Integral involving the logarithm












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Let $p$ and $q$ be two positive integers, I wonder if the following integral admits a closed-form and if yeah so in which way it should be worked out.



$$int_{0}^{1}frac{ln(1-x^p)ln(1+x^q)}{x^{p+q}}dx,=sum_{n,kgeq1}frac{(-1)^n}{nk(np+kq+1-p-q)}$$






integration definite-integrals power-series







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  • $begingroup$
    What have you tried by yourself?
    $endgroup$
    – mrtaurho
    Dec 25 '18 at 19:30










  • $begingroup$
    using Tylor series of $ln(1-x^q)$ and $ln(1+x^p)$
    $endgroup$
    – Kays Tomy
    Dec 25 '18 at 19:36










  • $begingroup$
    What have you found so far from your attempts?
    $endgroup$
    – clathratus
    Dec 25 '18 at 22:38










  • $begingroup$
    Alternating series as it is shown above
    $endgroup$
    – Kays Tomy
    Dec 25 '18 at 22:48
















0












$begingroup$


Let $p$ and $q$ be two positive integers, I wonder if the following integral admits a closed-form and if yeah so in which way it should be worked out.



$$int_{0}^{1}frac{ln(1-x^p)ln(1+x^q)}{x^{p+q}}dx,=sum_{n,kgeq1}frac{(-1)^n}{nk(np+kq+1-p-q)}$$






integration definite-integrals power-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    What have you tried by yourself?
    $endgroup$
    – mrtaurho
    Dec 25 '18 at 19:30










  • $begingroup$
    using Tylor series of $ln(1-x^q)$ and $ln(1+x^p)$
    $endgroup$
    – Kays Tomy
    Dec 25 '18 at 19:36










  • $begingroup$
    What have you found so far from your attempts?
    $endgroup$
    – clathratus
    Dec 25 '18 at 22:38










  • $begingroup$
    Alternating series as it is shown above
    $endgroup$
    – Kays Tomy
    Dec 25 '18 at 22:48














0












0








0


0



$begingroup$


Let $p$ and $q$ be two positive integers, I wonder if the following integral admits a closed-form and if yeah so in which way it should be worked out.



$$int_{0}^{1}frac{ln(1-x^p)ln(1+x^q)}{x^{p+q}}dx,=sum_{n,kgeq1}frac{(-1)^n}{nk(np+kq+1-p-q)}$$






integration definite-integrals power-series







share|cite|improve this question











$endgroup$




Let $p$ and $q$ be two positive integers, I wonder if the following integral admits a closed-form and if yeah so in which way it should be worked out.



$$int_{0}^{1}frac{ln(1-x^p)ln(1+x^q)}{x^{p+q}}dx,=sum_{n,kgeq1}frac{(-1)^n}{nk(np+kq+1-p-q)}$$






integration definite-integrals power-series




integration definite-integrals power-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 26 '18 at 5:16









Masacroso

13.1k41747




13.1k41747










asked Dec 25 '18 at 19:03









Kays Tomy Kays Tomy

21017




21017












  • $begingroup$
    What have you tried by yourself?
    $endgroup$
    – mrtaurho
    Dec 25 '18 at 19:30










  • $begingroup$
    using Tylor series of $ln(1-x^q)$ and $ln(1+x^p)$
    $endgroup$
    – Kays Tomy
    Dec 25 '18 at 19:36










  • $begingroup$
    What have you found so far from your attempts?
    $endgroup$
    – clathratus
    Dec 25 '18 at 22:38










  • $begingroup$
    Alternating series as it is shown above
    $endgroup$
    – Kays Tomy
    Dec 25 '18 at 22:48


















  • $begingroup$
    What have you tried by yourself?
    $endgroup$
    – mrtaurho
    Dec 25 '18 at 19:30










  • $begingroup$
    using Tylor series of $ln(1-x^q)$ and $ln(1+x^p)$
    $endgroup$
    – Kays Tomy
    Dec 25 '18 at 19:36










  • $begingroup$
    What have you found so far from your attempts?
    $endgroup$
    – clathratus
    Dec 25 '18 at 22:38










  • $begingroup$
    Alternating series as it is shown above
    $endgroup$
    – Kays Tomy
    Dec 25 '18 at 22:48
















$begingroup$
What have you tried by yourself?
$endgroup$
– mrtaurho
Dec 25 '18 at 19:30




$begingroup$
What have you tried by yourself?
$endgroup$
– mrtaurho
Dec 25 '18 at 19:30












$begingroup$
using Tylor series of $ln(1-x^q)$ and $ln(1+x^p)$
$endgroup$
– Kays Tomy
Dec 25 '18 at 19:36




$begingroup$
using Tylor series of $ln(1-x^q)$ and $ln(1+x^p)$
$endgroup$
– Kays Tomy
Dec 25 '18 at 19:36












$begingroup$
What have you found so far from your attempts?
$endgroup$
– clathratus
Dec 25 '18 at 22:38




$begingroup$
What have you found so far from your attempts?
$endgroup$
– clathratus
Dec 25 '18 at 22:38












$begingroup$
Alternating series as it is shown above
$endgroup$
– Kays Tomy
Dec 25 '18 at 22:48




$begingroup$
Alternating series as it is shown above
$endgroup$
– Kays Tomy
Dec 25 '18 at 22:48










1 Answer
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$begingroup$

Here is a partial answer in terms of an infinite series. After expanding both logarithms into their respective Maclaurin series, integrating term by term and applying a certain series representation of the digamma function, I've obtained this series:
$$int_0^1 frac{log(1-x^p)log(1+x^q)}{x^{p+q}},dx=sum_{kge 0} (-1)^{k+1} frac{psi^{(0)}left(frac{kq+1}{p}right)+gamma}{(k+1)(kq-p+1)}$$
Where $psi^{(0)}$ $(=Gamma'/Gamma )$ is the digamma function and $gamma$ ($=lim_{ntoinfty} H_n-log n$) is the Euler-Mascheroni constant. I'm not sure how to simplify this sum, which appears especially challenging because of the appearance of digamma function.



Checking the basic case $q=0$ yields
$$int_0^1 frac{log(1-x^p)}{x^p},dx=frac{psi^{(0)}(1/p)+gamma}{p-1}$$
Which seems to agree with some of the computational verifications I've done.






share|cite|improve this answer









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    active

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    0












    $begingroup$

    Here is a partial answer in terms of an infinite series. After expanding both logarithms into their respective Maclaurin series, integrating term by term and applying a certain series representation of the digamma function, I've obtained this series:
    $$int_0^1 frac{log(1-x^p)log(1+x^q)}{x^{p+q}},dx=sum_{kge 0} (-1)^{k+1} frac{psi^{(0)}left(frac{kq+1}{p}right)+gamma}{(k+1)(kq-p+1)}$$
    Where $psi^{(0)}$ $(=Gamma'/Gamma )$ is the digamma function and $gamma$ ($=lim_{ntoinfty} H_n-log n$) is the Euler-Mascheroni constant. I'm not sure how to simplify this sum, which appears especially challenging because of the appearance of digamma function.



    Checking the basic case $q=0$ yields
    $$int_0^1 frac{log(1-x^p)}{x^p},dx=frac{psi^{(0)}(1/p)+gamma}{p-1}$$
    Which seems to agree with some of the computational verifications I've done.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Here is a partial answer in terms of an infinite series. After expanding both logarithms into their respective Maclaurin series, integrating term by term and applying a certain series representation of the digamma function, I've obtained this series:
      $$int_0^1 frac{log(1-x^p)log(1+x^q)}{x^{p+q}},dx=sum_{kge 0} (-1)^{k+1} frac{psi^{(0)}left(frac{kq+1}{p}right)+gamma}{(k+1)(kq-p+1)}$$
      Where $psi^{(0)}$ $(=Gamma'/Gamma )$ is the digamma function and $gamma$ ($=lim_{ntoinfty} H_n-log n$) is the Euler-Mascheroni constant. I'm not sure how to simplify this sum, which appears especially challenging because of the appearance of digamma function.



      Checking the basic case $q=0$ yields
      $$int_0^1 frac{log(1-x^p)}{x^p},dx=frac{psi^{(0)}(1/p)+gamma}{p-1}$$
      Which seems to agree with some of the computational verifications I've done.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Here is a partial answer in terms of an infinite series. After expanding both logarithms into their respective Maclaurin series, integrating term by term and applying a certain series representation of the digamma function, I've obtained this series:
        $$int_0^1 frac{log(1-x^p)log(1+x^q)}{x^{p+q}},dx=sum_{kge 0} (-1)^{k+1} frac{psi^{(0)}left(frac{kq+1}{p}right)+gamma}{(k+1)(kq-p+1)}$$
        Where $psi^{(0)}$ $(=Gamma'/Gamma )$ is the digamma function and $gamma$ ($=lim_{ntoinfty} H_n-log n$) is the Euler-Mascheroni constant. I'm not sure how to simplify this sum, which appears especially challenging because of the appearance of digamma function.



        Checking the basic case $q=0$ yields
        $$int_0^1 frac{log(1-x^p)}{x^p},dx=frac{psi^{(0)}(1/p)+gamma}{p-1}$$
        Which seems to agree with some of the computational verifications I've done.






        share|cite|improve this answer









        $endgroup$



        Here is a partial answer in terms of an infinite series. After expanding both logarithms into their respective Maclaurin series, integrating term by term and applying a certain series representation of the digamma function, I've obtained this series:
        $$int_0^1 frac{log(1-x^p)log(1+x^q)}{x^{p+q}},dx=sum_{kge 0} (-1)^{k+1} frac{psi^{(0)}left(frac{kq+1}{p}right)+gamma}{(k+1)(kq-p+1)}$$
        Where $psi^{(0)}$ $(=Gamma'/Gamma )$ is the digamma function and $gamma$ ($=lim_{ntoinfty} H_n-log n$) is the Euler-Mascheroni constant. I'm not sure how to simplify this sum, which appears especially challenging because of the appearance of digamma function.



        Checking the basic case $q=0$ yields
        $$int_0^1 frac{log(1-x^p)}{x^p},dx=frac{psi^{(0)}(1/p)+gamma}{p-1}$$
        Which seems to agree with some of the computational verifications I've done.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 22:43









        ZacharyZachary

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