Mixing
Evaluate $sumlimits_{n=1}^{infty} (-1)^n frac{ln{n}}{n} $
$begingroup$
$$sumlimits_{n=1}^{infty} (-1)^n frac{ln{n}}{n} $$
Hint : $$ x_n = frac{ln{2}}{2} + frac{ln{3}}{3} + cdots frac{ln{n}}{n} - frac{ln^2{2}}{2} $$
Which converges, if we calculate it's limit we should get $ln2(gamma-frac{ln{2}}{2})$.
I don't understand where this hint comes from and how it helps us solve the series.
calculus sequences-and-series limits
asked Jan 28 at 14:19
SADBOYSSADBOYS
51119
$endgroup$
add a comment |
$begingroup$
$$sumlimits_{n=1}^{infty} (-1)^n frac{ln{n}}{n} $$
Hint : $$ x_n = frac{ln{2}}{2} + frac{ln{3}}{3} + cdots frac{ln{n}}{n} - frac{ln^2{2}}{2} $$
Which converges, if we calculate it's limit we should get $ln2(gamma-frac{ln{2}}{2})$.
I don't understand where this hint comes from and how it helps us solve the series.
calculus sequences-and-series limits
asked Jan 28 at 14:19
SADBOYSSADBOYS
51119
$endgroup$
$begingroup$
How did you solve it? Using $eta'(1)$?
$endgroup$
– Diger
Jan 28 at 15:01
$begingroup$
@Diger See my development of $eta'(1)$ using only the Euler-Maclaurin Summation Formula.
$endgroup$
– Mark Viola
Jan 28 at 16:27
$begingroup$
@SADBOYS The "Hint" as written, is not much of a hint since $x_n$, as written, fails to converge. I've posted a solution that I hope you find useful. ;-)
$endgroup$
– Mark Viola
Jan 28 at 16:36
$begingroup$
@Mark: I actually meant the OP, since it seemed he had solved it another way, but was specifically asking for how to use the hint. But thanks anyway.
$endgroup$
– Diger
Jan 28 at 17:31
add a comment |
$begingroup$
$$sumlimits_{n=1}^{infty} (-1)^n frac{ln{n}}{n} $$
Hint : $$ x_n = frac{ln{2}}{2} + frac{ln{3}}{3} + cdots frac{ln{n}}{n} - frac{ln^2{2}}{2} $$
Which converges, if we calculate it's limit we should get $ln2(gamma-frac{ln{2}}{2})$.
I don't understand where this hint comes from and how it helps us solve the series.
calculus sequences-and-series limits
asked Jan 28 at 14:19
SADBOYSSADBOYS
51119
$endgroup$
$$sumlimits_{n=1}^{infty} (-1)^n frac{ln{n}}{n} $$
Hint : $$ x_n = frac{ln{2}}{2} + frac{ln{3}}{3} + cdots frac{ln{n}}{n} - frac{ln^2{2}}{2} $$
Which converges, if we calculate it's limit we should get $ln2(gamma-frac{ln{2}}{2})$.
I don't understand where this hint comes from and how it helps us solve the series.
calculus sequences-and-series limits
calculus sequences-and-series limits
asked Jan 28 at 14:19
SADBOYSSADBOYS
51119
asked Jan 28 at 14:19
SADBOYSSADBOYS
51119
asked Jan 28 at 14:19
SADBOYSSADBOYS
51119
asked Jan 28 at 14:19
SADBOYSSADBOYS
51119
asked Jan 28 at 14:19
SADBOYSSADBOYS
51119
51119
$begingroup$
How did you solve it? Using $eta'(1)$?
$endgroup$
– Diger
Jan 28 at 15:01
$begingroup$
@Diger See my development of $eta'(1)$ using only the Euler-Maclaurin Summation Formula.
$endgroup$
– Mark Viola
Jan 28 at 16:27
$begingroup$
@SADBOYS The "Hint" as written, is not much of a hint since $x_n$, as written, fails to converge. I've posted a solution that I hope you find useful. ;-)
$endgroup$
– Mark Viola
Jan 28 at 16:36
$begingroup$
@Mark: I actually meant the OP, since it seemed he had solved it another way, but was specifically asking for how to use the hint. But thanks anyway.
$endgroup$
– Diger
Jan 28 at 17:31
add a comment |
$begingroup$
How did you solve it? Using $eta'(1)$?
$endgroup$
– Diger
Jan 28 at 15:01
$begingroup$
@Diger See my development of $eta'(1)$ using only the Euler-Maclaurin Summation Formula.
$endgroup$
– Mark Viola
Jan 28 at 16:27
$begingroup$
@SADBOYS The "Hint" as written, is not much of a hint since $x_n$, as written, fails to converge. I've posted a solution that I hope you find useful. ;-)
$endgroup$
– Mark Viola
Jan 28 at 16:36
$begingroup$
@Mark: I actually meant the OP, since it seemed he had solved it another way, but was specifically asking for how to use the hint. But thanks anyway.
$endgroup$
– Diger
Jan 28 at 17:31
$begingroup$
How did you solve it? Using $eta'(1)$?
$endgroup$
– Diger
Jan 28 at 15:01
$begingroup$
How did you solve it? Using $eta'(1)$?
$endgroup$
– Diger
Jan 28 at 15:01
$begingroup$
@Diger See my development of $eta'(1)$ using only the Euler-Maclaurin Summation Formula.
$endgroup$
– Mark Viola
Jan 28 at 16:27
$begingroup$
@Diger See my development of $eta'(1)$ using only the Euler-Maclaurin Summation Formula.
$endgroup$
– Mark Viola
Jan 28 at 16:27
$begingroup$
@SADBOYS The "Hint" as written, is not much of a hint since $x_n$, as written, fails to converge. I've posted a solution that I hope you find useful. ;-)
$endgroup$
– Mark Viola
Jan 28 at 16:36
$begingroup$
@SADBOYS The "Hint" as written, is not much of a hint since $x_n$, as written, fails to converge. I've posted a solution that I hope you find useful. ;-)
$endgroup$
– Mark Viola
Jan 28 at 16:36
$begingroup$
@Mark: I actually meant the OP, since it seemed he had solved it another way, but was specifically asking for how to use the hint. But thanks anyway.
$endgroup$
– Diger
Jan 28 at 17:31
$begingroup$
@Mark: I actually meant the OP, since it seemed he had solved it another way, but was specifically asking for how to use the hint. But thanks anyway.
$endgroup$
– Diger
Jan 28 at 17:31
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
I thought that it would be instructive to present a straightforward way to evaluate the series of interest using the Euler Maclaurin Summation Formula. To that end we proceed.
Note that we can write any alternating sum $sum_{n=1}^{2N}(-1)^na_n$ as
$$sum_{n=1}^{2N}(-1)^na_n=2sum_{n=1}^N a_{2n}-sum_{n=1}^{2N}a_ntag1$$
Using $(1)$, we see that
$$begin{align}
sum_{n=1}^{2N}(-1)^n frac{log(n)}{n}&=2sum_{n=1}^N frac{log(2n)}{2n}-sum_{n=1}^{2N}frac{log(n)}{n}\\
&=log(2)sum_{n=1}^Nfrac1n-sum_{n=N+1}^{2N}frac{log(n)}{n}tag2
end{align}$$
Applying the Euler Maclaurin Summation Formula to the second summation on the right-hand side of $(2)$ reveals
$$begin{align}
sum_{n=N+1}^{2N}frac{log(n)}{n}&=int_N^{2N}frac{log(x)}{x},dx+Oleft(frac{log(N)}{N}right)\\
&=frac12 log^2(2N)-frac12log^2(N)+Oleft(frac{log(N)}{N}right)\\
&=frac12log^2(2)+log(2)log(N)+Oleft(frac{log(N)}{N}right)tag3
end{align}$$
Substitution of $(3)$ into $(2)$ yields
$$begin{align}
sum_{n=1}^{2N}(-1)^n frac{log(n)}{n}&=log(2)left(-log(N)+sum_{n=1}^N frac1nright)-frac12log^2(2)+Oleft(frac{log(N)}{N}right)
end{align}$$
Finally, using the limit definition of the Euler-Mascheroni constant
$$gammaequivlim_{Ntoinfty}left(-log(N)+sum_{n=1}^Nfrac1nright)$$
we arrive at the coveted limit
$$sum_{n=1}^inftyfrac{(-1)^nlog(n)}{n}=gammalog(2)-frac12log^2(2)$$
answered Jan 28 at 16:24
Mark ViolaMark Viola
134k1278176
$endgroup$
add a comment |
$begingroup$
Here is another method using analytic regularization.
We have $eta(s)=left(1-2^{1-s}right) zeta(s)$, and so about $s=1$
$$
eta(s)=left(log(2)(s-1) - frac{log^2(2)}{2} , (s-1)^2 + {cal O}left((s-1)^3right)right) zeta(s) \
zeta(s) = -frac{1}{2pi i} int_{-iinfty}^{iinfty} {rm d}lambda , lambda^{-s} , frac{rm d}{{rm d}lambda} log left( frac{sin(pilambda)}{pi lambda} right)
$$
where for $s>1$ the contour can be closed to the right and the residue theorem is used. For regularity at $lambda=0$, $s<2$ is required also. Substituting $lambda=it$
$$
zeta(s) = frac{sinleft(frac{pi s}{2}right)}{pi} int_{0}^{infty} {rm d}t , t^{-s} , frac{rm d}{{rm d}t} log left( frac{sinh(pi t)}{pi t} right) \
stackrel{{rm P.I. | s>1}}{=} frac{sinleft(frac{pi s}{2}right)}{pi (s-1)} int_{0}^{infty} {rm d}t , t^{1-s} , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right)
$$
where the second line now converges for $0<s<2$ and hence
$$
eta(s) = left(log(2) - frac{log^2(2)}{2} , (s-1) + {cal O}left((s-1)^2right)right) frac{sinleft(frac{pi s}{2}right)}{pi} int_{0}^{infty} {rm d}t , t^{1-s} , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) , .
$$
Deriving with respect to $s$ and setting $s=1$
$$
eta'(1)=-frac{log(2)}{pi} int_0^infty {rm d}t log(t) , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) - frac{log^2(2)}{2pi} int_0^infty {rm d}t , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) \
=-{log(2)} int_0^infty {rm d}t log(t) , frac{rm d}{{rm d}t} left( coth(pi t) - frac{1}{pi t}right) - frac{log^2(2)}{2}
$$
we have
$$
coth(pi t)-frac{1}{pi t} = frac{2t}{pi} sum_{k=1}^infty frac{1}{k^2+t^2} , .
$$
When interchanging summation and integration order we acquire divergencies, because $coth(infty)=1$, but each summand vanishes for $trightarrow infty$. Due to the uniqueness of the result, it does not change up to some divergent part though:
$$
-{log(2)} int_0^infty {rm d}t log(t) , frac{rm d}{{rm d}t} left( coth(pi t) - frac{1}{pi t}right) \
sim -log(2) sum_{k=1}^N int_0^infty {rm d}t , log(t) frac{rm d}{{rm d t}} frac{2t/pi}{k^2+t^2} \
=log(2) sum_{k=1}^N int_0^infty {rm d}t , frac{2/pi}{k^2+t^2} \
=log(2) sum_{k=1}^N frac{1}{k} \
= log(2) left{ log(N) + gamma + {cal O}(1/N) right}
$$
and therefore
$$
eta'(1)=gamma log(2) - frac{log^2(2)}{2} , .
$$
answered Jan 28 at 22:22
DigerDiger
1,8101514
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
I thought that it would be instructive to present a straightforward way to evaluate the series of interest using the Euler Maclaurin Summation Formula. To that end we proceed.
Note that we can write any alternating sum $sum_{n=1}^{2N}(-1)^na_n$ as
$$sum_{n=1}^{2N}(-1)^na_n=2sum_{n=1}^N a_{2n}-sum_{n=1}^{2N}a_ntag1$$
Using $(1)$, we see that
$$begin{align}
sum_{n=1}^{2N}(-1)^n frac{log(n)}{n}&=2sum_{n=1}^N frac{log(2n)}{2n}-sum_{n=1}^{2N}frac{log(n)}{n}\\
&=log(2)sum_{n=1}^Nfrac1n-sum_{n=N+1}^{2N}frac{log(n)}{n}tag2
end{align}$$
Applying the Euler Maclaurin Summation Formula to the second summation on the right-hand side of $(2)$ reveals
$$begin{align}
sum_{n=N+1}^{2N}frac{log(n)}{n}&=int_N^{2N}frac{log(x)}{x},dx+Oleft(frac{log(N)}{N}right)\\
&=frac12 log^2(2N)-frac12log^2(N)+Oleft(frac{log(N)}{N}right)\\
&=frac12log^2(2)+log(2)log(N)+Oleft(frac{log(N)}{N}right)tag3
end{align}$$
Substitution of $(3)$ into $(2)$ yields
$$begin{align}
sum_{n=1}^{2N}(-1)^n frac{log(n)}{n}&=log(2)left(-log(N)+sum_{n=1}^N frac1nright)-frac12log^2(2)+Oleft(frac{log(N)}{N}right)
end{align}$$
Finally, using the limit definition of the Euler-Mascheroni constant
$$gammaequivlim_{Ntoinfty}left(-log(N)+sum_{n=1}^Nfrac1nright)$$
we arrive at the coveted limit
$$sum_{n=1}^inftyfrac{(-1)^nlog(n)}{n}=gammalog(2)-frac12log^2(2)$$
answered Jan 28 at 16:24
Mark ViolaMark Viola
134k1278176
$endgroup$
add a comment |
$begingroup$
I thought that it would be instructive to present a straightforward way to evaluate the series of interest using the Euler Maclaurin Summation Formula. To that end we proceed.
Note that we can write any alternating sum $sum_{n=1}^{2N}(-1)^na_n$ as
$$sum_{n=1}^{2N}(-1)^na_n=2sum_{n=1}^N a_{2n}-sum_{n=1}^{2N}a_ntag1$$
Using $(1)$, we see that
$$begin{align}
sum_{n=1}^{2N}(-1)^n frac{log(n)}{n}&=2sum_{n=1}^N frac{log(2n)}{2n}-sum_{n=1}^{2N}frac{log(n)}{n}\\
&=log(2)sum_{n=1}^Nfrac1n-sum_{n=N+1}^{2N}frac{log(n)}{n}tag2
end{align}$$
Applying the Euler Maclaurin Summation Formula to the second summation on the right-hand side of $(2)$ reveals
$$begin{align}
sum_{n=N+1}^{2N}frac{log(n)}{n}&=int_N^{2N}frac{log(x)}{x},dx+Oleft(frac{log(N)}{N}right)\\
&=frac12 log^2(2N)-frac12log^2(N)+Oleft(frac{log(N)}{N}right)\\
&=frac12log^2(2)+log(2)log(N)+Oleft(frac{log(N)}{N}right)tag3
end{align}$$
Substitution of $(3)$ into $(2)$ yields
$$begin{align}
sum_{n=1}^{2N}(-1)^n frac{log(n)}{n}&=log(2)left(-log(N)+sum_{n=1}^N frac1nright)-frac12log^2(2)+Oleft(frac{log(N)}{N}right)
end{align}$$
Finally, using the limit definition of the Euler-Mascheroni constant
$$gammaequivlim_{Ntoinfty}left(-log(N)+sum_{n=1}^Nfrac1nright)$$
we arrive at the coveted limit
$$sum_{n=1}^inftyfrac{(-1)^nlog(n)}{n}=gammalog(2)-frac12log^2(2)$$
answered Jan 28 at 16:24
Mark ViolaMark Viola
134k1278176
$endgroup$
add a comment |
$begingroup$
I thought that it would be instructive to present a straightforward way to evaluate the series of interest using the Euler Maclaurin Summation Formula. To that end we proceed.
Note that we can write any alternating sum $sum_{n=1}^{2N}(-1)^na_n$ as
$$sum_{n=1}^{2N}(-1)^na_n=2sum_{n=1}^N a_{2n}-sum_{n=1}^{2N}a_ntag1$$
Using $(1)$, we see that
$$begin{align}
sum_{n=1}^{2N}(-1)^n frac{log(n)}{n}&=2sum_{n=1}^N frac{log(2n)}{2n}-sum_{n=1}^{2N}frac{log(n)}{n}\\
&=log(2)sum_{n=1}^Nfrac1n-sum_{n=N+1}^{2N}frac{log(n)}{n}tag2
end{align}$$
Applying the Euler Maclaurin Summation Formula to the second summation on the right-hand side of $(2)$ reveals
$$begin{align}
sum_{n=N+1}^{2N}frac{log(n)}{n}&=int_N^{2N}frac{log(x)}{x},dx+Oleft(frac{log(N)}{N}right)\\
&=frac12 log^2(2N)-frac12log^2(N)+Oleft(frac{log(N)}{N}right)\\
&=frac12log^2(2)+log(2)log(N)+Oleft(frac{log(N)}{N}right)tag3
end{align}$$
Substitution of $(3)$ into $(2)$ yields
$$begin{align}
sum_{n=1}^{2N}(-1)^n frac{log(n)}{n}&=log(2)left(-log(N)+sum_{n=1}^N frac1nright)-frac12log^2(2)+Oleft(frac{log(N)}{N}right)
end{align}$$
Finally, using the limit definition of the Euler-Mascheroni constant
$$gammaequivlim_{Ntoinfty}left(-log(N)+sum_{n=1}^Nfrac1nright)$$
we arrive at the coveted limit
$$sum_{n=1}^inftyfrac{(-1)^nlog(n)}{n}=gammalog(2)-frac12log^2(2)$$
answered Jan 28 at 16:24
Mark ViolaMark Viola
134k1278176
$endgroup$
I thought that it would be instructive to present a straightforward way to evaluate the series of interest using the Euler Maclaurin Summation Formula. To that end we proceed.
Note that we can write any alternating sum $sum_{n=1}^{2N}(-1)^na_n$ as
$$sum_{n=1}^{2N}(-1)^na_n=2sum_{n=1}^N a_{2n}-sum_{n=1}^{2N}a_ntag1$$
Using $(1)$, we see that
$$begin{align}
sum_{n=1}^{2N}(-1)^n frac{log(n)}{n}&=2sum_{n=1}^N frac{log(2n)}{2n}-sum_{n=1}^{2N}frac{log(n)}{n}\\
&=log(2)sum_{n=1}^Nfrac1n-sum_{n=N+1}^{2N}frac{log(n)}{n}tag2
end{align}$$
Applying the Euler Maclaurin Summation Formula to the second summation on the right-hand side of $(2)$ reveals
$$begin{align}
sum_{n=N+1}^{2N}frac{log(n)}{n}&=int_N^{2N}frac{log(x)}{x},dx+Oleft(frac{log(N)}{N}right)\\
&=frac12 log^2(2N)-frac12log^2(N)+Oleft(frac{log(N)}{N}right)\\
&=frac12log^2(2)+log(2)log(N)+Oleft(frac{log(N)}{N}right)tag3
end{align}$$
Substitution of $(3)$ into $(2)$ yields
$$begin{align}
sum_{n=1}^{2N}(-1)^n frac{log(n)}{n}&=log(2)left(-log(N)+sum_{n=1}^N frac1nright)-frac12log^2(2)+Oleft(frac{log(N)}{N}right)
end{align}$$
Finally, using the limit definition of the Euler-Mascheroni constant
$$gammaequivlim_{Ntoinfty}left(-log(N)+sum_{n=1}^Nfrac1nright)$$
we arrive at the coveted limit
$$sum_{n=1}^inftyfrac{(-1)^nlog(n)}{n}=gammalog(2)-frac12log^2(2)$$
answered Jan 28 at 16:24
Mark ViolaMark Viola
134k1278176
edited Jan 28 at 16:38
answered Jan 28 at 16:24
Mark ViolaMark Viola
134k1278176
answered Jan 28 at 16:24
Mark ViolaMark Viola
134k1278176
answered Jan 28 at 16:24
Mark ViolaMark Viola
134k1278176
134k1278176
add a comment |
add a comment |
$begingroup$
Here is another method using analytic regularization.
We have $eta(s)=left(1-2^{1-s}right) zeta(s)$, and so about $s=1$
$$
eta(s)=left(log(2)(s-1) - frac{log^2(2)}{2} , (s-1)^2 + {cal O}left((s-1)^3right)right) zeta(s) \
zeta(s) = -frac{1}{2pi i} int_{-iinfty}^{iinfty} {rm d}lambda , lambda^{-s} , frac{rm d}{{rm d}lambda} log left( frac{sin(pilambda)}{pi lambda} right)
$$
where for $s>1$ the contour can be closed to the right and the residue theorem is used. For regularity at $lambda=0$, $s<2$ is required also. Substituting $lambda=it$
$$
zeta(s) = frac{sinleft(frac{pi s}{2}right)}{pi} int_{0}^{infty} {rm d}t , t^{-s} , frac{rm d}{{rm d}t} log left( frac{sinh(pi t)}{pi t} right) \
stackrel{{rm P.I. | s>1}}{=} frac{sinleft(frac{pi s}{2}right)}{pi (s-1)} int_{0}^{infty} {rm d}t , t^{1-s} , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right)
$$
where the second line now converges for $0<s<2$ and hence
$$
eta(s) = left(log(2) - frac{log^2(2)}{2} , (s-1) + {cal O}left((s-1)^2right)right) frac{sinleft(frac{pi s}{2}right)}{pi} int_{0}^{infty} {rm d}t , t^{1-s} , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) , .
$$
Deriving with respect to $s$ and setting $s=1$
$$
eta'(1)=-frac{log(2)}{pi} int_0^infty {rm d}t log(t) , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) - frac{log^2(2)}{2pi} int_0^infty {rm d}t , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) \
=-{log(2)} int_0^infty {rm d}t log(t) , frac{rm d}{{rm d}t} left( coth(pi t) - frac{1}{pi t}right) - frac{log^2(2)}{2}
$$
we have
$$
coth(pi t)-frac{1}{pi t} = frac{2t}{pi} sum_{k=1}^infty frac{1}{k^2+t^2} , .
$$
When interchanging summation and integration order we acquire divergencies, because $coth(infty)=1$, but each summand vanishes for $trightarrow infty$. Due to the uniqueness of the result, it does not change up to some divergent part though:
$$
-{log(2)} int_0^infty {rm d}t log(t) , frac{rm d}{{rm d}t} left( coth(pi t) - frac{1}{pi t}right) \
sim -log(2) sum_{k=1}^N int_0^infty {rm d}t , log(t) frac{rm d}{{rm d t}} frac{2t/pi}{k^2+t^2} \
=log(2) sum_{k=1}^N int_0^infty {rm d}t , frac{2/pi}{k^2+t^2} \
=log(2) sum_{k=1}^N frac{1}{k} \
= log(2) left{ log(N) + gamma + {cal O}(1/N) right}
$$
and therefore
$$
eta'(1)=gamma log(2) - frac{log^2(2)}{2} , .
$$
answered Jan 28 at 22:22
DigerDiger
1,8101514
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add a comment |
$begingroup$
Here is another method using analytic regularization.
We have $eta(s)=left(1-2^{1-s}right) zeta(s)$, and so about $s=1$
$$
eta(s)=left(log(2)(s-1) - frac{log^2(2)}{2} , (s-1)^2 + {cal O}left((s-1)^3right)right) zeta(s) \
zeta(s) = -frac{1}{2pi i} int_{-iinfty}^{iinfty} {rm d}lambda , lambda^{-s} , frac{rm d}{{rm d}lambda} log left( frac{sin(pilambda)}{pi lambda} right)
$$
where for $s>1$ the contour can be closed to the right and the residue theorem is used. For regularity at $lambda=0$, $s<2$ is required also. Substituting $lambda=it$
$$
zeta(s) = frac{sinleft(frac{pi s}{2}right)}{pi} int_{0}^{infty} {rm d}t , t^{-s} , frac{rm d}{{rm d}t} log left( frac{sinh(pi t)}{pi t} right) \
stackrel{{rm P.I. | s>1}}{=} frac{sinleft(frac{pi s}{2}right)}{pi (s-1)} int_{0}^{infty} {rm d}t , t^{1-s} , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right)
$$
where the second line now converges for $0<s<2$ and hence
$$
eta(s) = left(log(2) - frac{log^2(2)}{2} , (s-1) + {cal O}left((s-1)^2right)right) frac{sinleft(frac{pi s}{2}right)}{pi} int_{0}^{infty} {rm d}t , t^{1-s} , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) , .
$$
Deriving with respect to $s$ and setting $s=1$
$$
eta'(1)=-frac{log(2)}{pi} int_0^infty {rm d}t log(t) , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) - frac{log^2(2)}{2pi} int_0^infty {rm d}t , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) \
=-{log(2)} int_0^infty {rm d}t log(t) , frac{rm d}{{rm d}t} left( coth(pi t) - frac{1}{pi t}right) - frac{log^2(2)}{2}
$$
we have
$$
coth(pi t)-frac{1}{pi t} = frac{2t}{pi} sum_{k=1}^infty frac{1}{k^2+t^2} , .
$$
When interchanging summation and integration order we acquire divergencies, because $coth(infty)=1$, but each summand vanishes for $trightarrow infty$. Due to the uniqueness of the result, it does not change up to some divergent part though:
$$
-{log(2)} int_0^infty {rm d}t log(t) , frac{rm d}{{rm d}t} left( coth(pi t) - frac{1}{pi t}right) \
sim -log(2) sum_{k=1}^N int_0^infty {rm d}t , log(t) frac{rm d}{{rm d t}} frac{2t/pi}{k^2+t^2} \
=log(2) sum_{k=1}^N int_0^infty {rm d}t , frac{2/pi}{k^2+t^2} \
=log(2) sum_{k=1}^N frac{1}{k} \
= log(2) left{ log(N) + gamma + {cal O}(1/N) right}
$$
and therefore
$$
eta'(1)=gamma log(2) - frac{log^2(2)}{2} , .
$$
answered Jan 28 at 22:22
DigerDiger
1,8101514
$endgroup$
add a comment |
$begingroup$
Here is another method using analytic regularization.
We have $eta(s)=left(1-2^{1-s}right) zeta(s)$, and so about $s=1$
$$
eta(s)=left(log(2)(s-1) - frac{log^2(2)}{2} , (s-1)^2 + {cal O}left((s-1)^3right)right) zeta(s) \
zeta(s) = -frac{1}{2pi i} int_{-iinfty}^{iinfty} {rm d}lambda , lambda^{-s} , frac{rm d}{{rm d}lambda} log left( frac{sin(pilambda)}{pi lambda} right)
$$
where for $s>1$ the contour can be closed to the right and the residue theorem is used. For regularity at $lambda=0$, $s<2$ is required also. Substituting $lambda=it$
$$
zeta(s) = frac{sinleft(frac{pi s}{2}right)}{pi} int_{0}^{infty} {rm d}t , t^{-s} , frac{rm d}{{rm d}t} log left( frac{sinh(pi t)}{pi t} right) \
stackrel{{rm P.I. | s>1}}{=} frac{sinleft(frac{pi s}{2}right)}{pi (s-1)} int_{0}^{infty} {rm d}t , t^{1-s} , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right)
$$
where the second line now converges for $0<s<2$ and hence
$$
eta(s) = left(log(2) - frac{log^2(2)}{2} , (s-1) + {cal O}left((s-1)^2right)right) frac{sinleft(frac{pi s}{2}right)}{pi} int_{0}^{infty} {rm d}t , t^{1-s} , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) , .
$$
Deriving with respect to $s$ and setting $s=1$
$$
eta'(1)=-frac{log(2)}{pi} int_0^infty {rm d}t log(t) , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) - frac{log^2(2)}{2pi} int_0^infty {rm d}t , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) \
=-{log(2)} int_0^infty {rm d}t log(t) , frac{rm d}{{rm d}t} left( coth(pi t) - frac{1}{pi t}right) - frac{log^2(2)}{2}
$$
we have
$$
coth(pi t)-frac{1}{pi t} = frac{2t}{pi} sum_{k=1}^infty frac{1}{k^2+t^2} , .
$$
When interchanging summation and integration order we acquire divergencies, because $coth(infty)=1$, but each summand vanishes for $trightarrow infty$. Due to the uniqueness of the result, it does not change up to some divergent part though:
$$
-{log(2)} int_0^infty {rm d}t log(t) , frac{rm d}{{rm d}t} left( coth(pi t) - frac{1}{pi t}right) \
sim -log(2) sum_{k=1}^N int_0^infty {rm d}t , log(t) frac{rm d}{{rm d t}} frac{2t/pi}{k^2+t^2} \
=log(2) sum_{k=1}^N int_0^infty {rm d}t , frac{2/pi}{k^2+t^2} \
=log(2) sum_{k=1}^N frac{1}{k} \
= log(2) left{ log(N) + gamma + {cal O}(1/N) right}
$$
and therefore
$$
eta'(1)=gamma log(2) - frac{log^2(2)}{2} , .
$$
answered Jan 28 at 22:22
DigerDiger
1,8101514
$endgroup$
Here is another method using analytic regularization.
We have $eta(s)=left(1-2^{1-s}right) zeta(s)$, and so about $s=1$
$$
eta(s)=left(log(2)(s-1) - frac{log^2(2)}{2} , (s-1)^2 + {cal O}left((s-1)^3right)right) zeta(s) \
zeta(s) = -frac{1}{2pi i} int_{-iinfty}^{iinfty} {rm d}lambda , lambda^{-s} , frac{rm d}{{rm d}lambda} log left( frac{sin(pilambda)}{pi lambda} right)
$$
where for $s>1$ the contour can be closed to the right and the residue theorem is used. For regularity at $lambda=0$, $s<2$ is required also. Substituting $lambda=it$
$$
zeta(s) = frac{sinleft(frac{pi s}{2}right)}{pi} int_{0}^{infty} {rm d}t , t^{-s} , frac{rm d}{{rm d}t} log left( frac{sinh(pi t)}{pi t} right) \
stackrel{{rm P.I. | s>1}}{=} frac{sinleft(frac{pi s}{2}right)}{pi (s-1)} int_{0}^{infty} {rm d}t , t^{1-s} , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right)
$$
where the second line now converges for $0<s<2$ and hence
$$
eta(s) = left(log(2) - frac{log^2(2)}{2} , (s-1) + {cal O}left((s-1)^2right)right) frac{sinleft(frac{pi s}{2}right)}{pi} int_{0}^{infty} {rm d}t , t^{1-s} , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) , .
$$
Deriving with respect to $s$ and setting $s=1$
$$
eta'(1)=-frac{log(2)}{pi} int_0^infty {rm d}t log(t) , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) - frac{log^2(2)}{2pi} int_0^infty {rm d}t , frac{rm d^2}{{rm d}t^2} log left( frac{sinh(pi t)}{pi t} right) \
=-{log(2)} int_0^infty {rm d}t log(t) , frac{rm d}{{rm d}t} left( coth(pi t) - frac{1}{pi t}right) - frac{log^2(2)}{2}
$$
we have
$$
coth(pi t)-frac{1}{pi t} = frac{2t}{pi} sum_{k=1}^infty frac{1}{k^2+t^2} , .
$$
When interchanging summation and integration order we acquire divergencies, because $coth(infty)=1$, but each summand vanishes for $trightarrow infty$. Due to the uniqueness of the result, it does not change up to some divergent part though:
$$
-{log(2)} int_0^infty {rm d}t log(t) , frac{rm d}{{rm d}t} left( coth(pi t) - frac{1}{pi t}right) \
sim -log(2) sum_{k=1}^N int_0^infty {rm d}t , log(t) frac{rm d}{{rm d t}} frac{2t/pi}{k^2+t^2} \
=log(2) sum_{k=1}^N int_0^infty {rm d}t , frac{2/pi}{k^2+t^2} \
=log(2) sum_{k=1}^N frac{1}{k} \
= log(2) left{ log(N) + gamma + {cal O}(1/N) right}
$$
and therefore
$$
eta'(1)=gamma log(2) - frac{log^2(2)}{2} , .
$$
answered Jan 28 at 22:22
DigerDiger
1,8101514
edited Jan 28 at 22:28
answered Jan 28 at 22:22
DigerDiger
1,8101514
answered Jan 28 at 22:22
DigerDiger
1,8101514
answered Jan 28 at 22:22
DigerDiger
1,8101514
1,8101514
add a comment |
add a comment |
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$begingroup$
How did you solve it? Using $eta'(1)$?
$endgroup$
– Diger
Jan 28 at 15:01
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@Diger See my development of $eta'(1)$ using only the Euler-Maclaurin Summation Formula.
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– Mark Viola
Jan 28 at 16:27
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@SADBOYS The "Hint" as written, is not much of a hint since $x_n$, as written, fails to converge. I've posted a solution that I hope you find useful. ;-)
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– Mark Viola
Jan 28 at 16:36
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@Mark: I actually meant the OP, since it seemed he had solved it another way, but was specifically asking for how to use the hint. But thanks anyway.
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– Diger
Jan 28 at 17:31