcolorblue=frac 12operatornameLi_2left(-frac 13right)+frac 14log^23+frac 12log^22-frac pi^236$$Confirmed by Wolfram Alpha.
Call the integral $mathfrakI$ and make the transformation $xmapstotfrac 1-x1+x$. What's left is$$mathfrakI=log^22+intlimits_0^1mathrm dx,frac log(1+3x^2)1+x-3intlimits_0^1mathrm dx,frac log(1+x)1+x$$
The last integral is trivially equal to$$intlimits_0^1mathrm dx,frac log(1+x)1+xcolorred=frac 12log^22$$The middle integral is difficult and I am still not aware of a full solution to the integral by hand. Replace the three with a parameter say, $a$, and differentiate with respect to $a$$$mathfrakI(a)=intlimits_0^1mathrm dxspacefrac log(1+ax^2)1+x$$Thus$$beginalign*mathfrakI'(a) & =intlimits_0^1mathrm dxspaceleft[frac 1(1+a)(1+x)+frac x(1+a)(1+ax^2)-frac 1(1+a)(1+ax^2)right]\ & =frac 11+alog 2+frac log(1+a)2a(1+a)-frac 1sqrt a(1+a)arctansqrt aendalign*$$Integrate both sides with respect to $a$ and from zero to three to get (I used Wolfram Alpha on this step)$$intlimits_0^1mathrm dxspacefrac log(1+3x^2)1+xcolorbrown=log^22+frac 12operatornameLi_2left(-frac 13right)+frac 14log^23-frac pi^236$$Now add everything together and you should get the result I stated at the beginning of my answer.
answered 1 hour ago
Frank W.
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up vote
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up vote
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$$intlimits_0^1mathrm dx,frac log(1+x^3)1+xcolorblue=frac 12operatornameLi_2left(-frac 13right)+frac 14log^23+frac 12log^22-frac pi^236$$Confirmed by Wolfram Alpha.
Call the integral $mathfrakI$ and make the transformation $xmapstotfrac 1-x1+x$. What's left is$$mathfrakI=log^22+intlimits_0^1mathrm dx,frac log(1+3x^2)1+x-3intlimits_0^1mathrm dx,frac log(1+x)1+x$$
The last integral is trivially equal to$$intlimits_0^1mathrm dx,frac log(1+x)1+xcolorred=frac 12log^22$$The middle integral is difficult and I am still not aware of a full solution to the integral by hand. Replace the three with a parameter say, $a$, and differentiate with respect to $a$$$mathfrakI(a)=intlimits_0^1mathrm dxspacefrac log(1+ax^2)1+x$$Thus$$beginalign*mathfrakI'(a) & =intlimits_0^1mathrm dxspaceleft[frac 1(1+a)(1+x)+frac x(1+a)(1+ax^2)-frac 1(1+a)(1+ax^2)right]\ & =frac 11+alog 2+frac log(1+a)2a(1+a)-frac 1sqrt a(1+a)arctansqrt aendalign*$$Integrate both sides with respect to $a$ and from zero to three to get (I used Wolfram Alpha on this step)$$intlimits_0^1mathrm dxspacefrac log(1+3x^2)1+xcolorbrown=log^22+frac 12operatornameLi_2left(-frac 13right)+frac 14log^23-frac pi^236$$Now add everything together and you should get the result I stated at the beginning of my answer.
answered 1 hour ago
Frank W.
2,1461315
$$intlimits_0^1mathrm dx,frac log(1+x^3)1+xcolorblue=frac 12operatornameLi_2left(-frac 13right)+frac 14log^23+frac 12log^22-frac pi^236$$Confirmed by Wolfram Alpha.
Call the integral $mathfrakI$ and make the transformation $xmapstotfrac 1-x1+x$. What's left is$$mathfrakI=log^22+intlimits_0^1mathrm dx,frac log(1+3x^2)1+x-3intlimits_0^1mathrm dx,frac log(1+x)1+x$$
The last integral is trivially equal to$$intlimits_0^1mathrm dx,frac log(1+x)1+xcolorred=frac 12log^22$$The middle integral is difficult and I am still not aware of a full solution to the integral by hand. Replace the three with a parameter say, $a$, and differentiate with respect to $a$$$mathfrakI(a)=intlimits_0^1mathrm dxspacefrac log(1+ax^2)1+x$$Thus$$beginalign*mathfrakI'(a) & =intlimits_0^1mathrm dxspaceleft[frac 1(1+a)(1+x)+frac x(1+a)(1+ax^2)-frac 1(1+a)(1+ax^2)right]\ & =frac 11+alog 2+frac log(1+a)2a(1+a)-frac 1sqrt a(1+a)arctansqrt aendalign*$$Integrate both sides with respect to $a$ and from zero to three to get (I used Wolfram Alpha on this step)$$intlimits_0^1mathrm dxspacefrac log(1+3x^2)1+xcolorbrown=log^22+frac 12operatornameLi_2left(-frac 13right)+frac 14log^23-frac pi^236$$Now add everything together and you should get the result I stated at the beginning of my answer.
answered 1 hour ago
Frank W.
2,1461315
answered 1 hour ago
Frank W.
2,1461315
answered 1 hour ago
Frank W.
2,1461315
answered 1 hour ago
Frank W.
2,1461315
2,1461315
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