On the logarithm of the fractional part Integral












4












$begingroup$


Let ${}$ denote the fractional part function, then does the following integral admit a closed-form ?



$$int_{0}^{1}xlnbigg(bigg{frac{1}{x}bigg}bigg)dx$$






calculus integration zeta-functions fractional-part







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Highly doubt it. Playing a bit with this in Wolfram alpha, it seems as though this will be an infinite series of very complicated terms.
    $endgroup$
    – SmileyCraft
    Dec 20 '18 at 19:06






  • 1




    $begingroup$
    @SmileyCraft: I actually don't think the terms are terribly complicated, but I do agree outside of an infinite series, there isn't much hope for an expression involving only elementary functions.
    $endgroup$
    – Clayton
    Dec 20 '18 at 19:11
















4












$begingroup$


Let ${}$ denote the fractional part function, then does the following integral admit a closed-form ?



$$int_{0}^{1}xlnbigg(bigg{frac{1}{x}bigg}bigg)dx$$






calculus integration zeta-functions fractional-part







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Highly doubt it. Playing a bit with this in Wolfram alpha, it seems as though this will be an infinite series of very complicated terms.
    $endgroup$
    – SmileyCraft
    Dec 20 '18 at 19:06






  • 1




    $begingroup$
    @SmileyCraft: I actually don't think the terms are terribly complicated, but I do agree outside of an infinite series, there isn't much hope for an expression involving only elementary functions.
    $endgroup$
    – Clayton
    Dec 20 '18 at 19:11














4












4








4


0



$begingroup$


Let ${}$ denote the fractional part function, then does the following integral admit a closed-form ?



$$int_{0}^{1}xlnbigg(bigg{frac{1}{x}bigg}bigg)dx$$






calculus integration zeta-functions fractional-part







share|cite|improve this question









$endgroup$




Let ${}$ denote the fractional part function, then does the following integral admit a closed-form ?



$$int_{0}^{1}xlnbigg(bigg{frac{1}{x}bigg}bigg)dx$$






calculus integration zeta-functions fractional-part




calculus integration zeta-functions fractional-part






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 20 '18 at 18:49









Kays Tomy Kays Tomy

19717




19717








  • 1




    $begingroup$
    Highly doubt it. Playing a bit with this in Wolfram alpha, it seems as though this will be an infinite series of very complicated terms.
    $endgroup$
    – SmileyCraft
    Dec 20 '18 at 19:06






  • 1




    $begingroup$
    @SmileyCraft: I actually don't think the terms are terribly complicated, but I do agree outside of an infinite series, there isn't much hope for an expression involving only elementary functions.
    $endgroup$
    – Clayton
    Dec 20 '18 at 19:11














  • 1




    $begingroup$
    Highly doubt it. Playing a bit with this in Wolfram alpha, it seems as though this will be an infinite series of very complicated terms.
    $endgroup$
    – SmileyCraft
    Dec 20 '18 at 19:06






  • 1




    $begingroup$
    @SmileyCraft: I actually don't think the terms are terribly complicated, but I do agree outside of an infinite series, there isn't much hope for an expression involving only elementary functions.
    $endgroup$
    – Clayton
    Dec 20 '18 at 19:11








1




1




$begingroup$
Highly doubt it. Playing a bit with this in Wolfram alpha, it seems as though this will be an infinite series of very complicated terms.
$endgroup$
– SmileyCraft
Dec 20 '18 at 19:06




$begingroup$
Highly doubt it. Playing a bit with this in Wolfram alpha, it seems as though this will be an infinite series of very complicated terms.
$endgroup$
– SmileyCraft
Dec 20 '18 at 19:06




1




1




$begingroup$
@SmileyCraft: I actually don't think the terms are terribly complicated, but I do agree outside of an infinite series, there isn't much hope for an expression involving only elementary functions.
$endgroup$
– Clayton
Dec 20 '18 at 19:11




$begingroup$
@SmileyCraft: I actually don't think the terms are terribly complicated, but I do agree outside of an infinite series, there isn't much hope for an expression involving only elementary functions.
$endgroup$
– Clayton
Dec 20 '18 at 19:11










1 Answer
1






active

oldest

votes


















5












$begingroup$

This is not a complete answer but provides an infinite series expression.



Upon using the $u$-substitution $u=1/x$, we have



$$
int_0^1 xlnleft(left{frac1xright}right),dx=int_1^inftyfrac{ln({u})}{u^3},du=sum_{n=1}^inftyint_n^{n+1}frac{ln(u-n)}{u^3},du.
$$
Now using the substitution $x=ln(u-n)$, the series becomes
$$
sum_{n=1}^inftyint_{-infty}^0frac{x e^x}{(e^x+n)^3},dx.
$$
Evaluating this last integral via integration by parts gives
$$
-frac{1+(n+1) ln left(1+frac{1}{n}right)}{ 2n^2 (n+1)}.
$$
Inserting this in for the integral and after doing a little bit of algebra and simplifying, the last sum is equivalent to
$$
-frac12left(-1+frac{pi^2}{6}+sum_{n=1}^inftyfrac{ln(1+frac1n)}{n^2}right).
$$



Altogether, this gives
$$
int_0^1 xlnleft(left{frac1xright}right),dx=-frac12left(-1+frac{pi^2}{6}+sum_{n=1}^inftyfrac{ln(1+frac1n)}{n^2}right)approx-0.754071
$$
where I used Mathematica for the last approximation. I don't think there is much hope for the last sum to simplify because you would have to deal with a sum of $zeta(s)$ evaluated at odd positive integers, for which there is no known elementary expression.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have got the same result, thank you indeed
    $endgroup$
    – Kays Tomy
    Dec 20 '18 at 20:43










  • $begingroup$
    @KaysTomy: You are welcome.
    $endgroup$
    – Clayton
    Dec 20 '18 at 20:50












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

This is not a complete answer but provides an infinite series expression.



Upon using the $u$-substitution $u=1/x$, we have



$$
int_0^1 xlnleft(left{frac1xright}right),dx=int_1^inftyfrac{ln({u})}{u^3},du=sum_{n=1}^inftyint_n^{n+1}frac{ln(u-n)}{u^3},du.
$$
Now using the substitution $x=ln(u-n)$, the series becomes
$$
sum_{n=1}^inftyint_{-infty}^0frac{x e^x}{(e^x+n)^3},dx.
$$
Evaluating this last integral via integration by parts gives
$$
-frac{1+(n+1) ln left(1+frac{1}{n}right)}{ 2n^2 (n+1)}.
$$
Inserting this in for the integral and after doing a little bit of algebra and simplifying, the last sum is equivalent to
$$
-frac12left(-1+frac{pi^2}{6}+sum_{n=1}^inftyfrac{ln(1+frac1n)}{n^2}right).
$$



Altogether, this gives
$$
int_0^1 xlnleft(left{frac1xright}right),dx=-frac12left(-1+frac{pi^2}{6}+sum_{n=1}^inftyfrac{ln(1+frac1n)}{n^2}right)approx-0.754071
$$
where I used Mathematica for the last approximation. I don't think there is much hope for the last sum to simplify because you would have to deal with a sum of $zeta(s)$ evaluated at odd positive integers, for which there is no known elementary expression.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have got the same result, thank you indeed
    $endgroup$
    – Kays Tomy
    Dec 20 '18 at 20:43










  • $begingroup$
    @KaysTomy: You are welcome.
    $endgroup$
    – Clayton
    Dec 20 '18 at 20:50
















5












$begingroup$

This is not a complete answer but provides an infinite series expression.



Upon using the $u$-substitution $u=1/x$, we have



$$
int_0^1 xlnleft(left{frac1xright}right),dx=int_1^inftyfrac{ln({u})}{u^3},du=sum_{n=1}^inftyint_n^{n+1}frac{ln(u-n)}{u^3},du.
$$
Now using the substitution $x=ln(u-n)$, the series becomes
$$
sum_{n=1}^inftyint_{-infty}^0frac{x e^x}{(e^x+n)^3},dx.
$$
Evaluating this last integral via integration by parts gives
$$
-frac{1+(n+1) ln left(1+frac{1}{n}right)}{ 2n^2 (n+1)}.
$$
Inserting this in for the integral and after doing a little bit of algebra and simplifying, the last sum is equivalent to
$$
-frac12left(-1+frac{pi^2}{6}+sum_{n=1}^inftyfrac{ln(1+frac1n)}{n^2}right).
$$



Altogether, this gives
$$
int_0^1 xlnleft(left{frac1xright}right),dx=-frac12left(-1+frac{pi^2}{6}+sum_{n=1}^inftyfrac{ln(1+frac1n)}{n^2}right)approx-0.754071
$$
where I used Mathematica for the last approximation. I don't think there is much hope for the last sum to simplify because you would have to deal with a sum of $zeta(s)$ evaluated at odd positive integers, for which there is no known elementary expression.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have got the same result, thank you indeed
    $endgroup$
    – Kays Tomy
    Dec 20 '18 at 20:43










  • $begingroup$
    @KaysTomy: You are welcome.
    $endgroup$
    – Clayton
    Dec 20 '18 at 20:50














5












5








5





$begingroup$

This is not a complete answer but provides an infinite series expression.



Upon using the $u$-substitution $u=1/x$, we have



$$
int_0^1 xlnleft(left{frac1xright}right),dx=int_1^inftyfrac{ln({u})}{u^3},du=sum_{n=1}^inftyint_n^{n+1}frac{ln(u-n)}{u^3},du.
$$
Now using the substitution $x=ln(u-n)$, the series becomes
$$
sum_{n=1}^inftyint_{-infty}^0frac{x e^x}{(e^x+n)^3},dx.
$$
Evaluating this last integral via integration by parts gives
$$
-frac{1+(n+1) ln left(1+frac{1}{n}right)}{ 2n^2 (n+1)}.
$$
Inserting this in for the integral and after doing a little bit of algebra and simplifying, the last sum is equivalent to
$$
-frac12left(-1+frac{pi^2}{6}+sum_{n=1}^inftyfrac{ln(1+frac1n)}{n^2}right).
$$



Altogether, this gives
$$
int_0^1 xlnleft(left{frac1xright}right),dx=-frac12left(-1+frac{pi^2}{6}+sum_{n=1}^inftyfrac{ln(1+frac1n)}{n^2}right)approx-0.754071
$$
where I used Mathematica for the last approximation. I don't think there is much hope for the last sum to simplify because you would have to deal with a sum of $zeta(s)$ evaluated at odd positive integers, for which there is no known elementary expression.






share|cite|improve this answer









$endgroup$



This is not a complete answer but provides an infinite series expression.



Upon using the $u$-substitution $u=1/x$, we have



$$
int_0^1 xlnleft(left{frac1xright}right),dx=int_1^inftyfrac{ln({u})}{u^3},du=sum_{n=1}^inftyint_n^{n+1}frac{ln(u-n)}{u^3},du.
$$
Now using the substitution $x=ln(u-n)$, the series becomes
$$
sum_{n=1}^inftyint_{-infty}^0frac{x e^x}{(e^x+n)^3},dx.
$$
Evaluating this last integral via integration by parts gives
$$
-frac{1+(n+1) ln left(1+frac{1}{n}right)}{ 2n^2 (n+1)}.
$$
Inserting this in for the integral and after doing a little bit of algebra and simplifying, the last sum is equivalent to
$$
-frac12left(-1+frac{pi^2}{6}+sum_{n=1}^inftyfrac{ln(1+frac1n)}{n^2}right).
$$



Altogether, this gives
$$
int_0^1 xlnleft(left{frac1xright}right),dx=-frac12left(-1+frac{pi^2}{6}+sum_{n=1}^inftyfrac{ln(1+frac1n)}{n^2}right)approx-0.754071
$$
where I used Mathematica for the last approximation. I don't think there is much hope for the last sum to simplify because you would have to deal with a sum of $zeta(s)$ evaluated at odd positive integers, for which there is no known elementary expression.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 20:14









ClaytonClayton

19.6k33288




19.6k33288












  • $begingroup$
    I have got the same result, thank you indeed
    $endgroup$
    – Kays Tomy
    Dec 20 '18 at 20:43










  • $begingroup$
    @KaysTomy: You are welcome.
    $endgroup$
    – Clayton
    Dec 20 '18 at 20:50


















  • $begingroup$
    I have got the same result, thank you indeed
    $endgroup$
    – Kays Tomy
    Dec 20 '18 at 20:43










  • $begingroup$
    @KaysTomy: You are welcome.
    $endgroup$
    – Clayton
    Dec 20 '18 at 20:50
















$begingroup$
I have got the same result, thank you indeed
$endgroup$
– Kays Tomy
Dec 20 '18 at 20:43




$begingroup$
I have got the same result, thank you indeed
$endgroup$
– Kays Tomy
Dec 20 '18 at 20:43












$begingroup$
@KaysTomy: You are welcome.
$endgroup$
– Clayton
Dec 20 '18 at 20:50




$begingroup$
@KaysTomy: You are welcome.
$endgroup$
– Clayton
Dec 20 '18 at 20:50


















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