On the integral $int_1^inftybig({x}^n-frac1{n+1}big)frac{dx}x$
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According to Dirichlet's test (integral version),
$$
I_n=int_1^inftybig({x}^n-frac1{n+1}big)frac{dx}x
$$
converges, where $n$ is a positive integer and ${x}$ denotes the fractional part of $x$. Using series, I found out the values of $I_1$ and $I_2$. $I_1=frac12ln(2pi)-1$ and $I_2=frac12ln(2pi)-frac12-2ln A$, where $A$ denotes Glaisher's constant.
My Attempt to Generalize $I_n$
$$I_n=sum_{m=1}^inftyint_0^1big(t^n-frac1{n+1}big)frac{dt}{t+m}\
=sum_{m=1}^infty P_n(m)-frac1{n+1}lnbig(1+frac1mbig)+m^n(-1)^nlnbig(1+frac1mbig)\
=sum_{m=1}^infty-frac1{n+1}lnbig(1+frac1mbig)+frac1{(n+1)m}-frac1{(n+2)m^2}+cdots\
=fracgamma{n+1}+sum_{k=2}^inftyfrac{(-1)^{k-1}zeta(k)}{n+k}$$where $P_n$ is a polynomial with $deg P_n=n-1$ and $gamma$ denotes Euler's constant.
My questions are:
(i) Is my answer right?
(ii) If my answer is right, can I make it a little bit more
simplified?
(iii) How to find the value of $I_3$?
Edit: the convergence test of $I_n$
Denote $F(x)=int_1^x{t}^n-frac1{n+1}dt$, we have
$$F(x+1)-F(x)=int_x^{x+1}{t}^ndt-frac1{n+1}=0.$$
Obviously, $F(x)$ is bounded in $[0,1]$. So $F(x)$ is bounded in $mathbb{R}$. Also, $1/x$ is a decreasing function in $[1,+infty)$ and $lim_{xtoinfty}1/x=0$ Hence $I_n$ converges.
calculus definite-integrals summation riemann-zeta
edited Nov 14 at 14:34
Flermat
1,24511129
asked Jul 12 at 11:01
Kemono Chen
1,599330
add a comment |
up vote
4
down vote
favorite
According to Dirichlet's test (integral version),
$$
I_n=int_1^inftybig({x}^n-frac1{n+1}big)frac{dx}x
$$
converges, where $n$ is a positive integer and ${x}$ denotes the fractional part of $x$. Using series, I found out the values of $I_1$ and $I_2$. $I_1=frac12ln(2pi)-1$ and $I_2=frac12ln(2pi)-frac12-2ln A$, where $A$ denotes Glaisher's constant.
My Attempt to Generalize $I_n$
$$I_n=sum_{m=1}^inftyint_0^1big(t^n-frac1{n+1}big)frac{dt}{t+m}\
=sum_{m=1}^infty P_n(m)-frac1{n+1}lnbig(1+frac1mbig)+m^n(-1)^nlnbig(1+frac1mbig)\
=sum_{m=1}^infty-frac1{n+1}lnbig(1+frac1mbig)+frac1{(n+1)m}-frac1{(n+2)m^2}+cdots\
=fracgamma{n+1}+sum_{k=2}^inftyfrac{(-1)^{k-1}zeta(k)}{n+k}$$where $P_n$ is a polynomial with $deg P_n=n-1$ and $gamma$ denotes Euler's constant.
My questions are:
(i) Is my answer right?
(ii) If my answer is right, can I make it a little bit more
simplified?
(iii) How to find the value of $I_3$?
Edit: the convergence test of $I_n$
Denote $F(x)=int_1^x{t}^n-frac1{n+1}dt$, we have
$$F(x+1)-F(x)=int_x^{x+1}{t}^ndt-frac1{n+1}=0.$$
Obviously, $F(x)$ is bounded in $[0,1]$. So $F(x)$ is bounded in $mathbb{R}$. Also, $1/x$ is a decreasing function in $[1,+infty)$ and $lim_{xtoinfty}1/x=0$ Hence $I_n$ converges.
calculus definite-integrals summation riemann-zeta
edited Nov 14 at 14:34
Flermat
1,24511129
asked Jul 12 at 11:01
Kemono Chen
1,599330
Can you please explain - in detail - how you deduced the convergence of $I_n$ from Dirichlet's test?
– uniquesolution
Jul 12 at 11:23
@uniquesolution Edited. Please check.
– Kemono Chen
Jul 12 at 11:56
$$I_3=frac{3zeta(3)}{4pi^2} + frac12 ln(2pi) - 3ln A-frac13$$ $$I_4 = frac{3zeta(3)}{2 pi^2} - frac{3 zeta'(4)}{pi^4} +frac{8}{15} ln(2 pi) - 4ln A+frac{gamma}{30}-frac14$$
– nospoon
Jul 12 at 20:22
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
According to Dirichlet's test (integral version),
$$
I_n=int_1^inftybig({x}^n-frac1{n+1}big)frac{dx}x
$$
converges, where $n$ is a positive integer and ${x}$ denotes the fractional part of $x$. Using series, I found out the values of $I_1$ and $I_2$. $I_1=frac12ln(2pi)-1$ and $I_2=frac12ln(2pi)-frac12-2ln A$, where $A$ denotes Glaisher's constant.
My Attempt to Generalize $I_n$
$$I_n=sum_{m=1}^inftyint_0^1big(t^n-frac1{n+1}big)frac{dt}{t+m}\
=sum_{m=1}^infty P_n(m)-frac1{n+1}lnbig(1+frac1mbig)+m^n(-1)^nlnbig(1+frac1mbig)\
=sum_{m=1}^infty-frac1{n+1}lnbig(1+frac1mbig)+frac1{(n+1)m}-frac1{(n+2)m^2}+cdots\
=fracgamma{n+1}+sum_{k=2}^inftyfrac{(-1)^{k-1}zeta(k)}{n+k}$$where $P_n$ is a polynomial with $deg P_n=n-1$ and $gamma$ denotes Euler's constant.
My questions are:
(i) Is my answer right?
(ii) If my answer is right, can I make it a little bit more
simplified?
(iii) How to find the value of $I_3$?
Edit: the convergence test of $I_n$
Denote $F(x)=int_1^x{t}^n-frac1{n+1}dt$, we have
$$F(x+1)-F(x)=int_x^{x+1}{t}^ndt-frac1{n+1}=0.$$
Obviously, $F(x)$ is bounded in $[0,1]$. So $F(x)$ is bounded in $mathbb{R}$. Also, $1/x$ is a decreasing function in $[1,+infty)$ and $lim_{xtoinfty}1/x=0$ Hence $I_n$ converges.
calculus definite-integrals summation riemann-zeta
edited Nov 14 at 14:34
Flermat
1,24511129
asked Jul 12 at 11:01
Kemono Chen
1,599330
According to Dirichlet's test (integral version),
$$
I_n=int_1^inftybig({x}^n-frac1{n+1}big)frac{dx}x
$$
converges, where $n$ is a positive integer and ${x}$ denotes the fractional part of $x$. Using series, I found out the values of $I_1$ and $I_2$. $I_1=frac12ln(2pi)-1$ and $I_2=frac12ln(2pi)-frac12-2ln A$, where $A$ denotes Glaisher's constant.
My Attempt to Generalize $I_n$
$$I_n=sum_{m=1}^inftyint_0^1big(t^n-frac1{n+1}big)frac{dt}{t+m}\
=sum_{m=1}^infty P_n(m)-frac1{n+1}lnbig(1+frac1mbig)+m^n(-1)^nlnbig(1+frac1mbig)\
=sum_{m=1}^infty-frac1{n+1}lnbig(1+frac1mbig)+frac1{(n+1)m}-frac1{(n+2)m^2}+cdots\
=fracgamma{n+1}+sum_{k=2}^inftyfrac{(-1)^{k-1}zeta(k)}{n+k}$$where $P_n$ is a polynomial with $deg P_n=n-1$ and $gamma$ denotes Euler's constant.
My questions are:
(i) Is my answer right?
(ii) If my answer is right, can I make it a little bit more
simplified?
(iii) How to find the value of $I_3$?
Edit: the convergence test of $I_n$
Denote $F(x)=int_1^x{t}^n-frac1{n+1}dt$, we have
$$F(x+1)-F(x)=int_x^{x+1}{t}^ndt-frac1{n+1}=0.$$
Obviously, $F(x)$ is bounded in $[0,1]$. So $F(x)$ is bounded in $mathbb{R}$. Also, $1/x$ is a decreasing function in $[1,+infty)$ and $lim_{xtoinfty}1/x=0$ Hence $I_n$ converges.
calculus definite-integrals summation riemann-zeta
calculus definite-integrals summation riemann-zeta
edited Nov 14 at 14:34
Flermat
1,24511129
asked Jul 12 at 11:01
Kemono Chen
1,599330
edited Nov 14 at 14:34
Flermat
1,24511129
asked Jul 12 at 11:01
Kemono Chen
1,599330
edited Nov 14 at 14:34
Flermat
1,24511129
edited Nov 14 at 14:34
Flermat
1,24511129
edited Nov 14 at 14:34
Flermat
1,24511129
1,24511129
asked Jul 12 at 11:01
Kemono Chen
1,599330
asked Jul 12 at 11:01
Kemono Chen
1,599330
asked Jul 12 at 11:01
Kemono Chen
1,599330
1,599330
Can you please explain - in detail - how you deduced the convergence of $I_n$ from Dirichlet's test?
– uniquesolution
Jul 12 at 11:23
@uniquesolution Edited. Please check.
– Kemono Chen
Jul 12 at 11:56
$$I_3=frac{3zeta(3)}{4pi^2} + frac12 ln(2pi) - 3ln A-frac13$$ $$I_4 = frac{3zeta(3)}{2 pi^2} - frac{3 zeta'(4)}{pi^4} +frac{8}{15} ln(2 pi) - 4ln A+frac{gamma}{30}-frac14$$
– nospoon
Jul 12 at 20:22
add a comment |
Can you please explain - in detail - how you deduced the convergence of $I_n$ from Dirichlet's test?
– uniquesolution
Jul 12 at 11:23
@uniquesolution Edited. Please check.
– Kemono Chen
Jul 12 at 11:56
$$I_3=frac{3zeta(3)}{4pi^2} + frac12 ln(2pi) - 3ln A-frac13$$ $$I_4 = frac{3zeta(3)}{2 pi^2} - frac{3 zeta'(4)}{pi^4} +frac{8}{15} ln(2 pi) - 4ln A+frac{gamma}{30}-frac14$$
– nospoon
Jul 12 at 20:22
Can you please explain - in detail - how you deduced the convergence of $I_n$ from Dirichlet's test?
– uniquesolution
Jul 12 at 11:23
Can you please explain - in detail - how you deduced the convergence of $I_n$ from Dirichlet's test?
– uniquesolution
Jul 12 at 11:23
@uniquesolution Edited. Please check.
– Kemono Chen
Jul 12 at 11:56
@uniquesolution Edited. Please check.
– Kemono Chen
Jul 12 at 11:56
$$I_3=frac{3zeta(3)}{4pi^2} + frac12 ln(2pi) - 3ln A-frac13$$ $$I_4 = frac{3zeta(3)}{2 pi^2} - frac{3 zeta'(4)}{pi^4} +frac{8}{15} ln(2 pi) - 4ln A+frac{gamma}{30}-frac14$$
– nospoon
Jul 12 at 20:22
$$I_3=frac{3zeta(3)}{4pi^2} + frac12 ln(2pi) - 3ln A-frac13$$ $$I_4 = frac{3zeta(3)}{2 pi^2} - frac{3 zeta'(4)}{pi^4} +frac{8}{15} ln(2 pi) - 4ln A+frac{gamma}{30}-frac14$$
– nospoon
Jul 12 at 20:22
add a comment |
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Can you please explain - in detail - how you deduced the convergence of $I_n$ from Dirichlet's test?
– uniquesolution
Jul 12 at 11:23
@uniquesolution Edited. Please check.
– Kemono Chen
Jul 12 at 11:56
$$I_3=frac{3zeta(3)}{4pi^2} + frac12 ln(2pi) - 3ln A-frac13$$ $$I_4 = frac{3zeta(3)}{2 pi^2} - frac{3 zeta'(4)}{pi^4} +frac{8}{15} ln(2 pi) - 4ln A+frac{gamma}{30}-frac14$$
– nospoon
Jul 12 at 20:22