Mixing
On the zero of $f(x)=int_0^x ln(sinh(z))dz$
$begingroup$
$$f(x)=int_0^x ln(sinh(z))dz$$
I have observed that the only zero, besides $x=0$, of $f(x)$ is $x=x_0approx2.146$. Is it possible that $x_0$ has a closed form in terms of $pi$ or some other constant? If not, is there a possible series expansion for it? Much thanks and any help is appreciated.
definite-integrals roots
asked Dec 21 '18 at 16:33
aledenaleden
2,5051511
$endgroup$
add a comment |
$begingroup$
$$f(x)=int_0^x ln(sinh(z))dz$$
I have observed that the only zero, besides $x=0$, of $f(x)$ is $x=x_0approx2.146$. Is it possible that $x_0$ has a closed form in terms of $pi$ or some other constant? If not, is there a possible series expansion for it? Much thanks and any help is appreciated.
definite-integrals roots
asked Dec 21 '18 at 16:33
aledenaleden
2,5051511
$endgroup$
1
$begingroup$
This integral gives $f(x) =frac{1}{12} left(6 text{Li}_2left(e^{-2 x}right)-6 x left(x+2 log left(1-e^{-2 x}right)-2 log (sinh (x))right)-pi ^2right)$
$endgroup$
– Cesareo
Dec 21 '18 at 17:13
1
$begingroup$
$x_0approx2.146306122947278$
$endgroup$
– James Arathoon
Dec 21 '18 at 17:24
add a comment |
$begingroup$
$$f(x)=int_0^x ln(sinh(z))dz$$
I have observed that the only zero, besides $x=0$, of $f(x)$ is $x=x_0approx2.146$. Is it possible that $x_0$ has a closed form in terms of $pi$ or some other constant? If not, is there a possible series expansion for it? Much thanks and any help is appreciated.
definite-integrals roots
asked Dec 21 '18 at 16:33
aledenaleden
2,5051511
$endgroup$
$$f(x)=int_0^x ln(sinh(z))dz$$
I have observed that the only zero, besides $x=0$, of $f(x)$ is $x=x_0approx2.146$. Is it possible that $x_0$ has a closed form in terms of $pi$ or some other constant? If not, is there a possible series expansion for it? Much thanks and any help is appreciated.
definite-integrals roots
definite-integrals roots
asked Dec 21 '18 at 16:33
aledenaleden
2,5051511
asked Dec 21 '18 at 16:33
aledenaleden
2,5051511
asked Dec 21 '18 at 16:33
aledenaleden
2,5051511
asked Dec 21 '18 at 16:33
aledenaleden
2,5051511
asked Dec 21 '18 at 16:33
aledenaleden
2,5051511
2,5051511
1
$begingroup$
This integral gives $f(x) =frac{1}{12} left(6 text{Li}_2left(e^{-2 x}right)-6 x left(x+2 log left(1-e^{-2 x}right)-2 log (sinh (x))right)-pi ^2right)$
$endgroup$
– Cesareo
Dec 21 '18 at 17:13
1
$begingroup$
$x_0approx2.146306122947278$
$endgroup$
– James Arathoon
Dec 21 '18 at 17:24
add a comment |
1
$begingroup$
This integral gives $f(x) =frac{1}{12} left(6 text{Li}_2left(e^{-2 x}right)-6 x left(x+2 log left(1-e^{-2 x}right)-2 log (sinh (x))right)-pi ^2right)$
$endgroup$
– Cesareo
Dec 21 '18 at 17:13
1
$begingroup$
$x_0approx2.146306122947278$
$endgroup$
– James Arathoon
Dec 21 '18 at 17:24
1
1
$begingroup$
This integral gives $f(x) =frac{1}{12} left(6 text{Li}_2left(e^{-2 x}right)-6 x left(x+2 log left(1-e^{-2 x}right)-2 log (sinh (x))right)-pi ^2right)$
$endgroup$
– Cesareo
Dec 21 '18 at 17:13
$begingroup$
This integral gives $f(x) =frac{1}{12} left(6 text{Li}_2left(e^{-2 x}right)-6 x left(x+2 log left(1-e^{-2 x}right)-2 log (sinh (x))right)-pi ^2right)$
$endgroup$
– Cesareo
Dec 21 '18 at 17:13
1
1
$begingroup$
$x_0approx2.146306122947278$
$endgroup$
– James Arathoon
Dec 21 '18 at 17:24
$begingroup$
$x_0approx2.146306122947278$
$endgroup$
– James Arathoon
Dec 21 '18 at 17:24
add a comment |
1 Answer
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$begingroup$
This is a limited scope answer to address your subsidiary questions, since no other answers have been forthcoming so far:
Infinite Series for $log(sinh(z))$
$$log(sinh(z))=log (z)-sum _{k=1}^{infty } frac{ (-1)^k zeta (2 k)}{k pi ^{2 k}}z^{2 k}$$
which can easily be integrated term-wise when convergent to give
$$int_0^xlog(sinh(z)); dz=-x+ xlog (x)-sum _{k=1}^{infty } frac{ (-1)^k zeta (2 k)}{k (2k+1)pi ^{2 k}}x^{2 k+1}$$
There also appears to be a hyperbolic fourier type infinite series (analogous to the cos series for $log(sin z)$) which I am not sure how to prove (I've raised a separate question here):
$$log(sinh(z))=-frac{1}{2} (i pi )-log (2)-sum _{k=1}^{infty } frac{cosh (2 k z)}{k}$$
which integrates to
$$int_0^xlog(sinh(z)); dz=-frac{x}{2} (i pi )-x log (2)-sum _{k=1}^{infty } frac{sinh (2 k x)}{2 k^2}$$
Using this second formula Mathematica gives at the zero $x_0approx2.14631$:
$$frac{text{Li}_2left(e^{-2 (2.14631)}right)-text{Li}_2left(e^{2 (2.14631)}right)}{2.14631}=4 log (2)+2 i pi$$
which gives a strong indication I think that you won't find a closed form for $x_0$.
answered Jan 24 at 22:27
James ArathoonJames Arathoon
1,568423
$endgroup$
add a comment |
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1 Answer
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$begingroup$
This is a limited scope answer to address your subsidiary questions, since no other answers have been forthcoming so far:
Infinite Series for $log(sinh(z))$
$$log(sinh(z))=log (z)-sum _{k=1}^{infty } frac{ (-1)^k zeta (2 k)}{k pi ^{2 k}}z^{2 k}$$
which can easily be integrated term-wise when convergent to give
$$int_0^xlog(sinh(z)); dz=-x+ xlog (x)-sum _{k=1}^{infty } frac{ (-1)^k zeta (2 k)}{k (2k+1)pi ^{2 k}}x^{2 k+1}$$
There also appears to be a hyperbolic fourier type infinite series (analogous to the cos series for $log(sin z)$) which I am not sure how to prove (I've raised a separate question here):
$$log(sinh(z))=-frac{1}{2} (i pi )-log (2)-sum _{k=1}^{infty } frac{cosh (2 k z)}{k}$$
which integrates to
$$int_0^xlog(sinh(z)); dz=-frac{x}{2} (i pi )-x log (2)-sum _{k=1}^{infty } frac{sinh (2 k x)}{2 k^2}$$
Using this second formula Mathematica gives at the zero $x_0approx2.14631$:
$$frac{text{Li}_2left(e^{-2 (2.14631)}right)-text{Li}_2left(e^{2 (2.14631)}right)}{2.14631}=4 log (2)+2 i pi$$
which gives a strong indication I think that you won't find a closed form for $x_0$.
answered Jan 24 at 22:27
James ArathoonJames Arathoon
1,568423
$endgroup$
add a comment |
$begingroup$
This is a limited scope answer to address your subsidiary questions, since no other answers have been forthcoming so far:
Infinite Series for $log(sinh(z))$
$$log(sinh(z))=log (z)-sum _{k=1}^{infty } frac{ (-1)^k zeta (2 k)}{k pi ^{2 k}}z^{2 k}$$
which can easily be integrated term-wise when convergent to give
$$int_0^xlog(sinh(z)); dz=-x+ xlog (x)-sum _{k=1}^{infty } frac{ (-1)^k zeta (2 k)}{k (2k+1)pi ^{2 k}}x^{2 k+1}$$
There also appears to be a hyperbolic fourier type infinite series (analogous to the cos series for $log(sin z)$) which I am not sure how to prove (I've raised a separate question here):
$$log(sinh(z))=-frac{1}{2} (i pi )-log (2)-sum _{k=1}^{infty } frac{cosh (2 k z)}{k}$$
which integrates to
$$int_0^xlog(sinh(z)); dz=-frac{x}{2} (i pi )-x log (2)-sum _{k=1}^{infty } frac{sinh (2 k x)}{2 k^2}$$
Using this second formula Mathematica gives at the zero $x_0approx2.14631$:
$$frac{text{Li}_2left(e^{-2 (2.14631)}right)-text{Li}_2left(e^{2 (2.14631)}right)}{2.14631}=4 log (2)+2 i pi$$
which gives a strong indication I think that you won't find a closed form for $x_0$.
answered Jan 24 at 22:27
James ArathoonJames Arathoon
1,568423
$endgroup$
add a comment |
$begingroup$
This is a limited scope answer to address your subsidiary questions, since no other answers have been forthcoming so far:
Infinite Series for $log(sinh(z))$
$$log(sinh(z))=log (z)-sum _{k=1}^{infty } frac{ (-1)^k zeta (2 k)}{k pi ^{2 k}}z^{2 k}$$
which can easily be integrated term-wise when convergent to give
$$int_0^xlog(sinh(z)); dz=-x+ xlog (x)-sum _{k=1}^{infty } frac{ (-1)^k zeta (2 k)}{k (2k+1)pi ^{2 k}}x^{2 k+1}$$
There also appears to be a hyperbolic fourier type infinite series (analogous to the cos series for $log(sin z)$) which I am not sure how to prove (I've raised a separate question here):
$$log(sinh(z))=-frac{1}{2} (i pi )-log (2)-sum _{k=1}^{infty } frac{cosh (2 k z)}{k}$$
which integrates to
$$int_0^xlog(sinh(z)); dz=-frac{x}{2} (i pi )-x log (2)-sum _{k=1}^{infty } frac{sinh (2 k x)}{2 k^2}$$
Using this second formula Mathematica gives at the zero $x_0approx2.14631$:
$$frac{text{Li}_2left(e^{-2 (2.14631)}right)-text{Li}_2left(e^{2 (2.14631)}right)}{2.14631}=4 log (2)+2 i pi$$
which gives a strong indication I think that you won't find a closed form for $x_0$.
answered Jan 24 at 22:27
James ArathoonJames Arathoon
1,568423
$endgroup$
This is a limited scope answer to address your subsidiary questions, since no other answers have been forthcoming so far:
Infinite Series for $log(sinh(z))$
$$log(sinh(z))=log (z)-sum _{k=1}^{infty } frac{ (-1)^k zeta (2 k)}{k pi ^{2 k}}z^{2 k}$$
which can easily be integrated term-wise when convergent to give
$$int_0^xlog(sinh(z)); dz=-x+ xlog (x)-sum _{k=1}^{infty } frac{ (-1)^k zeta (2 k)}{k (2k+1)pi ^{2 k}}x^{2 k+1}$$
There also appears to be a hyperbolic fourier type infinite series (analogous to the cos series for $log(sin z)$) which I am not sure how to prove (I've raised a separate question here):
$$log(sinh(z))=-frac{1}{2} (i pi )-log (2)-sum _{k=1}^{infty } frac{cosh (2 k z)}{k}$$
which integrates to
$$int_0^xlog(sinh(z)); dz=-frac{x}{2} (i pi )-x log (2)-sum _{k=1}^{infty } frac{sinh (2 k x)}{2 k^2}$$
Using this second formula Mathematica gives at the zero $x_0approx2.14631$:
$$frac{text{Li}_2left(e^{-2 (2.14631)}right)-text{Li}_2left(e^{2 (2.14631)}right)}{2.14631}=4 log (2)+2 i pi$$
which gives a strong indication I think that you won't find a closed form for $x_0$.
answered Jan 24 at 22:27
James ArathoonJames Arathoon
1,568423
answered Jan 24 at 22:27
James ArathoonJames Arathoon
1,568423
answered Jan 24 at 22:27
James ArathoonJames Arathoon
1,568423
answered Jan 24 at 22:27
James ArathoonJames Arathoon
1,568423
1,568423
add a comment |
add a comment |
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1
$begingroup$
This integral gives $f(x) =frac{1}{12} left(6 text{Li}_2left(e^{-2 x}right)-6 x left(x+2 log left(1-e^{-2 x}right)-2 log (sinh (x))right)-pi ^2right)$
$endgroup$
– Cesareo
Dec 21 '18 at 17:13
1
$begingroup$
$x_0approx2.146306122947278$
$endgroup$
– James Arathoon
Dec 21 '18 at 17:24