Mixing
An AMM-like integral $int_0^1frac{arctan x}xlnfrac{(1+x^2)^3}{(1+x)^2}dx$
How can we evaluate $$I=int_0^1frac{arctan x}xlnfrac{(1+x^2)^3}{(1+x)^2}dx=0?$$
I tried substitution $x=frac{1-t}{1+t}$ and got
$$I=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} arctan frac{t-1}{t+1}}{t^2-1}dt\
=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} (arctan t-fracpi4)}{t^2-1}dt$$
I'm able to evaluate $$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$
But I have no idea where to start with the rest one.
calculus integration definite-integrals
asked Nov 20 '18 at 9:20
Kemono Chen
2,569437
add a comment |
How can we evaluate $$I=int_0^1frac{arctan x}xlnfrac{(1+x^2)^3}{(1+x)^2}dx=0?$$
I tried substitution $x=frac{1-t}{1+t}$ and got
$$I=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} arctan frac{t-1}{t+1}}{t^2-1}dt\
=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} (arctan t-fracpi4)}{t^2-1}dt$$
I'm able to evaluate $$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$
But I have no idea where to start with the rest one.
calculus integration definite-integrals
asked Nov 20 '18 at 9:20
Kemono Chen
2,569437
Just wondering: Why do you want to calculate this (by hand)
– klirk
Nov 20 '18 at 20:38
@klirk Just an interest.
– Kemono Chen
Nov 21 '18 at 0:18
$$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt=frac{pi^2}{48}$$
– user178256
Nov 21 '18 at 11:00
add a comment |
How can we evaluate $$I=int_0^1frac{arctan x}xlnfrac{(1+x^2)^3}{(1+x)^2}dx=0?$$
I tried substitution $x=frac{1-t}{1+t}$ and got
$$I=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} arctan frac{t-1}{t+1}}{t^2-1}dt\
=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} (arctan t-fracpi4)}{t^2-1}dt$$
I'm able to evaluate $$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$
But I have no idea where to start with the rest one.
calculus integration definite-integrals
asked Nov 20 '18 at 9:20
Kemono Chen
2,569437
How can we evaluate $$I=int_0^1frac{arctan x}xlnfrac{(1+x^2)^3}{(1+x)^2}dx=0?$$
I tried substitution $x=frac{1-t}{1+t}$ and got
$$I=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} arctan frac{t-1}{t+1}}{t^2-1}dt\
=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} (arctan t-fracpi4)}{t^2-1}dt$$
I'm able to evaluate $$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$
But I have no idea where to start with the rest one.
calculus integration definite-integrals
calculus integration definite-integrals
asked Nov 20 '18 at 9:20
Kemono Chen
2,569437
asked Nov 20 '18 at 9:20
Kemono Chen
2,569437
asked Nov 20 '18 at 9:20
Kemono Chen
2,569437
asked Nov 20 '18 at 9:20
Kemono Chen
2,569437
asked Nov 20 '18 at 9:20
Kemono Chen
2,569437
2,569437
Just wondering: Why do you want to calculate this (by hand)
– klirk
Nov 20 '18 at 20:38
@klirk Just an interest.
– Kemono Chen
Nov 21 '18 at 0:18
$$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt=frac{pi^2}{48}$$
– user178256
Nov 21 '18 at 11:00
add a comment |
Just wondering: Why do you want to calculate this (by hand)
– klirk
Nov 20 '18 at 20:38
@klirk Just an interest.
– Kemono Chen
Nov 21 '18 at 0:18
$$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt=frac{pi^2}{48}$$
– user178256
Nov 21 '18 at 11:00
Just wondering: Why do you want to calculate this (by hand)
– klirk
Nov 20 '18 at 20:38
Just wondering: Why do you want to calculate this (by hand)
– klirk
Nov 20 '18 at 20:38
@klirk Just an interest.
– Kemono Chen
Nov 21 '18 at 0:18
@klirk Just an interest.
– Kemono Chen
Nov 21 '18 at 0:18
$$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt=frac{pi^2}{48}$$
– user178256
Nov 21 '18 at 11:00
$$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt=frac{pi^2}{48}$$
– user178256
Nov 21 '18 at 11:00
add a comment |
2 Answers
2
active
oldest
votes
$text{A solution by Cornel Ioan Valean.}$ The problem is similar to the problem $textbf{AMM 12054}.$ Using the well-known result in $textbf{4.535.1}$ from $text{Table of Integrals, Series and Products}$ by I.S. Gradshteyn and I.M. Ryzhik:
begin{equation*}
int_0^1 frac{arctan(y x)}{1+y^2x}textrm{d}x=frac{1}{2y^2}arctan(y)log(1+y^2),
end{equation*}
We have:
begin{equation*}
frac{1}{2}int_0^1frac{arctan(y)log(1+y^2)}{y}dy=int_0^1left(int_0^1 frac{yarctan(y x)}{1+y^2x}textrm{d}xright)textrm{d}yoverset{yx=t}{=}int_0^1left(int_0^y frac{arctan(t)}{1+y t}textrm{d}tright)textrm{d}y\
end{equation*}
begin{equation*}
=int_0^1left(int_t^1 frac{arctan(t)}{1+y t}textrm{d}yright)textrm{d}t=int_0^1frac{displaystyle arctan(y)logleft(frac{1+y}{1+y^2}right)}{y} textrm{d}y,
end{equation*}
And the result is proved.
answered Nov 21 '18 at 11:52
Zacky
4,9061750
1
(+1) Very neat.
– nospoon
Nov 21 '18 at 12:24
add a comment |
Through the dilogarithm/trilogarithm machinery it can be shown that
$$ int_{0}^{1}frac{log(1+i x)log(1+x)}{x},dx=\frac{pi K}{2}-frac{9ipi^3}{64}+3iKlog(2)-frac{3pi i}{16}log^2(2)+frac{5pi^2}{32}log(2)-frac{log^3(2)}{8}-frac{69}{16}zeta(3)+6,text{Li}_3left(tfrac{1+i}{2}right) $$
$$ int_{0}^{1}frac{log^2(1+i x)}{x},dx=\
-frac{pi K}{2}-frac{3ipi^3}{64}+iKlog(2)-frac{pi i}{16}log^2(2)+frac{5pi^2}{96}log(2)-frac{log^3(2)}{24}-frac{3}{16}zeta(3)+2,text{Li}_3left(tfrac{1+i}{2}right) $$
$$ int_{0}^{1}frac{log(1+ix)log(1-ix)}{x},dx= frac{pi K}{2}-frac{27}{32}zeta(3)$$
hence the claim follows by $arctan x=text{Im},log(1+ix)$ and $log(1+x^2)=log(1+ix)+log(1-ix)$.
answered Nov 20 '18 at 20:20
Jack D'Aurizio
286k33279656
Thank you for the great answer. :) But I prefer a solution without brute force.
– Kemono Chen
Nov 21 '18 at 3:02
add a comment |
Your Answer
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2 Answers
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2 Answers
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$text{A solution by Cornel Ioan Valean.}$ The problem is similar to the problem $textbf{AMM 12054}.$ Using the well-known result in $textbf{4.535.1}$ from $text{Table of Integrals, Series and Products}$ by I.S. Gradshteyn and I.M. Ryzhik:
begin{equation*}
int_0^1 frac{arctan(y x)}{1+y^2x}textrm{d}x=frac{1}{2y^2}arctan(y)log(1+y^2),
end{equation*}
We have:
begin{equation*}
frac{1}{2}int_0^1frac{arctan(y)log(1+y^2)}{y}dy=int_0^1left(int_0^1 frac{yarctan(y x)}{1+y^2x}textrm{d}xright)textrm{d}yoverset{yx=t}{=}int_0^1left(int_0^y frac{arctan(t)}{1+y t}textrm{d}tright)textrm{d}y\
end{equation*}
begin{equation*}
=int_0^1left(int_t^1 frac{arctan(t)}{1+y t}textrm{d}yright)textrm{d}t=int_0^1frac{displaystyle arctan(y)logleft(frac{1+y}{1+y^2}right)}{y} textrm{d}y,
end{equation*}
And the result is proved.
answered Nov 21 '18 at 11:52
Zacky
4,9061750
1
(+1) Very neat.
– nospoon
Nov 21 '18 at 12:24
add a comment |
$text{A solution by Cornel Ioan Valean.}$ The problem is similar to the problem $textbf{AMM 12054}.$ Using the well-known result in $textbf{4.535.1}$ from $text{Table of Integrals, Series and Products}$ by I.S. Gradshteyn and I.M. Ryzhik:
begin{equation*}
int_0^1 frac{arctan(y x)}{1+y^2x}textrm{d}x=frac{1}{2y^2}arctan(y)log(1+y^2),
end{equation*}
We have:
begin{equation*}
frac{1}{2}int_0^1frac{arctan(y)log(1+y^2)}{y}dy=int_0^1left(int_0^1 frac{yarctan(y x)}{1+y^2x}textrm{d}xright)textrm{d}yoverset{yx=t}{=}int_0^1left(int_0^y frac{arctan(t)}{1+y t}textrm{d}tright)textrm{d}y\
end{equation*}
begin{equation*}
=int_0^1left(int_t^1 frac{arctan(t)}{1+y t}textrm{d}yright)textrm{d}t=int_0^1frac{displaystyle arctan(y)logleft(frac{1+y}{1+y^2}right)}{y} textrm{d}y,
end{equation*}
And the result is proved.
answered Nov 21 '18 at 11:52
Zacky
4,9061750
1
(+1) Very neat.
– nospoon
Nov 21 '18 at 12:24
add a comment |
$text{A solution by Cornel Ioan Valean.}$ The problem is similar to the problem $textbf{AMM 12054}.$ Using the well-known result in $textbf{4.535.1}$ from $text{Table of Integrals, Series and Products}$ by I.S. Gradshteyn and I.M. Ryzhik:
begin{equation*}
int_0^1 frac{arctan(y x)}{1+y^2x}textrm{d}x=frac{1}{2y^2}arctan(y)log(1+y^2),
end{equation*}
We have:
begin{equation*}
frac{1}{2}int_0^1frac{arctan(y)log(1+y^2)}{y}dy=int_0^1left(int_0^1 frac{yarctan(y x)}{1+y^2x}textrm{d}xright)textrm{d}yoverset{yx=t}{=}int_0^1left(int_0^y frac{arctan(t)}{1+y t}textrm{d}tright)textrm{d}y\
end{equation*}
begin{equation*}
=int_0^1left(int_t^1 frac{arctan(t)}{1+y t}textrm{d}yright)textrm{d}t=int_0^1frac{displaystyle arctan(y)logleft(frac{1+y}{1+y^2}right)}{y} textrm{d}y,
end{equation*}
And the result is proved.
answered Nov 21 '18 at 11:52
Zacky
4,9061750
$text{A solution by Cornel Ioan Valean.}$ The problem is similar to the problem $textbf{AMM 12054}.$ Using the well-known result in $textbf{4.535.1}$ from $text{Table of Integrals, Series and Products}$ by I.S. Gradshteyn and I.M. Ryzhik:
begin{equation*}
int_0^1 frac{arctan(y x)}{1+y^2x}textrm{d}x=frac{1}{2y^2}arctan(y)log(1+y^2),
end{equation*}
We have:
begin{equation*}
frac{1}{2}int_0^1frac{arctan(y)log(1+y^2)}{y}dy=int_0^1left(int_0^1 frac{yarctan(y x)}{1+y^2x}textrm{d}xright)textrm{d}yoverset{yx=t}{=}int_0^1left(int_0^y frac{arctan(t)}{1+y t}textrm{d}tright)textrm{d}y\
end{equation*}
begin{equation*}
=int_0^1left(int_t^1 frac{arctan(t)}{1+y t}textrm{d}yright)textrm{d}t=int_0^1frac{displaystyle arctan(y)logleft(frac{1+y}{1+y^2}right)}{y} textrm{d}y,
end{equation*}
And the result is proved.
answered Nov 21 '18 at 11:52
Zacky
4,9061750
edited Dec 23 '18 at 21:09
answered Nov 21 '18 at 11:52
Zacky
4,9061750
answered Nov 21 '18 at 11:52
Zacky
4,9061750
answered Nov 21 '18 at 11:52
Zacky
4,9061750
4,9061750
1
(+1) Very neat.
– nospoon
Nov 21 '18 at 12:24
add a comment |
1
(+1) Very neat.
– nospoon
Nov 21 '18 at 12:24
1
1
(+1) Very neat.
– nospoon
Nov 21 '18 at 12:24
(+1) Very neat.
– nospoon
Nov 21 '18 at 12:24
add a comment |
Through the dilogarithm/trilogarithm machinery it can be shown that
$$ int_{0}^{1}frac{log(1+i x)log(1+x)}{x},dx=\frac{pi K}{2}-frac{9ipi^3}{64}+3iKlog(2)-frac{3pi i}{16}log^2(2)+frac{5pi^2}{32}log(2)-frac{log^3(2)}{8}-frac{69}{16}zeta(3)+6,text{Li}_3left(tfrac{1+i}{2}right) $$
$$ int_{0}^{1}frac{log^2(1+i x)}{x},dx=\
-frac{pi K}{2}-frac{3ipi^3}{64}+iKlog(2)-frac{pi i}{16}log^2(2)+frac{5pi^2}{96}log(2)-frac{log^3(2)}{24}-frac{3}{16}zeta(3)+2,text{Li}_3left(tfrac{1+i}{2}right) $$
$$ int_{0}^{1}frac{log(1+ix)log(1-ix)}{x},dx= frac{pi K}{2}-frac{27}{32}zeta(3)$$
hence the claim follows by $arctan x=text{Im},log(1+ix)$ and $log(1+x^2)=log(1+ix)+log(1-ix)$.
answered Nov 20 '18 at 20:20
Jack D'Aurizio
286k33279656
Thank you for the great answer. :) But I prefer a solution without brute force.
– Kemono Chen
Nov 21 '18 at 3:02
add a comment |
Through the dilogarithm/trilogarithm machinery it can be shown that
$$ int_{0}^{1}frac{log(1+i x)log(1+x)}{x},dx=\frac{pi K}{2}-frac{9ipi^3}{64}+3iKlog(2)-frac{3pi i}{16}log^2(2)+frac{5pi^2}{32}log(2)-frac{log^3(2)}{8}-frac{69}{16}zeta(3)+6,text{Li}_3left(tfrac{1+i}{2}right) $$
$$ int_{0}^{1}frac{log^2(1+i x)}{x},dx=\
-frac{pi K}{2}-frac{3ipi^3}{64}+iKlog(2)-frac{pi i}{16}log^2(2)+frac{5pi^2}{96}log(2)-frac{log^3(2)}{24}-frac{3}{16}zeta(3)+2,text{Li}_3left(tfrac{1+i}{2}right) $$
$$ int_{0}^{1}frac{log(1+ix)log(1-ix)}{x},dx= frac{pi K}{2}-frac{27}{32}zeta(3)$$
hence the claim follows by $arctan x=text{Im},log(1+ix)$ and $log(1+x^2)=log(1+ix)+log(1-ix)$.
answered Nov 20 '18 at 20:20
Jack D'Aurizio
286k33279656
Thank you for the great answer. :) But I prefer a solution without brute force.
– Kemono Chen
Nov 21 '18 at 3:02
add a comment |
Through the dilogarithm/trilogarithm machinery it can be shown that
$$ int_{0}^{1}frac{log(1+i x)log(1+x)}{x},dx=\frac{pi K}{2}-frac{9ipi^3}{64}+3iKlog(2)-frac{3pi i}{16}log^2(2)+frac{5pi^2}{32}log(2)-frac{log^3(2)}{8}-frac{69}{16}zeta(3)+6,text{Li}_3left(tfrac{1+i}{2}right) $$
$$ int_{0}^{1}frac{log^2(1+i x)}{x},dx=\
-frac{pi K}{2}-frac{3ipi^3}{64}+iKlog(2)-frac{pi i}{16}log^2(2)+frac{5pi^2}{96}log(2)-frac{log^3(2)}{24}-frac{3}{16}zeta(3)+2,text{Li}_3left(tfrac{1+i}{2}right) $$
$$ int_{0}^{1}frac{log(1+ix)log(1-ix)}{x},dx= frac{pi K}{2}-frac{27}{32}zeta(3)$$
hence the claim follows by $arctan x=text{Im},log(1+ix)$ and $log(1+x^2)=log(1+ix)+log(1-ix)$.
answered Nov 20 '18 at 20:20
Jack D'Aurizio
286k33279656
Through the dilogarithm/trilogarithm machinery it can be shown that
$$ int_{0}^{1}frac{log(1+i x)log(1+x)}{x},dx=\frac{pi K}{2}-frac{9ipi^3}{64}+3iKlog(2)-frac{3pi i}{16}log^2(2)+frac{5pi^2}{32}log(2)-frac{log^3(2)}{8}-frac{69}{16}zeta(3)+6,text{Li}_3left(tfrac{1+i}{2}right) $$
$$ int_{0}^{1}frac{log^2(1+i x)}{x},dx=\
-frac{pi K}{2}-frac{3ipi^3}{64}+iKlog(2)-frac{pi i}{16}log^2(2)+frac{5pi^2}{96}log(2)-frac{log^3(2)}{24}-frac{3}{16}zeta(3)+2,text{Li}_3left(tfrac{1+i}{2}right) $$
$$ int_{0}^{1}frac{log(1+ix)log(1-ix)}{x},dx= frac{pi K}{2}-frac{27}{32}zeta(3)$$
hence the claim follows by $arctan x=text{Im},log(1+ix)$ and $log(1+x^2)=log(1+ix)+log(1-ix)$.
answered Nov 20 '18 at 20:20
Jack D'Aurizio
286k33279656
answered Nov 20 '18 at 20:20
Jack D'Aurizio
286k33279656
answered Nov 20 '18 at 20:20
Jack D'Aurizio
286k33279656
answered Nov 20 '18 at 20:20
Jack D'Aurizio
286k33279656
286k33279656
Thank you for the great answer. :) But I prefer a solution without brute force.
– Kemono Chen
Nov 21 '18 at 3:02
add a comment |
Thank you for the great answer. :) But I prefer a solution without brute force.
– Kemono Chen
Nov 21 '18 at 3:02
Thank you for the great answer. :) But I prefer a solution without brute force.
– Kemono Chen
Nov 21 '18 at 3:02
Thank you for the great answer. :) But I prefer a solution without brute force.
– Kemono Chen
Nov 21 '18 at 3:02
add a comment |
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Just wondering: Why do you want to calculate this (by hand)
– klirk
Nov 20 '18 at 20:38
@klirk Just an interest.
– Kemono Chen
Nov 21 '18 at 0:18
$$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt=frac{pi^2}{48}$$
– user178256
Nov 21 '18 at 11:00