Mixing
Evaluating $int_0^{tfrac{pi}{4}}ln(cos x-sin x)ln(cos x) dx - int_0^{tfrac{pi}{4}}ln(cos x+sin x)ln(sin...
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In order to compute, in an elementary way,
$displaystyle int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$
(see Evaluating $int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$ )
i need to show, in a simple way, that:
$displaystyle int_0^{tfrac{pi}{4}}ln(cos x-sin x)ln(cos x) dx - int_0^{tfrac{pi}{4}}ln(cos x+sin x)ln(sin x)dx=dfrac{Gln 2}{2}$
$G$ being the Catalan constant.
The reason of my interest for this question is, if i am right, that this formula permits to find out a relation between integrals:
$displaystyle int_0^1 dfrac{ln(1+x)ln(1+x^2)}{1+x^2}dx$
$displaystyle int_0^1 dfrac{ln(1+x)^2}{1+x^2}dx$
$displaystyle int_0^{tfrac{pi}{4}}big(ln(cos x)big)^2 dfrac{}{}dx$
and some constants.
integration definite-integrals
edited Jan 15 at 13:23
Lorenzo B.
1,8402520
asked Jul 6 '16 at 11:32
FDPFDP
5,45211524
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add a comment |
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In order to compute, in an elementary way,
$displaystyle int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$
(see Evaluating $int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$ )
i need to show, in a simple way, that:
$displaystyle int_0^{tfrac{pi}{4}}ln(cos x-sin x)ln(cos x) dx - int_0^{tfrac{pi}{4}}ln(cos x+sin x)ln(sin x)dx=dfrac{Gln 2}{2}$
$G$ being the Catalan constant.
The reason of my interest for this question is, if i am right, that this formula permits to find out a relation between integrals:
$displaystyle int_0^1 dfrac{ln(1+x)ln(1+x^2)}{1+x^2}dx$
$displaystyle int_0^1 dfrac{ln(1+x)^2}{1+x^2}dx$
$displaystyle int_0^{tfrac{pi}{4}}big(ln(cos x)big)^2 dfrac{}{}dx$
and some constants.
integration definite-integrals
edited Jan 15 at 13:23
Lorenzo B.
1,8402520
asked Jul 6 '16 at 11:32
FDPFDP
5,45211524
$endgroup$
add a comment |
$begingroup$
In order to compute, in an elementary way,
$displaystyle int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$
(see Evaluating $int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$ )
i need to show, in a simple way, that:
$displaystyle int_0^{tfrac{pi}{4}}ln(cos x-sin x)ln(cos x) dx - int_0^{tfrac{pi}{4}}ln(cos x+sin x)ln(sin x)dx=dfrac{Gln 2}{2}$
$G$ being the Catalan constant.
The reason of my interest for this question is, if i am right, that this formula permits to find out a relation between integrals:
$displaystyle int_0^1 dfrac{ln(1+x)ln(1+x^2)}{1+x^2}dx$
$displaystyle int_0^1 dfrac{ln(1+x)^2}{1+x^2}dx$
$displaystyle int_0^{tfrac{pi}{4}}big(ln(cos x)big)^2 dfrac{}{}dx$
and some constants.
integration definite-integrals
edited Jan 15 at 13:23
Lorenzo B.
1,8402520
asked Jul 6 '16 at 11:32
FDPFDP
5,45211524
$endgroup$
In order to compute, in an elementary way,
$displaystyle int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$
(see Evaluating $int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$ )
i need to show, in a simple way, that:
$displaystyle int_0^{tfrac{pi}{4}}ln(cos x-sin x)ln(cos x) dx - int_0^{tfrac{pi}{4}}ln(cos x+sin x)ln(sin x)dx=dfrac{Gln 2}{2}$
$G$ being the Catalan constant.
The reason of my interest for this question is, if i am right, that this formula permits to find out a relation between integrals:
$displaystyle int_0^1 dfrac{ln(1+x)ln(1+x^2)}{1+x^2}dx$
$displaystyle int_0^1 dfrac{ln(1+x)^2}{1+x^2}dx$
$displaystyle int_0^{tfrac{pi}{4}}big(ln(cos x)big)^2 dfrac{}{}dx$
and some constants.
integration definite-integrals
integration definite-integrals
edited Jan 15 at 13:23
Lorenzo B.
1,8402520
asked Jul 6 '16 at 11:32
FDPFDP
5,45211524
edited Jan 15 at 13:23
Lorenzo B.
1,8402520
asked Jul 6 '16 at 11:32
FDPFDP
5,45211524
edited Jan 15 at 13:23
Lorenzo B.
1,8402520
edited Jan 15 at 13:23
Lorenzo B.
1,8402520
edited Jan 15 at 13:23
Lorenzo B.
1,8402520
1,8402520
asked Jul 6 '16 at 11:32
FDPFDP
5,45211524
asked Jul 6 '16 at 11:32
FDPFDP
5,45211524
asked Jul 6 '16 at 11:32
FDPFDP
5,45211524
5,45211524
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
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Using Simpsons rule we know
$$cos(x)-sin(x)=sqrt{2}sinleft(frac{pi}{4}-xright)$$ and
$$cos(x)+sin(x)=sqrt{2}cosleft(frac{pi}{4}-xright)$$
Plugging this in the original integrals yields:
$$int_0^{frac{pi}{4}}lnleft(sqrt{2}sinleft(frac{pi}{4}-xright)right)lnleft(cos(x)right)mathrm{d}x-
int_0^{frac{pi}{4}}lnleft(sqrt{2}cosleft(frac{pi}{4}-xright)right)lnleft(sin(x)right)mathrm{d}x$$
Splitting the logarithms and rearranging yields:
$$int_0^{frac{pi}{4}}ln(sqrt{2})left(ln(cos(x))-ln(sin(x)right)mathrm{d}x+\
+left(int_0^{frac{pi}{4}}lnleft(sinleft(frac{pi}{4}-x right)right)ln(cos(x)) mathrm{d}x-int_0^{frac{pi}{4}}lnleft(cosleft(frac{pi}{4}-x right)right)ln(sin(x)) mathrm{d}x right)$$
The first part yields
$$int_0^{frac{pi}{4}}ln(sqrt{2})left(ln(cos(x))-ln(sin(x)right)mathrm{d}x = ln(sqrt{2})G=frac{Gln(2)}{2}$$
and because $$int_0^bf(x)mathrm{d}x=int_0^bf(b-x)mathrm{d}x$$
the second part yields
$$int_0^{frac{pi}{4}}lnleft(sinleft(frac{pi}{4}-x right)right)ln(cos(x)) mathrm{d}x-int_0^{frac{pi}{4}}lnleft(cosleft(frac{pi}{4}-x right)right)ln(sin(x)) mathrm{d}x=0$$
answered Jul 6 '16 at 18:52
Erik JurriënErik Jurriën
29617
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Very nice ! Thank you
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– FDP
Jul 6 '16 at 20:07
add a comment |
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begin{align}
I&:=int_0^{tfrac{pi}{4}}ln(cos x-sin x)ln(cos x),dx - int_0^{tfrac{pi}{4}}ln(cos x+sin x)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},cosleft(x+frac {pi}4right)right)ln(cos x),dx - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
& quadtext{setting}; x=frac {pi}4-y; text{in the first integral gives}\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},cosleft(frac {pi}2-yright)right)lnleft(cosleft(frac {pi}2-frac {pi}4-yright)right),dy - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(yright)right)lnleft(sinleft(y+frac {pi}4right)right),dy - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}left(ln(sqrt{2})+lnleft(sin xright)right)lnleft(sinleft(x+frac {pi}4right)right) - left(ln(sqrt{2})+lnleft(sinleft(x+frac {pi}4right)right)right)ln(sin x);dx\
&=ln(sqrt{2})int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x);dx\
&=dfrac{ln 2}{2}G\
end{align}
It remains to prove that $displaystyle int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x);dx=G$.
This is detailed around $(16)$ in this interesting paper by Jameson and Lord or using :
begin{align}
int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x),dx&=int_{tfrac{pi}{4}}^{tfrac{pi}{2}}lnleft(sin xright),dx-int_0^{tfrac{pi}{4}}ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}ln(cos x),dx-int_0^{tfrac{pi}{4}}ln(sin x),dx\
&=-int_0^{tfrac{pi}{4}}ln(tan x),dx\
&=-int_0^1frac{ln t}{1+t^2},dt,quadtext{integrated by parts}\
&=-left.ln(t);arctan(t)right|_0^1+int_0^1frac{arctan(t)}{t},dt\
&=int_0^1sum_{n=0}^infty (-1)^nfrac{t^{2n}}{2n+1},dt\
&=left.sum_{n=0}^infty (-1)^nfrac{t^{2n+1}}{(2n+1)^2}right|_0^1\
&=G\
end{align}
answered Jul 6 '16 at 18:53
Raymond ManzoniRaymond Manzoni
37.2k563117
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Thank you ! I was expecting such tricky proof. To say the truth i was close to find it. I hope it will help me to terminate my very lengthy computation
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– FDP
Jul 6 '16 at 19:57
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Thanks for the link to this paper.
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– FDP
Jul 6 '16 at 21:51
add a comment |
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Relationship between the proposed integrals.
$displaystyle int_0^1 dfrac{ln(1+x)^2}{1+x^2}dx$-$displaystyle int_0^1 dfrac{ln(1+x)ln(1+x^2)}{1+x^2}dx=-frac{3pi}{16}ln^2{2}+dfrac{Gln 2}{2}$
answered Jul 6 '16 at 21:24
user178256user178256
2,1401832
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Have you an elementary proof of this?
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– FDP
Jul 6 '16 at 21:54
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Anyway I think i don't need such precise value to compute $int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$
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– FDP
Jul 6 '16 at 21:57
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I think i see how to prove this. It relies on the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ using Simpson's rule and some properties of $cos$ and $sin$. Thank you, it will simplify my computation :)
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– FDP
Jul 6 '16 at 23:37
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Consider $displaystyle int_0^1 dfrac{ln(1+x^2)^2}{1+x^2}dx$ $$x=frac{1-y}{1+y}$$ we obtain the relationship
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– user178256
Jul 7 '16 at 6:16
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Direct computation using the change of variable $x=tan(y)$ is ok too. it leads, if i'm correct to the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ but to make the things easier you need to substract to both integrals $int_0^{frac{pi}{4}} (ln(cos x)^2dx$. Thanks again, i haven't noticed this relation
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– FDP
Jul 7 '16 at 8:29
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3 Answers
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3 Answers
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$begingroup$
Using Simpsons rule we know
$$cos(x)-sin(x)=sqrt{2}sinleft(frac{pi}{4}-xright)$$ and
$$cos(x)+sin(x)=sqrt{2}cosleft(frac{pi}{4}-xright)$$
Plugging this in the original integrals yields:
$$int_0^{frac{pi}{4}}lnleft(sqrt{2}sinleft(frac{pi}{4}-xright)right)lnleft(cos(x)right)mathrm{d}x-
int_0^{frac{pi}{4}}lnleft(sqrt{2}cosleft(frac{pi}{4}-xright)right)lnleft(sin(x)right)mathrm{d}x$$
Splitting the logarithms and rearranging yields:
$$int_0^{frac{pi}{4}}ln(sqrt{2})left(ln(cos(x))-ln(sin(x)right)mathrm{d}x+\
+left(int_0^{frac{pi}{4}}lnleft(sinleft(frac{pi}{4}-x right)right)ln(cos(x)) mathrm{d}x-int_0^{frac{pi}{4}}lnleft(cosleft(frac{pi}{4}-x right)right)ln(sin(x)) mathrm{d}x right)$$
The first part yields
$$int_0^{frac{pi}{4}}ln(sqrt{2})left(ln(cos(x))-ln(sin(x)right)mathrm{d}x = ln(sqrt{2})G=frac{Gln(2)}{2}$$
and because $$int_0^bf(x)mathrm{d}x=int_0^bf(b-x)mathrm{d}x$$
the second part yields
$$int_0^{frac{pi}{4}}lnleft(sinleft(frac{pi}{4}-x right)right)ln(cos(x)) mathrm{d}x-int_0^{frac{pi}{4}}lnleft(cosleft(frac{pi}{4}-x right)right)ln(sin(x)) mathrm{d}x=0$$
answered Jul 6 '16 at 18:52
Erik JurriënErik Jurriën
29617
$endgroup$
$begingroup$
Very nice ! Thank you
$endgroup$
– FDP
Jul 6 '16 at 20:07
add a comment |
$begingroup$
Using Simpsons rule we know
$$cos(x)-sin(x)=sqrt{2}sinleft(frac{pi}{4}-xright)$$ and
$$cos(x)+sin(x)=sqrt{2}cosleft(frac{pi}{4}-xright)$$
Plugging this in the original integrals yields:
$$int_0^{frac{pi}{4}}lnleft(sqrt{2}sinleft(frac{pi}{4}-xright)right)lnleft(cos(x)right)mathrm{d}x-
int_0^{frac{pi}{4}}lnleft(sqrt{2}cosleft(frac{pi}{4}-xright)right)lnleft(sin(x)right)mathrm{d}x$$
Splitting the logarithms and rearranging yields:
$$int_0^{frac{pi}{4}}ln(sqrt{2})left(ln(cos(x))-ln(sin(x)right)mathrm{d}x+\
+left(int_0^{frac{pi}{4}}lnleft(sinleft(frac{pi}{4}-x right)right)ln(cos(x)) mathrm{d}x-int_0^{frac{pi}{4}}lnleft(cosleft(frac{pi}{4}-x right)right)ln(sin(x)) mathrm{d}x right)$$
The first part yields
$$int_0^{frac{pi}{4}}ln(sqrt{2})left(ln(cos(x))-ln(sin(x)right)mathrm{d}x = ln(sqrt{2})G=frac{Gln(2)}{2}$$
and because $$int_0^bf(x)mathrm{d}x=int_0^bf(b-x)mathrm{d}x$$
the second part yields
$$int_0^{frac{pi}{4}}lnleft(sinleft(frac{pi}{4}-x right)right)ln(cos(x)) mathrm{d}x-int_0^{frac{pi}{4}}lnleft(cosleft(frac{pi}{4}-x right)right)ln(sin(x)) mathrm{d}x=0$$
answered Jul 6 '16 at 18:52
Erik JurriënErik Jurriën
29617
$endgroup$
$begingroup$
Very nice ! Thank you
$endgroup$
– FDP
Jul 6 '16 at 20:07
add a comment |
$begingroup$
Using Simpsons rule we know
$$cos(x)-sin(x)=sqrt{2}sinleft(frac{pi}{4}-xright)$$ and
$$cos(x)+sin(x)=sqrt{2}cosleft(frac{pi}{4}-xright)$$
Plugging this in the original integrals yields:
$$int_0^{frac{pi}{4}}lnleft(sqrt{2}sinleft(frac{pi}{4}-xright)right)lnleft(cos(x)right)mathrm{d}x-
int_0^{frac{pi}{4}}lnleft(sqrt{2}cosleft(frac{pi}{4}-xright)right)lnleft(sin(x)right)mathrm{d}x$$
Splitting the logarithms and rearranging yields:
$$int_0^{frac{pi}{4}}ln(sqrt{2})left(ln(cos(x))-ln(sin(x)right)mathrm{d}x+\
+left(int_0^{frac{pi}{4}}lnleft(sinleft(frac{pi}{4}-x right)right)ln(cos(x)) mathrm{d}x-int_0^{frac{pi}{4}}lnleft(cosleft(frac{pi}{4}-x right)right)ln(sin(x)) mathrm{d}x right)$$
The first part yields
$$int_0^{frac{pi}{4}}ln(sqrt{2})left(ln(cos(x))-ln(sin(x)right)mathrm{d}x = ln(sqrt{2})G=frac{Gln(2)}{2}$$
and because $$int_0^bf(x)mathrm{d}x=int_0^bf(b-x)mathrm{d}x$$
the second part yields
$$int_0^{frac{pi}{4}}lnleft(sinleft(frac{pi}{4}-x right)right)ln(cos(x)) mathrm{d}x-int_0^{frac{pi}{4}}lnleft(cosleft(frac{pi}{4}-x right)right)ln(sin(x)) mathrm{d}x=0$$
answered Jul 6 '16 at 18:52
Erik JurriënErik Jurriën
29617
$endgroup$
Using Simpsons rule we know
$$cos(x)-sin(x)=sqrt{2}sinleft(frac{pi}{4}-xright)$$ and
$$cos(x)+sin(x)=sqrt{2}cosleft(frac{pi}{4}-xright)$$
Plugging this in the original integrals yields:
$$int_0^{frac{pi}{4}}lnleft(sqrt{2}sinleft(frac{pi}{4}-xright)right)lnleft(cos(x)right)mathrm{d}x-
int_0^{frac{pi}{4}}lnleft(sqrt{2}cosleft(frac{pi}{4}-xright)right)lnleft(sin(x)right)mathrm{d}x$$
Splitting the logarithms and rearranging yields:
$$int_0^{frac{pi}{4}}ln(sqrt{2})left(ln(cos(x))-ln(sin(x)right)mathrm{d}x+\
+left(int_0^{frac{pi}{4}}lnleft(sinleft(frac{pi}{4}-x right)right)ln(cos(x)) mathrm{d}x-int_0^{frac{pi}{4}}lnleft(cosleft(frac{pi}{4}-x right)right)ln(sin(x)) mathrm{d}x right)$$
The first part yields
$$int_0^{frac{pi}{4}}ln(sqrt{2})left(ln(cos(x))-ln(sin(x)right)mathrm{d}x = ln(sqrt{2})G=frac{Gln(2)}{2}$$
and because $$int_0^bf(x)mathrm{d}x=int_0^bf(b-x)mathrm{d}x$$
the second part yields
$$int_0^{frac{pi}{4}}lnleft(sinleft(frac{pi}{4}-x right)right)ln(cos(x)) mathrm{d}x-int_0^{frac{pi}{4}}lnleft(cosleft(frac{pi}{4}-x right)right)ln(sin(x)) mathrm{d}x=0$$
answered Jul 6 '16 at 18:52
Erik JurriënErik Jurriën
29617
edited Jul 6 '16 at 19:30
answered Jul 6 '16 at 18:52
Erik JurriënErik Jurriën
29617
answered Jul 6 '16 at 18:52
Erik JurriënErik Jurriën
29617
answered Jul 6 '16 at 18:52
Erik JurriënErik Jurriën
29617
29617
$begingroup$
Very nice ! Thank you
$endgroup$
– FDP
Jul 6 '16 at 20:07
add a comment |
$begingroup$
Very nice ! Thank you
$endgroup$
– FDP
Jul 6 '16 at 20:07
$begingroup$
Very nice ! Thank you
$endgroup$
– FDP
Jul 6 '16 at 20:07
$begingroup$
Very nice ! Thank you
$endgroup$
– FDP
Jul 6 '16 at 20:07
add a comment |
$begingroup$
begin{align}
I&:=int_0^{tfrac{pi}{4}}ln(cos x-sin x)ln(cos x),dx - int_0^{tfrac{pi}{4}}ln(cos x+sin x)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},cosleft(x+frac {pi}4right)right)ln(cos x),dx - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
& quadtext{setting}; x=frac {pi}4-y; text{in the first integral gives}\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},cosleft(frac {pi}2-yright)right)lnleft(cosleft(frac {pi}2-frac {pi}4-yright)right),dy - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(yright)right)lnleft(sinleft(y+frac {pi}4right)right),dy - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}left(ln(sqrt{2})+lnleft(sin xright)right)lnleft(sinleft(x+frac {pi}4right)right) - left(ln(sqrt{2})+lnleft(sinleft(x+frac {pi}4right)right)right)ln(sin x);dx\
&=ln(sqrt{2})int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x);dx\
&=dfrac{ln 2}{2}G\
end{align}
It remains to prove that $displaystyle int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x);dx=G$.
This is detailed around $(16)$ in this interesting paper by Jameson and Lord or using :
begin{align}
int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x),dx&=int_{tfrac{pi}{4}}^{tfrac{pi}{2}}lnleft(sin xright),dx-int_0^{tfrac{pi}{4}}ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}ln(cos x),dx-int_0^{tfrac{pi}{4}}ln(sin x),dx\
&=-int_0^{tfrac{pi}{4}}ln(tan x),dx\
&=-int_0^1frac{ln t}{1+t^2},dt,quadtext{integrated by parts}\
&=-left.ln(t);arctan(t)right|_0^1+int_0^1frac{arctan(t)}{t},dt\
&=int_0^1sum_{n=0}^infty (-1)^nfrac{t^{2n}}{2n+1},dt\
&=left.sum_{n=0}^infty (-1)^nfrac{t^{2n+1}}{(2n+1)^2}right|_0^1\
&=G\
end{align}
answered Jul 6 '16 at 18:53
Raymond ManzoniRaymond Manzoni
37.2k563117
$endgroup$
$begingroup$
Thank you ! I was expecting such tricky proof. To say the truth i was close to find it. I hope it will help me to terminate my very lengthy computation
$endgroup$
– FDP
Jul 6 '16 at 19:57
$begingroup$
Thanks for the link to this paper.
$endgroup$
– FDP
Jul 6 '16 at 21:51
add a comment |
$begingroup$
begin{align}
I&:=int_0^{tfrac{pi}{4}}ln(cos x-sin x)ln(cos x),dx - int_0^{tfrac{pi}{4}}ln(cos x+sin x)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},cosleft(x+frac {pi}4right)right)ln(cos x),dx - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
& quadtext{setting}; x=frac {pi}4-y; text{in the first integral gives}\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},cosleft(frac {pi}2-yright)right)lnleft(cosleft(frac {pi}2-frac {pi}4-yright)right),dy - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(yright)right)lnleft(sinleft(y+frac {pi}4right)right),dy - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}left(ln(sqrt{2})+lnleft(sin xright)right)lnleft(sinleft(x+frac {pi}4right)right) - left(ln(sqrt{2})+lnleft(sinleft(x+frac {pi}4right)right)right)ln(sin x);dx\
&=ln(sqrt{2})int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x);dx\
&=dfrac{ln 2}{2}G\
end{align}
It remains to prove that $displaystyle int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x);dx=G$.
This is detailed around $(16)$ in this interesting paper by Jameson and Lord or using :
begin{align}
int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x),dx&=int_{tfrac{pi}{4}}^{tfrac{pi}{2}}lnleft(sin xright),dx-int_0^{tfrac{pi}{4}}ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}ln(cos x),dx-int_0^{tfrac{pi}{4}}ln(sin x),dx\
&=-int_0^{tfrac{pi}{4}}ln(tan x),dx\
&=-int_0^1frac{ln t}{1+t^2},dt,quadtext{integrated by parts}\
&=-left.ln(t);arctan(t)right|_0^1+int_0^1frac{arctan(t)}{t},dt\
&=int_0^1sum_{n=0}^infty (-1)^nfrac{t^{2n}}{2n+1},dt\
&=left.sum_{n=0}^infty (-1)^nfrac{t^{2n+1}}{(2n+1)^2}right|_0^1\
&=G\
end{align}
answered Jul 6 '16 at 18:53
Raymond ManzoniRaymond Manzoni
37.2k563117
$endgroup$
$begingroup$
Thank you ! I was expecting such tricky proof. To say the truth i was close to find it. I hope it will help me to terminate my very lengthy computation
$endgroup$
– FDP
Jul 6 '16 at 19:57
$begingroup$
Thanks for the link to this paper.
$endgroup$
– FDP
Jul 6 '16 at 21:51
add a comment |
$begingroup$
begin{align}
I&:=int_0^{tfrac{pi}{4}}ln(cos x-sin x)ln(cos x),dx - int_0^{tfrac{pi}{4}}ln(cos x+sin x)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},cosleft(x+frac {pi}4right)right)ln(cos x),dx - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
& quadtext{setting}; x=frac {pi}4-y; text{in the first integral gives}\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},cosleft(frac {pi}2-yright)right)lnleft(cosleft(frac {pi}2-frac {pi}4-yright)right),dy - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(yright)right)lnleft(sinleft(y+frac {pi}4right)right),dy - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}left(ln(sqrt{2})+lnleft(sin xright)right)lnleft(sinleft(x+frac {pi}4right)right) - left(ln(sqrt{2})+lnleft(sinleft(x+frac {pi}4right)right)right)ln(sin x);dx\
&=ln(sqrt{2})int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x);dx\
&=dfrac{ln 2}{2}G\
end{align}
It remains to prove that $displaystyle int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x);dx=G$.
This is detailed around $(16)$ in this interesting paper by Jameson and Lord or using :
begin{align}
int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x),dx&=int_{tfrac{pi}{4}}^{tfrac{pi}{2}}lnleft(sin xright),dx-int_0^{tfrac{pi}{4}}ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}ln(cos x),dx-int_0^{tfrac{pi}{4}}ln(sin x),dx\
&=-int_0^{tfrac{pi}{4}}ln(tan x),dx\
&=-int_0^1frac{ln t}{1+t^2},dt,quadtext{integrated by parts}\
&=-left.ln(t);arctan(t)right|_0^1+int_0^1frac{arctan(t)}{t},dt\
&=int_0^1sum_{n=0}^infty (-1)^nfrac{t^{2n}}{2n+1},dt\
&=left.sum_{n=0}^infty (-1)^nfrac{t^{2n+1}}{(2n+1)^2}right|_0^1\
&=G\
end{align}
answered Jul 6 '16 at 18:53
Raymond ManzoniRaymond Manzoni
37.2k563117
$endgroup$
begin{align}
I&:=int_0^{tfrac{pi}{4}}ln(cos x-sin x)ln(cos x),dx - int_0^{tfrac{pi}{4}}ln(cos x+sin x)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},cosleft(x+frac {pi}4right)right)ln(cos x),dx - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
& quadtext{setting}; x=frac {pi}4-y; text{in the first integral gives}\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},cosleft(frac {pi}2-yright)right)lnleft(cosleft(frac {pi}2-frac {pi}4-yright)right),dy - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(yright)right)lnleft(sinleft(y+frac {pi}4right)right),dy - int_0^{tfrac{pi}{4}}lnleft(sqrt{2},sinleft(x+frac {pi}4right)right)ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}left(ln(sqrt{2})+lnleft(sin xright)right)lnleft(sinleft(x+frac {pi}4right)right) - left(ln(sqrt{2})+lnleft(sinleft(x+frac {pi}4right)right)right)ln(sin x);dx\
&=ln(sqrt{2})int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x);dx\
&=dfrac{ln 2}{2}G\
end{align}
It remains to prove that $displaystyle int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x);dx=G$.
This is detailed around $(16)$ in this interesting paper by Jameson and Lord or using :
begin{align}
int_0^{tfrac{pi}{4}}lnleft(sinleft(x+frac {pi}4right)right)-ln(sin x),dx&=int_{tfrac{pi}{4}}^{tfrac{pi}{2}}lnleft(sin xright),dx-int_0^{tfrac{pi}{4}}ln(sin x),dx\
&=int_0^{tfrac{pi}{4}}ln(cos x),dx-int_0^{tfrac{pi}{4}}ln(sin x),dx\
&=-int_0^{tfrac{pi}{4}}ln(tan x),dx\
&=-int_0^1frac{ln t}{1+t^2},dt,quadtext{integrated by parts}\
&=-left.ln(t);arctan(t)right|_0^1+int_0^1frac{arctan(t)}{t},dt\
&=int_0^1sum_{n=0}^infty (-1)^nfrac{t^{2n}}{2n+1},dt\
&=left.sum_{n=0}^infty (-1)^nfrac{t^{2n+1}}{(2n+1)^2}right|_0^1\
&=G\
end{align}
answered Jul 6 '16 at 18:53
Raymond ManzoniRaymond Manzoni
37.2k563117
edited Jul 6 '16 at 23:16
answered Jul 6 '16 at 18:53
Raymond ManzoniRaymond Manzoni
37.2k563117
answered Jul 6 '16 at 18:53
Raymond ManzoniRaymond Manzoni
37.2k563117
answered Jul 6 '16 at 18:53
Raymond ManzoniRaymond Manzoni
37.2k563117
37.2k563117
$begingroup$
Thank you ! I was expecting such tricky proof. To say the truth i was close to find it. I hope it will help me to terminate my very lengthy computation
$endgroup$
– FDP
Jul 6 '16 at 19:57
$begingroup$
Thanks for the link to this paper.
$endgroup$
– FDP
Jul 6 '16 at 21:51
add a comment |
$begingroup$
Thank you ! I was expecting such tricky proof. To say the truth i was close to find it. I hope it will help me to terminate my very lengthy computation
$endgroup$
– FDP
Jul 6 '16 at 19:57
$begingroup$
Thanks for the link to this paper.
$endgroup$
– FDP
Jul 6 '16 at 21:51
$begingroup$
Thank you ! I was expecting such tricky proof. To say the truth i was close to find it. I hope it will help me to terminate my very lengthy computation
$endgroup$
– FDP
Jul 6 '16 at 19:57
$begingroup$
Thank you ! I was expecting such tricky proof. To say the truth i was close to find it. I hope it will help me to terminate my very lengthy computation
$endgroup$
– FDP
Jul 6 '16 at 19:57
$begingroup$
Thanks for the link to this paper.
$endgroup$
– FDP
Jul 6 '16 at 21:51
$begingroup$
Thanks for the link to this paper.
$endgroup$
– FDP
Jul 6 '16 at 21:51
add a comment |
$begingroup$
Relationship between the proposed integrals.
$displaystyle int_0^1 dfrac{ln(1+x)^2}{1+x^2}dx$-$displaystyle int_0^1 dfrac{ln(1+x)ln(1+x^2)}{1+x^2}dx=-frac{3pi}{16}ln^2{2}+dfrac{Gln 2}{2}$
answered Jul 6 '16 at 21:24
user178256user178256
2,1401832
$endgroup$
$begingroup$
Have you an elementary proof of this?
$endgroup$
– FDP
Jul 6 '16 at 21:54
$begingroup$
Anyway I think i don't need such precise value to compute $int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$
$endgroup$
– FDP
Jul 6 '16 at 21:57
$begingroup$
I think i see how to prove this. It relies on the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ using Simpson's rule and some properties of $cos$ and $sin$. Thank you, it will simplify my computation :)
$endgroup$
– FDP
Jul 6 '16 at 23:37
$begingroup$
Consider $displaystyle int_0^1 dfrac{ln(1+x^2)^2}{1+x^2}dx$ $$x=frac{1-y}{1+y}$$ we obtain the relationship
$endgroup$
– user178256
Jul 7 '16 at 6:16
$begingroup$
Direct computation using the change of variable $x=tan(y)$ is ok too. it leads, if i'm correct to the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ but to make the things easier you need to substract to both integrals $int_0^{frac{pi}{4}} (ln(cos x)^2dx$. Thanks again, i haven't noticed this relation
$endgroup$
– FDP
Jul 7 '16 at 8:29
|
show 2 more comments
$begingroup$
Relationship between the proposed integrals.
$displaystyle int_0^1 dfrac{ln(1+x)^2}{1+x^2}dx$-$displaystyle int_0^1 dfrac{ln(1+x)ln(1+x^2)}{1+x^2}dx=-frac{3pi}{16}ln^2{2}+dfrac{Gln 2}{2}$
answered Jul 6 '16 at 21:24
user178256user178256
2,1401832
$endgroup$
$begingroup$
Have you an elementary proof of this?
$endgroup$
– FDP
Jul 6 '16 at 21:54
$begingroup$
Anyway I think i don't need such precise value to compute $int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$
$endgroup$
– FDP
Jul 6 '16 at 21:57
$begingroup$
I think i see how to prove this. It relies on the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ using Simpson's rule and some properties of $cos$ and $sin$. Thank you, it will simplify my computation :)
$endgroup$
– FDP
Jul 6 '16 at 23:37
$begingroup$
Consider $displaystyle int_0^1 dfrac{ln(1+x^2)^2}{1+x^2}dx$ $$x=frac{1-y}{1+y}$$ we obtain the relationship
$endgroup$
– user178256
Jul 7 '16 at 6:16
$begingroup$
Direct computation using the change of variable $x=tan(y)$ is ok too. it leads, if i'm correct to the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ but to make the things easier you need to substract to both integrals $int_0^{frac{pi}{4}} (ln(cos x)^2dx$. Thanks again, i haven't noticed this relation
$endgroup$
– FDP
Jul 7 '16 at 8:29
|
show 2 more comments
$begingroup$
Relationship between the proposed integrals.
$displaystyle int_0^1 dfrac{ln(1+x)^2}{1+x^2}dx$-$displaystyle int_0^1 dfrac{ln(1+x)ln(1+x^2)}{1+x^2}dx=-frac{3pi}{16}ln^2{2}+dfrac{Gln 2}{2}$
answered Jul 6 '16 at 21:24
user178256user178256
2,1401832
$endgroup$
Relationship between the proposed integrals.
$displaystyle int_0^1 dfrac{ln(1+x)^2}{1+x^2}dx$-$displaystyle int_0^1 dfrac{ln(1+x)ln(1+x^2)}{1+x^2}dx=-frac{3pi}{16}ln^2{2}+dfrac{Gln 2}{2}$
answered Jul 6 '16 at 21:24
user178256user178256
2,1401832
answered Jul 6 '16 at 21:24
user178256user178256
2,1401832
answered Jul 6 '16 at 21:24
user178256user178256
2,1401832
answered Jul 6 '16 at 21:24
user178256user178256
2,1401832
2,1401832
$begingroup$
Have you an elementary proof of this?
$endgroup$
– FDP
Jul 6 '16 at 21:54
$begingroup$
Anyway I think i don't need such precise value to compute $int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$
$endgroup$
– FDP
Jul 6 '16 at 21:57
$begingroup$
I think i see how to prove this. It relies on the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ using Simpson's rule and some properties of $cos$ and $sin$. Thank you, it will simplify my computation :)
$endgroup$
– FDP
Jul 6 '16 at 23:37
$begingroup$
Consider $displaystyle int_0^1 dfrac{ln(1+x^2)^2}{1+x^2}dx$ $$x=frac{1-y}{1+y}$$ we obtain the relationship
$endgroup$
– user178256
Jul 7 '16 at 6:16
$begingroup$
Direct computation using the change of variable $x=tan(y)$ is ok too. it leads, if i'm correct to the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ but to make the things easier you need to substract to both integrals $int_0^{frac{pi}{4}} (ln(cos x)^2dx$. Thanks again, i haven't noticed this relation
$endgroup$
– FDP
Jul 7 '16 at 8:29
|
show 2 more comments
$begingroup$
Have you an elementary proof of this?
$endgroup$
– FDP
Jul 6 '16 at 21:54
$begingroup$
Anyway I think i don't need such precise value to compute $int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$
$endgroup$
– FDP
Jul 6 '16 at 21:57
$begingroup$
I think i see how to prove this. It relies on the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ using Simpson's rule and some properties of $cos$ and $sin$. Thank you, it will simplify my computation :)
$endgroup$
– FDP
Jul 6 '16 at 23:37
$begingroup$
Consider $displaystyle int_0^1 dfrac{ln(1+x^2)^2}{1+x^2}dx$ $$x=frac{1-y}{1+y}$$ we obtain the relationship
$endgroup$
– user178256
Jul 7 '16 at 6:16
$begingroup$
Direct computation using the change of variable $x=tan(y)$ is ok too. it leads, if i'm correct to the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ but to make the things easier you need to substract to both integrals $int_0^{frac{pi}{4}} (ln(cos x)^2dx$. Thanks again, i haven't noticed this relation
$endgroup$
– FDP
Jul 7 '16 at 8:29
$begingroup$
Have you an elementary proof of this?
$endgroup$
– FDP
Jul 6 '16 at 21:54
$begingroup$
Have you an elementary proof of this?
$endgroup$
– FDP
Jul 6 '16 at 21:54
$begingroup$
Anyway I think i don't need such precise value to compute $int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$
$endgroup$
– FDP
Jul 6 '16 at 21:57
$begingroup$
Anyway I think i don't need such precise value to compute $int_0^1 frac{x arctan x log left( 1-x^2right)}{1+x^2}dx$
$endgroup$
– FDP
Jul 6 '16 at 21:57
$begingroup$
I think i see how to prove this. It relies on the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ using Simpson's rule and some properties of $cos$ and $sin$. Thank you, it will simplify my computation :)
$endgroup$
– FDP
Jul 6 '16 at 23:37
$begingroup$
I think i see how to prove this. It relies on the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ using Simpson's rule and some properties of $cos$ and $sin$. Thank you, it will simplify my computation :)
$endgroup$
– FDP
Jul 6 '16 at 23:37
$begingroup$
Consider $displaystyle int_0^1 dfrac{ln(1+x^2)^2}{1+x^2}dx$ $$x=frac{1-y}{1+y}$$ we obtain the relationship
$endgroup$
– user178256
Jul 7 '16 at 6:16
$begingroup$
Consider $displaystyle int_0^1 dfrac{ln(1+x^2)^2}{1+x^2}dx$ $$x=frac{1-y}{1+y}$$ we obtain the relationship
$endgroup$
– user178256
Jul 7 '16 at 6:16
$begingroup$
Direct computation using the change of variable $x=tan(y)$ is ok too. it leads, if i'm correct to the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ but to make the things easier you need to substract to both integrals $int_0^{frac{pi}{4}} (ln(cos x)^2dx$. Thanks again, i haven't noticed this relation
$endgroup$
– FDP
Jul 7 '16 at 8:29
$begingroup$
Direct computation using the change of variable $x=tan(y)$ is ok too. it leads, if i'm correct to the evaluation of $int_0^{frac{pi}{4}} (ln(cos x+sin x))^2dx$ but to make the things easier you need to substract to both integrals $int_0^{frac{pi}{4}} (ln(cos x)^2dx$. Thanks again, i haven't noticed this relation
$endgroup$
– FDP
Jul 7 '16 at 8:29
|
show 2 more comments
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