Proving $Imoperatorname{Li}_2(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1)$












3












$begingroup$


$newcommand{Li}{operatorname{Li}_2}$




I found, numerically, that $$ImLi(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1).$$
How can we prove it?




My attempt of proving this equation:
Using identity $$Li(x)=int_0^1frac{x}{xt-1}ln tdt,$$
we can deduce$$begin{align}ImLi(sqrt i(sqrt 2-1))&=frac1{2i}int_0^1left(frac{sqrt i(sqrt2-1)}{sqrt i(sqrt2-1)t-1}-frac{sqrt {-i}(sqrt2-1)}{sqrt {-i}(sqrt2-1)t-1}right)ln tdt\
&=int_0^1frac{2-sqrt{2}}{left(4 sqrt{2}-6right) t^2-2 left(sqrt{2}-2right) t-2}ln tdt\
&=int_0^{2-sqrt2}-frac{1}{u^2-2u+2}lnfrac u{2-sqrt 2}du\
&=frac18piln(2-sqrt2)-int_{-1}^{1-sqrt2}frac{ln(v+1)}{v^2+1}dv\
&=frac18piln(2-sqrt2)-int_{pi/8}^{pi/4}ln(1-tan x)dx\
&=frac18piln(sqrt2-1)-int_{pi/8}^{pi/4}lnsec x+lnsinleft(fracpi4-xright)dx\
end{align}$$
I have no idea how to deal with the log-trig integral.






calculus definite-integrals polylogarithm







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
    $endgroup$
    – Awe Kumar Jha
    Dec 7 '18 at 11:49












  • $begingroup$
    $logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 17:06










  • $begingroup$
    What is $sqrt{i}$?
    $endgroup$
    – FDP
    Dec 7 '18 at 18:14










  • $begingroup$
    @FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 0:55










  • $begingroup$
    why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
    $endgroup$
    – FDP
    Dec 8 '18 at 11:30


















3












$begingroup$


$newcommand{Li}{operatorname{Li}_2}$




I found, numerically, that $$ImLi(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1).$$
How can we prove it?




My attempt of proving this equation:
Using identity $$Li(x)=int_0^1frac{x}{xt-1}ln tdt,$$
we can deduce$$begin{align}ImLi(sqrt i(sqrt 2-1))&=frac1{2i}int_0^1left(frac{sqrt i(sqrt2-1)}{sqrt i(sqrt2-1)t-1}-frac{sqrt {-i}(sqrt2-1)}{sqrt {-i}(sqrt2-1)t-1}right)ln tdt\
&=int_0^1frac{2-sqrt{2}}{left(4 sqrt{2}-6right) t^2-2 left(sqrt{2}-2right) t-2}ln tdt\
&=int_0^{2-sqrt2}-frac{1}{u^2-2u+2}lnfrac u{2-sqrt 2}du\
&=frac18piln(2-sqrt2)-int_{-1}^{1-sqrt2}frac{ln(v+1)}{v^2+1}dv\
&=frac18piln(2-sqrt2)-int_{pi/8}^{pi/4}ln(1-tan x)dx\
&=frac18piln(sqrt2-1)-int_{pi/8}^{pi/4}lnsec x+lnsinleft(fracpi4-xright)dx\
end{align}$$
I have no idea how to deal with the log-trig integral.






calculus definite-integrals polylogarithm







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
    $endgroup$
    – Awe Kumar Jha
    Dec 7 '18 at 11:49












  • $begingroup$
    $logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 17:06










  • $begingroup$
    What is $sqrt{i}$?
    $endgroup$
    – FDP
    Dec 7 '18 at 18:14










  • $begingroup$
    @FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 0:55










  • $begingroup$
    why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
    $endgroup$
    – FDP
    Dec 8 '18 at 11:30
















3












3








3


0



$begingroup$


$newcommand{Li}{operatorname{Li}_2}$




I found, numerically, that $$ImLi(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1).$$
How can we prove it?




My attempt of proving this equation:
Using identity $$Li(x)=int_0^1frac{x}{xt-1}ln tdt,$$
we can deduce$$begin{align}ImLi(sqrt i(sqrt 2-1))&=frac1{2i}int_0^1left(frac{sqrt i(sqrt2-1)}{sqrt i(sqrt2-1)t-1}-frac{sqrt {-i}(sqrt2-1)}{sqrt {-i}(sqrt2-1)t-1}right)ln tdt\
&=int_0^1frac{2-sqrt{2}}{left(4 sqrt{2}-6right) t^2-2 left(sqrt{2}-2right) t-2}ln tdt\
&=int_0^{2-sqrt2}-frac{1}{u^2-2u+2}lnfrac u{2-sqrt 2}du\
&=frac18piln(2-sqrt2)-int_{-1}^{1-sqrt2}frac{ln(v+1)}{v^2+1}dv\
&=frac18piln(2-sqrt2)-int_{pi/8}^{pi/4}ln(1-tan x)dx\
&=frac18piln(sqrt2-1)-int_{pi/8}^{pi/4}lnsec x+lnsinleft(fracpi4-xright)dx\
end{align}$$
I have no idea how to deal with the log-trig integral.






calculus definite-integrals polylogarithm







share|cite|improve this question









$endgroup$




$newcommand{Li}{operatorname{Li}_2}$




I found, numerically, that $$ImLi(sqrt i(sqrt 2-1))=frac34G+frac18piln(sqrt2-1).$$
How can we prove it?




My attempt of proving this equation:
Using identity $$Li(x)=int_0^1frac{x}{xt-1}ln tdt,$$
we can deduce$$begin{align}ImLi(sqrt i(sqrt 2-1))&=frac1{2i}int_0^1left(frac{sqrt i(sqrt2-1)}{sqrt i(sqrt2-1)t-1}-frac{sqrt {-i}(sqrt2-1)}{sqrt {-i}(sqrt2-1)t-1}right)ln tdt\
&=int_0^1frac{2-sqrt{2}}{left(4 sqrt{2}-6right) t^2-2 left(sqrt{2}-2right) t-2}ln tdt\
&=int_0^{2-sqrt2}-frac{1}{u^2-2u+2}lnfrac u{2-sqrt 2}du\
&=frac18piln(2-sqrt2)-int_{-1}^{1-sqrt2}frac{ln(v+1)}{v^2+1}dv\
&=frac18piln(2-sqrt2)-int_{pi/8}^{pi/4}ln(1-tan x)dx\
&=frac18piln(sqrt2-1)-int_{pi/8}^{pi/4}lnsec x+lnsinleft(fracpi4-xright)dx\
end{align}$$
I have no idea how to deal with the log-trig integral.






calculus definite-integrals polylogarithm




calculus definite-integrals polylogarithm






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 11:17









Kemono ChenKemono Chen

3,1941844




3,1941844












  • $begingroup$
    Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
    $endgroup$
    – Awe Kumar Jha
    Dec 7 '18 at 11:49












  • $begingroup$
    $logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 17:06










  • $begingroup$
    What is $sqrt{i}$?
    $endgroup$
    – FDP
    Dec 7 '18 at 18:14










  • $begingroup$
    @FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 0:55










  • $begingroup$
    why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
    $endgroup$
    – FDP
    Dec 8 '18 at 11:30




















  • $begingroup$
    Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
    $endgroup$
    – Awe Kumar Jha
    Dec 7 '18 at 11:49












  • $begingroup$
    $logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
    $endgroup$
    – Jack D'Aurizio
    Dec 7 '18 at 17:06










  • $begingroup$
    What is $sqrt{i}$?
    $endgroup$
    – FDP
    Dec 7 '18 at 18:14










  • $begingroup$
    @FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
    $endgroup$
    – Kemono Chen
    Dec 8 '18 at 0:55










  • $begingroup$
    why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
    $endgroup$
    – FDP
    Dec 8 '18 at 11:30


















$begingroup$
Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
$endgroup$
– Awe Kumar Jha
Dec 7 '18 at 11:49






$begingroup$
Try reducing the lower limit of your integral to zero and then use the identity ; $$G = int_0^{π/4}log cot theta dtheta $$
$endgroup$
– Awe Kumar Jha
Dec 7 '18 at 11:49














$begingroup$
$logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 17:06




$begingroup$
$logsin x$ and $logcos x$ have well-known Fourier series, and you just have to perform a termwise integration of them.
$endgroup$
– Jack D'Aurizio
Dec 7 '18 at 17:06












$begingroup$
What is $sqrt{i}$?
$endgroup$
– FDP
Dec 7 '18 at 18:14




$begingroup$
What is $sqrt{i}$?
$endgroup$
– FDP
Dec 7 '18 at 18:14












$begingroup$
@FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
$endgroup$
– Kemono Chen
Dec 8 '18 at 0:55




$begingroup$
@FDP $sqrt i= e^{pi i/4}=frac{1+i}{sqrt 2}$.
$endgroup$
– Kemono Chen
Dec 8 '18 at 0:55












$begingroup$
why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
$endgroup$
– FDP
Dec 8 '18 at 11:30






$begingroup$
why not begin{align}sqrt i= -e^{pi i/4}?end{align} if $x_0^2=a$ then $(-x_0)^2=a$.
$endgroup$
– FDP
Dec 8 '18 at 11:30












1 Answer
1






active

oldest

votes


















3












$begingroup$

$newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansions I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going all along the long way. Anyway, the only difficulty that remains after your attempt is the evaluation of the following integral




$$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$




First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and enforce the substitution $x+fracpi4mapsto x$ within the second integral to get



$$begin{align*}
mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
&=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
end{align*}$$



Now it is time to apply the first useful formula of the Clausen Function, namely




$$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$




Formula $(2)$ can be shown rather easy be utilizing the well-known Fourier Series Expansion of $log(cos x)$ combined with the series representation of the $Cl_2(z)$ function. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get



$$begin{align*}
mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
&~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
&=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
end{align*}$$



We are almost done! It is time to throw some more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula




$$begin{align*}
Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
end{align*}$$




From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value



$$begin{align*}
mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
&=frac12left[G+frac G2right]
end{align*}$$




$$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$




I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029789%2fproving-im-operatornameli-2-sqrt-i-sqrt-2-1-frac34g-frac18-pi-ln-sqrt%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    $newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansions I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going all along the long way. Anyway, the only difficulty that remains after your attempt is the evaluation of the following integral




    $$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$




    First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and enforce the substitution $x+fracpi4mapsto x$ within the second integral to get



    $$begin{align*}
    mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
    &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
    &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
    &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
    end{align*}$$



    Now it is time to apply the first useful formula of the Clausen Function, namely




    $$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$




    Formula $(2)$ can be shown rather easy be utilizing the well-known Fourier Series Expansion of $log(cos x)$ combined with the series representation of the $Cl_2(z)$ function. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get



    $$begin{align*}
    mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
    &~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
    &=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
    end{align*}$$



    We are almost done! It is time to throw some more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula




    $$begin{align*}
    Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
    Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
    end{align*}$$




    From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value



    $$begin{align*}
    mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
    &=frac12left[G+frac G2right]
    end{align*}$$




    $$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$




    I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      $newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansions I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going all along the long way. Anyway, the only difficulty that remains after your attempt is the evaluation of the following integral




      $$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$




      First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and enforce the substitution $x+fracpi4mapsto x$ within the second integral to get



      $$begin{align*}
      mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
      &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
      &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
      &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
      end{align*}$$



      Now it is time to apply the first useful formula of the Clausen Function, namely




      $$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$




      Formula $(2)$ can be shown rather easy be utilizing the well-known Fourier Series Expansion of $log(cos x)$ combined with the series representation of the $Cl_2(z)$ function. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get



      $$begin{align*}
      mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
      &~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
      &=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
      end{align*}$$



      We are almost done! It is time to throw some more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula




      $$begin{align*}
      Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
      Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
      end{align*}$$




      From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value



      $$begin{align*}
      mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
      &=frac12left[G+frac G2right]
      end{align*}$$




      $$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$




      I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        $newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansions I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going all along the long way. Anyway, the only difficulty that remains after your attempt is the evaluation of the following integral




        $$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$




        First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and enforce the substitution $x+fracpi4mapsto x$ within the second integral to get



        $$begin{align*}
        mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
        &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
        &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
        &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
        end{align*}$$



        Now it is time to apply the first useful formula of the Clausen Function, namely




        $$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$




        Formula $(2)$ can be shown rather easy be utilizing the well-known Fourier Series Expansion of $log(cos x)$ combined with the series representation of the $Cl_2(z)$ function. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get



        $$begin{align*}
        mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
        &~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
        &=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
        end{align*}$$



        We are almost done! It is time to throw some more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula




        $$begin{align*}
        Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
        Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
        end{align*}$$




        From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value



        $$begin{align*}
        mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
        &=frac12left[G+frac G2right]
        end{align*}$$




        $$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$




        I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.






        share|cite|improve this answer











        $endgroup$



        $newcommand{Cl}{operatorname{Cl}}$In order to dodge the extensive usage of the Fourier Series Expansions I will use the Clausen Function $Cl_2(z)$ to shorten things up; nevertheless the result will remain the same as one could get going all along the long way. Anyway, the only difficulty that remains after your attempt is the evaluation of the following integral




        $$mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}xtag1$$




        First of all I will do a bit of reshaping to actually apply useful formulae involving the Clausen Function. Therefore split up the first integral and enforce the substitution $x+fracpi4mapsto x$ within the second integral to get



        $$begin{align*}
        mathfrak{I}&=-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x\
        &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{pi/8}^{pi/4}logleft(cos x+frac pi4right)mathrm{d}x\
        &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_{3pi/8}^{pi/2}log(cos x)mathrm{d}x\
        &=int_0^{pi/4}log(cos x)mathrm{d}x-int_0^{pi/8}log(cos x)mathrm{d}x-int_0^{pi/2}log(cos x)mathrm{d}x+int_0^{3pi/8}log(cos x)mathrm{d}x
        end{align*}$$



        Now it is time to apply the first useful formula of the Clausen Function, namely




        $$int_0^t log(cos x)mathrm{d}x~=~frac12Cl_2(pi-2t)-tlog(2)tag2$$




        Formula $(2)$ can be shown rather easy be utilizing the well-known Fourier Series Expansion of $log(cos x)$ combined with the series representation of the $Cl_2(z)$ function. However, with this knowledge we can rewrite the integrals from above in terms of the Clausen Function to get



        $$begin{align*}
        mathfrak{I}&=frac12Cl_2left(pi-2fracpi4right)-frac12Cl_2left(pi-2fracpi8right)-frac12Cl_2left(pi-2fracpi2right)+frac12Cl_2left(pi-2frac{3pi}8right)\
        &~~~underbrace{-fracpi4log(2)+fracpi8log(2)+fracpi2log(2)-frac{3pi}8log(2)}_{=0}\
        &=frac12left[Cl_2left(fracpi2right)-Cl_2left(0right)+Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)right]
        end{align*}$$



        We are almost done! It is time to throw some more important formulae in: firstly the already mentioned series representation and secondly the so-called Duplication Formula




        $$begin{align*}
        Cl_2(z)~&=~sum_{n=1}^infty frac{sin(nz)}{n^2}tag3\
        Cl_2(2z)~&=~2Cl_2(z)-2Cl_2(pi-z)tag4
        end{align*}$$




        From $(3)$ we can direcetly conclude that $Cl_2left(fracpi2right)=G$ and that $Cl_2(0)=0$ where $G$ denotes Catalan's Constant. Using the $(4)$ with $z=fracpi4$ we get a representation for the other terms from above. Putting this all together gives us the final value



        $$begin{align*}
        mathfrak{I}&=frac12left[underbrace{Cl_2left(fracpi2right)}_{=G}-underbrace{Cl_2left(0right)}_{=0}+underbrace{Cl_2left(fracpi4right)-Cl_2left(frac{3pi}4right)}_{=frac G2}right]\
        &=frac12left[G+frac G2right]
        end{align*}$$




        $$therefore~mathfrak{I}~=~-int_{pi/8}^{pi/4}log(sec x)+logleft(sin fracpi4-xright)mathrm{d}x~=~frac34 G$$




        I recommend to study the Clausen Function hence it reduces the number of caculations needed for linear logarithmo-trigonometric integrals tremendously. If you are feeling uncomfortable with a part of the proof let me know and I will try to clear your doubts.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 28 at 14:28

























        answered Jan 4 at 19:21









        mrtaurhomrtaurho

        5,75551540




        5,75551540






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029789%2fproving-im-operatornameli-2-sqrt-i-sqrt-2-1-frac34g-frac18-pi-ln-sqrt%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
            Read more