Integral involving the logarithm
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
edited 2 days ago
Masacroso
12.9k41746
asked 2 days ago
Kays Tomy
1977
add a comment |
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
edited 2 days ago
Masacroso
12.9k41746
asked 2 days ago
Kays Tomy
1977
What have you tried by yourself?
– mrtaurho
2 days ago
using Tylor series of $ln(1-x^q)$ and $ln(1+x^p)$
– Kays Tomy
2 days ago
What have you found so far from your attempts?
– clathratus
2 days ago
Alternating series as it is shown above
– Kays Tomy
2 days ago
add a comment |
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
edited 2 days ago
Masacroso
12.9k41746
asked 2 days ago
Kays Tomy
1977
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
edited 2 days ago
Masacroso
12.9k41746
asked 2 days ago
Kays Tomy
1977
edited 2 days ago
Masacroso
12.9k41746
asked 2 days ago
Kays Tomy
1977
edited 2 days ago
Masacroso
12.9k41746
edited 2 days ago
Masacroso
12.9k41746
edited 2 days ago
Masacroso
12.9k41746
12.9k41746
asked 2 days ago
Kays Tomy
1977
asked 2 days ago
Kays Tomy
1977
asked 2 days ago
Kays Tomy
1977
1977
What have you tried by yourself?
– mrtaurho
2 days ago
using Tylor series of $ln(1-x^q)$ and $ln(1+x^p)$
– Kays Tomy
2 days ago
What have you found so far from your attempts?
– clathratus
2 days ago
Alternating series as it is shown above
– Kays Tomy
2 days ago
add a comment |
What have you tried by yourself?
– mrtaurho
2 days ago
using Tylor series of $ln(1-x^q)$ and $ln(1+x^p)$
– Kays Tomy
2 days ago
What have you found so far from your attempts?
– clathratus
2 days ago
Alternating series as it is shown above
– Kays Tomy
2 days ago
What have you tried by yourself?
– mrtaurho
2 days ago
What have you tried by yourself?
– mrtaurho
2 days ago
using Tylor series of $ln(1-x^q)$ and $ln(1+x^p)$
– Kays Tomy
2 days ago
using Tylor series of $ln(1-x^q)$ and $ln(1+x^p)$
– Kays Tomy
2 days ago
What have you found so far from your attempts?
– clathratus
2 days ago
What have you found so far from your attempts?
– clathratus
2 days ago
Alternating series as it is shown above
– Kays Tomy
2 days ago
Alternating series as it is shown above
– Kays Tomy
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
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.
answered 2 days ago
Zachary
2,2891211
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 2 days ago
Zachary
2,2891211
add a comment |
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.
answered 2 days ago
Zachary
2,2891211
add a comment |
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.
answered 2 days ago
Zachary
2,2891211
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.
answered 2 days ago
Zachary
2,2891211
answered 2 days ago
Zachary
2,2891211
answered 2 days ago
Zachary
2,2891211
answered 2 days ago
Zachary
2,2891211
2,2891211
add a comment |
add a comment |
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What have you tried by yourself?
– mrtaurho
2 days ago
using Tylor series of $ln(1-x^q)$ and $ln(1+x^p)$
– Kays Tomy
2 days ago
What have you found so far from your attempts?
– clathratus
2 days ago
Alternating series as it is shown above
– Kays Tomy
2 days ago