An integral involving the argument of the Gamma function and the Riemann Hypothesis
Evaluate $$I=int_{0}^{infty} frac{targ
Gamma(frac{1}{4}+frac{it}{2})}{(frac{1}{4}+t^2)^2}mathrm{d}t$$
where $Gamma(s)=int_{0}^{infty}e^{-x}x^{s-1}mathrm{d}x.$
Note that $I$ converges since $Gamma(s)sim slog s$. I tried Wolfram Alpha, but it hasn't given me an answer after almost 90 minutes of computation, hence perhaps will never do.
PS: Migrated from https://math.stackexchange.com/q/3045147
cv.complex-variables integration riemann-zeta-function riemann-hypothesis
asked Dec 18 '18 at 14:51
OneTwoOne
403
|
show 6 more comments
Evaluate $$I=int_{0}^{infty} frac{targ
Gamma(frac{1}{4}+frac{it}{2})}{(frac{1}{4}+t^2)^2}mathrm{d}t$$
where $Gamma(s)=int_{0}^{infty}e^{-x}x^{s-1}mathrm{d}x.$
Note that $I$ converges since $Gamma(s)sim slog s$. I tried Wolfram Alpha, but it hasn't given me an answer after almost 90 minutes of computation, hence perhaps will never do.
PS: Migrated from https://math.stackexchange.com/q/3045147
cv.complex-variables integration riemann-zeta-function riemann-hypothesis
asked Dec 18 '18 at 14:51
OneTwoOne
403
2
@OneTwoOne No. This not disproof the Riemann hypothesis. when one uses $argzeta(1/2+ix)$ one refers to the "discontinuous" arg defined in a certain specific way. Your formula for $argzeta(1/2+it)$ is not correct. It is only an equality $bmod pi$
– juan
Dec 18 '18 at 22:00
3
@OneTwoOne It is not so easy to explain. It is defined in the book by Titchmarsh Section 9.3. $argzeta(1/2+iT)$ is obtained by continuous variation along the straight lines joining 2, $2+iT$, $1/2+iT$, starting with the value $0$. It is a discontinuous function. While $argGamma(1/4+it/2)$ is usually meaning as the continuous argument. So the relation you writes between then is not true.
– juan
Dec 18 '18 at 22:13
2
@OneTwoOne This discontinuous function that uses user 64494 is not the same as you consider. RH is much more difficult than you think.
– juan
Dec 18 '18 at 22:16
4
@OneTwoOne What you have done have nothing to do with the Riemann Hypothesis. $argzeta(1/2+it)$ contains many information about the zeros. $Gamma(1/4+it/2)$ almost nothing.
– juan
Dec 18 '18 at 22:19
2
@OneTwoOne $frac{x}{2}logpi-argGamma(1/4+it/2)$ is a very simple function, continuous and indefinitely differentiable. $argzeta(1/2+ix)$ is continuous except at zeros of zeta (to simplify I assume the Riemann Hypothesis here) the function has jumps at the zeros of zeta. The equality you write is only true $mod pi$. But if you consider the function of user 64494
– juan
Dec 18 '18 at 22:23
|
show 6 more comments
Evaluate $$I=int_{0}^{infty} frac{targ
Gamma(frac{1}{4}+frac{it}{2})}{(frac{1}{4}+t^2)^2}mathrm{d}t$$
where $Gamma(s)=int_{0}^{infty}e^{-x}x^{s-1}mathrm{d}x.$
Note that $I$ converges since $Gamma(s)sim slog s$. I tried Wolfram Alpha, but it hasn't given me an answer after almost 90 minutes of computation, hence perhaps will never do.
PS: Migrated from https://math.stackexchange.com/q/3045147
cv.complex-variables integration riemann-zeta-function riemann-hypothesis
asked Dec 18 '18 at 14:51
OneTwoOne
403
Evaluate $$I=int_{0}^{infty} frac{targ
Gamma(frac{1}{4}+frac{it}{2})}{(frac{1}{4}+t^2)^2}mathrm{d}t$$
where $Gamma(s)=int_{0}^{infty}e^{-x}x^{s-1}mathrm{d}x.$
Note that $I$ converges since $Gamma(s)sim slog s$. I tried Wolfram Alpha, but it hasn't given me an answer after almost 90 minutes of computation, hence perhaps will never do.
PS: Migrated from https://math.stackexchange.com/q/3045147
cv.complex-variables integration riemann-zeta-function riemann-hypothesis
cv.complex-variables integration riemann-zeta-function riemann-hypothesis
asked Dec 18 '18 at 14:51
OneTwoOne
403
asked Dec 18 '18 at 14:51
OneTwoOne
403
edited Dec 23 '18 at 6:55
asked Dec 18 '18 at 14:51
OneTwoOne
403
asked Dec 18 '18 at 14:51
OneTwoOne
403
asked Dec 18 '18 at 14:51
OneTwoOne
403
403
2
@OneTwoOne No. This not disproof the Riemann hypothesis. when one uses $argzeta(1/2+ix)$ one refers to the "discontinuous" arg defined in a certain specific way. Your formula for $argzeta(1/2+it)$ is not correct. It is only an equality $bmod pi$
– juan
Dec 18 '18 at 22:00
3
@OneTwoOne It is not so easy to explain. It is defined in the book by Titchmarsh Section 9.3. $argzeta(1/2+iT)$ is obtained by continuous variation along the straight lines joining 2, $2+iT$, $1/2+iT$, starting with the value $0$. It is a discontinuous function. While $argGamma(1/4+it/2)$ is usually meaning as the continuous argument. So the relation you writes between then is not true.
– juan
Dec 18 '18 at 22:13
2
@OneTwoOne This discontinuous function that uses user 64494 is not the same as you consider. RH is much more difficult than you think.
– juan
Dec 18 '18 at 22:16
4
@OneTwoOne What you have done have nothing to do with the Riemann Hypothesis. $argzeta(1/2+it)$ contains many information about the zeros. $Gamma(1/4+it/2)$ almost nothing.
– juan
Dec 18 '18 at 22:19
2
@OneTwoOne $frac{x}{2}logpi-argGamma(1/4+it/2)$ is a very simple function, continuous and indefinitely differentiable. $argzeta(1/2+ix)$ is continuous except at zeros of zeta (to simplify I assume the Riemann Hypothesis here) the function has jumps at the zeros of zeta. The equality you write is only true $mod pi$. But if you consider the function of user 64494
– juan
Dec 18 '18 at 22:23
|
show 6 more comments
2
@OneTwoOne No. This not disproof the Riemann hypothesis. when one uses $argzeta(1/2+ix)$ one refers to the "discontinuous" arg defined in a certain specific way. Your formula for $argzeta(1/2+it)$ is not correct. It is only an equality $bmod pi$
– juan
Dec 18 '18 at 22:00
3
@OneTwoOne It is not so easy to explain. It is defined in the book by Titchmarsh Section 9.3. $argzeta(1/2+iT)$ is obtained by continuous variation along the straight lines joining 2, $2+iT$, $1/2+iT$, starting with the value $0$. It is a discontinuous function. While $argGamma(1/4+it/2)$ is usually meaning as the continuous argument. So the relation you writes between then is not true.
– juan
Dec 18 '18 at 22:13
2
@OneTwoOne This discontinuous function that uses user 64494 is not the same as you consider. RH is much more difficult than you think.
– juan
Dec 18 '18 at 22:16
4
@OneTwoOne What you have done have nothing to do with the Riemann Hypothesis. $argzeta(1/2+it)$ contains many information about the zeros. $Gamma(1/4+it/2)$ almost nothing.
– juan
Dec 18 '18 at 22:19
2
@OneTwoOne $frac{x}{2}logpi-argGamma(1/4+it/2)$ is a very simple function, continuous and indefinitely differentiable. $argzeta(1/2+ix)$ is continuous except at zeros of zeta (to simplify I assume the Riemann Hypothesis here) the function has jumps at the zeros of zeta. The equality you write is only true $mod pi$. But if you consider the function of user 64494
– juan
Dec 18 '18 at 22:23
2
2
@OneTwoOne No. This not disproof the Riemann hypothesis. when one uses $argzeta(1/2+ix)$ one refers to the "discontinuous" arg defined in a certain specific way. Your formula for $argzeta(1/2+it)$ is not correct. It is only an equality $bmod pi$
– juan
Dec 18 '18 at 22:00
@OneTwoOne No. This not disproof the Riemann hypothesis. when one uses $argzeta(1/2+ix)$ one refers to the "discontinuous" arg defined in a certain specific way. Your formula for $argzeta(1/2+it)$ is not correct. It is only an equality $bmod pi$
– juan
Dec 18 '18 at 22:00
3
3
@OneTwoOne It is not so easy to explain. It is defined in the book by Titchmarsh Section 9.3. $argzeta(1/2+iT)$ is obtained by continuous variation along the straight lines joining 2, $2+iT$, $1/2+iT$, starting with the value $0$. It is a discontinuous function. While $argGamma(1/4+it/2)$ is usually meaning as the continuous argument. So the relation you writes between then is not true.
– juan
Dec 18 '18 at 22:13
@OneTwoOne It is not so easy to explain. It is defined in the book by Titchmarsh Section 9.3. $argzeta(1/2+iT)$ is obtained by continuous variation along the straight lines joining 2, $2+iT$, $1/2+iT$, starting with the value $0$. It is a discontinuous function. While $argGamma(1/4+it/2)$ is usually meaning as the continuous argument. So the relation you writes between then is not true.
– juan
Dec 18 '18 at 22:13
2
2
@OneTwoOne This discontinuous function that uses user 64494 is not the same as you consider. RH is much more difficult than you think.
– juan
Dec 18 '18 at 22:16
@OneTwoOne This discontinuous function that uses user 64494 is not the same as you consider. RH is much more difficult than you think.
– juan
Dec 18 '18 at 22:16
4
4
@OneTwoOne What you have done have nothing to do with the Riemann Hypothesis. $argzeta(1/2+it)$ contains many information about the zeros. $Gamma(1/4+it/2)$ almost nothing.
– juan
Dec 18 '18 at 22:19
@OneTwoOne What you have done have nothing to do with the Riemann Hypothesis. $argzeta(1/2+it)$ contains many information about the zeros. $Gamma(1/4+it/2)$ almost nothing.
– juan
Dec 18 '18 at 22:19
2
2
@OneTwoOne $frac{x}{2}logpi-argGamma(1/4+it/2)$ is a very simple function, continuous and indefinitely differentiable. $argzeta(1/2+ix)$ is continuous except at zeros of zeta (to simplify I assume the Riemann Hypothesis here) the function has jumps at the zeros of zeta. The equality you write is only true $mod pi$. But if you consider the function of user 64494
– juan
Dec 18 '18 at 22:23
@OneTwoOne $frac{x}{2}logpi-argGamma(1/4+it/2)$ is a very simple function, continuous and indefinitely differentiable. $argzeta(1/2+ix)$ is continuous except at zeros of zeta (to simplify I assume the Riemann Hypothesis here) the function has jumps at the zeros of zeta. The equality you write is only true $mod pi$. But if you consider the function of user 64494
– juan
Dec 18 '18 at 22:23
|
show 6 more comments
1 Answer
1
active
oldest
votes
We prove that
$$I=-frac{pi}{4}(gamma+log 4).$$
$$I=int_0^inftyfrac{targGamma(frac14+frac{it}{2})}{(frac14+t^2)^2},dt.$$
$I$ is the imaginary part of the complex integral
$$int_0^infty frac{tlogGamma(frac14+frac{it}{2})}{(frac14+t^2)^2},dt$$
using the usual branch of the logarithm of $Gamma(s)$. Integrating by parts
$$=frac12int_0^infty logGamma(tfrac14+tfrac{it}{2}),dBigl{-frac{1}{frac14+t^2}Bigr}=
Bigl.-frac{1}{2(frac14+t^2)}log Gamma(tfrac14+tfrac{it}{2})Bigr|_{t=0}^infty+frac12int_0^infty frac{1}{frac14+t^2}frac{i}{2}frac{Gamma'(frac14+frac{it}{2})}{Gamma(frac14+frac{it}{2})},dt$$
$$=2logGamma(1/4)+frac{i}{4}int_0^infty frac{Gamma'(frac14+frac{it}{2})}{Gamma(frac14+frac{it}{2})}frac{dt}{frac14+t^2}.$$
We have
$$-frac{Gamma'(s)}{Gamma(s)}=gamma+frac{1}{s}+sum_{n=1}^inftyBigl(frac{1}{s+n}-frac{1}{n}Bigr).$$
It is easy to justify the interchange here so that
$$=2logGamma(1/4)-frac{i}{4}Bigl{gammaint_0^infty frac{dt}{frac14+t^2}+
int_0^infty frac{1}{frac14+frac{it}{2}}frac{dt}{frac14+t^2}+sum_{n=1}^infty
int_0^infty Bigl(frac{1}{frac14+frac{it}{2}+n}-frac{1}{n}Bigr)frac{dt}{frac14+t^2}
Bigr}$$
With Mathematica we find
$$int_0^infty frac{dt}{frac14+t^2}=pi,quad int_0^infty frac{1}{frac14+frac{it}{2}}frac{dt}{frac14+t^2}=2pi-4i,$$
$$int_0^infty Bigl(frac{1}{frac14+frac{it}{2}+n}-frac{1}{n}Bigr)frac{dt}{frac14+t^2}=-frac{pi+ilog(1+4n)}{n(2n+1)}.$$
Taking the imaginary part we obtain
$$I=-frac14Bigl{pigamma+2pi-sum_{n=1}^infty frac{pi}{n(2n+1)}Bigr}$$
For the sum in $n$ we get with Mathematica
$$sum_{n=1}^inftyfrac{1}{n(2n+1)}=2-2log 2.$$
It follows that
$$I=-frac14 bigl{pigamma+2pi-2pi+2pilog2bigr}=-frac{pi}{4}(gamma+log 4).$$
The evaluations with Mathematica are not difficult to prove.
answered Dec 18 '18 at 17:37
juan
5,36512830
Unfortunately, the output of the Mathematica's code N[-Pi/4*(EulerGamma + Log[4])] equals $-1.54214$. This is not in accordance with the numeric value NIntegrate[t * Arg[Gamma[1/4 + I*t/2]]/(1/4 + t^2)^2, {t, 0, Infinity}, AccuracyGoal -> 3, WorkingPrecision -> 15], i. e. $ -1.62060929754175$.
– user64494
Dec 18 '18 at 18:30
2
@user64494 You can not use here Arg[Gamma] because this is not the continuous argument. use instead Im[LogGamma[ and you will get my value.
– juan
Dec 18 '18 at 18:39
Can you elaborate your comment? The plots Plot[Arg[Gamma[1/4 + I * t/2]], {t, 0, 10}] and Plot[Im[LogGamma[1/4 + I * t/2]], {t, 0, 10}] are identical
– user64494
Dec 18 '18 at 18:49
1
@user64494 Plot it from 0 to 20, the problem is that the argument of gamma can be defined as a continuous function that sometimes cross pi, 3pi and so on. Of course you can consider the integral with the discontinuous argument. But usually when one speaks about the argument of Gamma one uses the continuous one.
– juan
Dec 18 '18 at 18:53
The question is unclearly formulated: usually $arg z$ stands for the main value of the argument of a complex number $z$ and the Arg[z] command realizes it.
– user64494
Dec 18 '18 at 18:58
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
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We prove that
$$I=-frac{pi}{4}(gamma+log 4).$$
$$I=int_0^inftyfrac{targGamma(frac14+frac{it}{2})}{(frac14+t^2)^2},dt.$$
$I$ is the imaginary part of the complex integral
$$int_0^infty frac{tlogGamma(frac14+frac{it}{2})}{(frac14+t^2)^2},dt$$
using the usual branch of the logarithm of $Gamma(s)$. Integrating by parts
$$=frac12int_0^infty logGamma(tfrac14+tfrac{it}{2}),dBigl{-frac{1}{frac14+t^2}Bigr}=
Bigl.-frac{1}{2(frac14+t^2)}log Gamma(tfrac14+tfrac{it}{2})Bigr|_{t=0}^infty+frac12int_0^infty frac{1}{frac14+t^2}frac{i}{2}frac{Gamma'(frac14+frac{it}{2})}{Gamma(frac14+frac{it}{2})},dt$$
$$=2logGamma(1/4)+frac{i}{4}int_0^infty frac{Gamma'(frac14+frac{it}{2})}{Gamma(frac14+frac{it}{2})}frac{dt}{frac14+t^2}.$$
We have
$$-frac{Gamma'(s)}{Gamma(s)}=gamma+frac{1}{s}+sum_{n=1}^inftyBigl(frac{1}{s+n}-frac{1}{n}Bigr).$$
It is easy to justify the interchange here so that
$$=2logGamma(1/4)-frac{i}{4}Bigl{gammaint_0^infty frac{dt}{frac14+t^2}+
int_0^infty frac{1}{frac14+frac{it}{2}}frac{dt}{frac14+t^2}+sum_{n=1}^infty
int_0^infty Bigl(frac{1}{frac14+frac{it}{2}+n}-frac{1}{n}Bigr)frac{dt}{frac14+t^2}
Bigr}$$
With Mathematica we find
$$int_0^infty frac{dt}{frac14+t^2}=pi,quad int_0^infty frac{1}{frac14+frac{it}{2}}frac{dt}{frac14+t^2}=2pi-4i,$$
$$int_0^infty Bigl(frac{1}{frac14+frac{it}{2}+n}-frac{1}{n}Bigr)frac{dt}{frac14+t^2}=-frac{pi+ilog(1+4n)}{n(2n+1)}.$$
Taking the imaginary part we obtain
$$I=-frac14Bigl{pigamma+2pi-sum_{n=1}^infty frac{pi}{n(2n+1)}Bigr}$$
For the sum in $n$ we get with Mathematica
$$sum_{n=1}^inftyfrac{1}{n(2n+1)}=2-2log 2.$$
It follows that
$$I=-frac14 bigl{pigamma+2pi-2pi+2pilog2bigr}=-frac{pi}{4}(gamma+log 4).$$
The evaluations with Mathematica are not difficult to prove.
answered Dec 18 '18 at 17:37
juan
5,36512830
Unfortunately, the output of the Mathematica's code N[-Pi/4*(EulerGamma + Log[4])] equals $-1.54214$. This is not in accordance with the numeric value NIntegrate[t * Arg[Gamma[1/4 + I*t/2]]/(1/4 + t^2)^2, {t, 0, Infinity}, AccuracyGoal -> 3, WorkingPrecision -> 15], i. e. $ -1.62060929754175$.
– user64494
Dec 18 '18 at 18:30
2
@user64494 You can not use here Arg[Gamma] because this is not the continuous argument. use instead Im[LogGamma[ and you will get my value.
– juan
Dec 18 '18 at 18:39
Can you elaborate your comment? The plots Plot[Arg[Gamma[1/4 + I * t/2]], {t, 0, 10}] and Plot[Im[LogGamma[1/4 + I * t/2]], {t, 0, 10}] are identical
– user64494
Dec 18 '18 at 18:49
1
@user64494 Plot it from 0 to 20, the problem is that the argument of gamma can be defined as a continuous function that sometimes cross pi, 3pi and so on. Of course you can consider the integral with the discontinuous argument. But usually when one speaks about the argument of Gamma one uses the continuous one.
– juan
Dec 18 '18 at 18:53
The question is unclearly formulated: usually $arg z$ stands for the main value of the argument of a complex number $z$ and the Arg[z] command realizes it.
– user64494
Dec 18 '18 at 18:58
add a comment |
We prove that
$$I=-frac{pi}{4}(gamma+log 4).$$
$$I=int_0^inftyfrac{targGamma(frac14+frac{it}{2})}{(frac14+t^2)^2},dt.$$
$I$ is the imaginary part of the complex integral
$$int_0^infty frac{tlogGamma(frac14+frac{it}{2})}{(frac14+t^2)^2},dt$$
using the usual branch of the logarithm of $Gamma(s)$. Integrating by parts
$$=frac12int_0^infty logGamma(tfrac14+tfrac{it}{2}),dBigl{-frac{1}{frac14+t^2}Bigr}=
Bigl.-frac{1}{2(frac14+t^2)}log Gamma(tfrac14+tfrac{it}{2})Bigr|_{t=0}^infty+frac12int_0^infty frac{1}{frac14+t^2}frac{i}{2}frac{Gamma'(frac14+frac{it}{2})}{Gamma(frac14+frac{it}{2})},dt$$
$$=2logGamma(1/4)+frac{i}{4}int_0^infty frac{Gamma'(frac14+frac{it}{2})}{Gamma(frac14+frac{it}{2})}frac{dt}{frac14+t^2}.$$
We have
$$-frac{Gamma'(s)}{Gamma(s)}=gamma+frac{1}{s}+sum_{n=1}^inftyBigl(frac{1}{s+n}-frac{1}{n}Bigr).$$
It is easy to justify the interchange here so that
$$=2logGamma(1/4)-frac{i}{4}Bigl{gammaint_0^infty frac{dt}{frac14+t^2}+
int_0^infty frac{1}{frac14+frac{it}{2}}frac{dt}{frac14+t^2}+sum_{n=1}^infty
int_0^infty Bigl(frac{1}{frac14+frac{it}{2}+n}-frac{1}{n}Bigr)frac{dt}{frac14+t^2}
Bigr}$$
With Mathematica we find
$$int_0^infty frac{dt}{frac14+t^2}=pi,quad int_0^infty frac{1}{frac14+frac{it}{2}}frac{dt}{frac14+t^2}=2pi-4i,$$
$$int_0^infty Bigl(frac{1}{frac14+frac{it}{2}+n}-frac{1}{n}Bigr)frac{dt}{frac14+t^2}=-frac{pi+ilog(1+4n)}{n(2n+1)}.$$
Taking the imaginary part we obtain
$$I=-frac14Bigl{pigamma+2pi-sum_{n=1}^infty frac{pi}{n(2n+1)}Bigr}$$
For the sum in $n$ we get with Mathematica
$$sum_{n=1}^inftyfrac{1}{n(2n+1)}=2-2log 2.$$
It follows that
$$I=-frac14 bigl{pigamma+2pi-2pi+2pilog2bigr}=-frac{pi}{4}(gamma+log 4).$$
The evaluations with Mathematica are not difficult to prove.
answered Dec 18 '18 at 17:37
juan
5,36512830
Unfortunately, the output of the Mathematica's code N[-Pi/4*(EulerGamma + Log[4])] equals $-1.54214$. This is not in accordance with the numeric value NIntegrate[t * Arg[Gamma[1/4 + I*t/2]]/(1/4 + t^2)^2, {t, 0, Infinity}, AccuracyGoal -> 3, WorkingPrecision -> 15], i. e. $ -1.62060929754175$.
– user64494
Dec 18 '18 at 18:30
2
@user64494 You can not use here Arg[Gamma] because this is not the continuous argument. use instead Im[LogGamma[ and you will get my value.
– juan
Dec 18 '18 at 18:39
Can you elaborate your comment? The plots Plot[Arg[Gamma[1/4 + I * t/2]], {t, 0, 10}] and Plot[Im[LogGamma[1/4 + I * t/2]], {t, 0, 10}] are identical
– user64494
Dec 18 '18 at 18:49
1
@user64494 Plot it from 0 to 20, the problem is that the argument of gamma can be defined as a continuous function that sometimes cross pi, 3pi and so on. Of course you can consider the integral with the discontinuous argument. But usually when one speaks about the argument of Gamma one uses the continuous one.
– juan
Dec 18 '18 at 18:53
The question is unclearly formulated: usually $arg z$ stands for the main value of the argument of a complex number $z$ and the Arg[z] command realizes it.
– user64494
Dec 18 '18 at 18:58
add a comment |
We prove that
$$I=-frac{pi}{4}(gamma+log 4).$$
$$I=int_0^inftyfrac{targGamma(frac14+frac{it}{2})}{(frac14+t^2)^2},dt.$$
$I$ is the imaginary part of the complex integral
$$int_0^infty frac{tlogGamma(frac14+frac{it}{2})}{(frac14+t^2)^2},dt$$
using the usual branch of the logarithm of $Gamma(s)$. Integrating by parts
$$=frac12int_0^infty logGamma(tfrac14+tfrac{it}{2}),dBigl{-frac{1}{frac14+t^2}Bigr}=
Bigl.-frac{1}{2(frac14+t^2)}log Gamma(tfrac14+tfrac{it}{2})Bigr|_{t=0}^infty+frac12int_0^infty frac{1}{frac14+t^2}frac{i}{2}frac{Gamma'(frac14+frac{it}{2})}{Gamma(frac14+frac{it}{2})},dt$$
$$=2logGamma(1/4)+frac{i}{4}int_0^infty frac{Gamma'(frac14+frac{it}{2})}{Gamma(frac14+frac{it}{2})}frac{dt}{frac14+t^2}.$$
We have
$$-frac{Gamma'(s)}{Gamma(s)}=gamma+frac{1}{s}+sum_{n=1}^inftyBigl(frac{1}{s+n}-frac{1}{n}Bigr).$$
It is easy to justify the interchange here so that
$$=2logGamma(1/4)-frac{i}{4}Bigl{gammaint_0^infty frac{dt}{frac14+t^2}+
int_0^infty frac{1}{frac14+frac{it}{2}}frac{dt}{frac14+t^2}+sum_{n=1}^infty
int_0^infty Bigl(frac{1}{frac14+frac{it}{2}+n}-frac{1}{n}Bigr)frac{dt}{frac14+t^2}
Bigr}$$
With Mathematica we find
$$int_0^infty frac{dt}{frac14+t^2}=pi,quad int_0^infty frac{1}{frac14+frac{it}{2}}frac{dt}{frac14+t^2}=2pi-4i,$$
$$int_0^infty Bigl(frac{1}{frac14+frac{it}{2}+n}-frac{1}{n}Bigr)frac{dt}{frac14+t^2}=-frac{pi+ilog(1+4n)}{n(2n+1)}.$$
Taking the imaginary part we obtain
$$I=-frac14Bigl{pigamma+2pi-sum_{n=1}^infty frac{pi}{n(2n+1)}Bigr}$$
For the sum in $n$ we get with Mathematica
$$sum_{n=1}^inftyfrac{1}{n(2n+1)}=2-2log 2.$$
It follows that
$$I=-frac14 bigl{pigamma+2pi-2pi+2pilog2bigr}=-frac{pi}{4}(gamma+log 4).$$
The evaluations with Mathematica are not difficult to prove.
answered Dec 18 '18 at 17:37
juan
5,36512830
We prove that
$$I=-frac{pi}{4}(gamma+log 4).$$
$$I=int_0^inftyfrac{targGamma(frac14+frac{it}{2})}{(frac14+t^2)^2},dt.$$
$I$ is the imaginary part of the complex integral
$$int_0^infty frac{tlogGamma(frac14+frac{it}{2})}{(frac14+t^2)^2},dt$$
using the usual branch of the logarithm of $Gamma(s)$. Integrating by parts
$$=frac12int_0^infty logGamma(tfrac14+tfrac{it}{2}),dBigl{-frac{1}{frac14+t^2}Bigr}=
Bigl.-frac{1}{2(frac14+t^2)}log Gamma(tfrac14+tfrac{it}{2})Bigr|_{t=0}^infty+frac12int_0^infty frac{1}{frac14+t^2}frac{i}{2}frac{Gamma'(frac14+frac{it}{2})}{Gamma(frac14+frac{it}{2})},dt$$
$$=2logGamma(1/4)+frac{i}{4}int_0^infty frac{Gamma'(frac14+frac{it}{2})}{Gamma(frac14+frac{it}{2})}frac{dt}{frac14+t^2}.$$
We have
$$-frac{Gamma'(s)}{Gamma(s)}=gamma+frac{1}{s}+sum_{n=1}^inftyBigl(frac{1}{s+n}-frac{1}{n}Bigr).$$
It is easy to justify the interchange here so that
$$=2logGamma(1/4)-frac{i}{4}Bigl{gammaint_0^infty frac{dt}{frac14+t^2}+
int_0^infty frac{1}{frac14+frac{it}{2}}frac{dt}{frac14+t^2}+sum_{n=1}^infty
int_0^infty Bigl(frac{1}{frac14+frac{it}{2}+n}-frac{1}{n}Bigr)frac{dt}{frac14+t^2}
Bigr}$$
With Mathematica we find
$$int_0^infty frac{dt}{frac14+t^2}=pi,quad int_0^infty frac{1}{frac14+frac{it}{2}}frac{dt}{frac14+t^2}=2pi-4i,$$
$$int_0^infty Bigl(frac{1}{frac14+frac{it}{2}+n}-frac{1}{n}Bigr)frac{dt}{frac14+t^2}=-frac{pi+ilog(1+4n)}{n(2n+1)}.$$
Taking the imaginary part we obtain
$$I=-frac14Bigl{pigamma+2pi-sum_{n=1}^infty frac{pi}{n(2n+1)}Bigr}$$
For the sum in $n$ we get with Mathematica
$$sum_{n=1}^inftyfrac{1}{n(2n+1)}=2-2log 2.$$
It follows that
$$I=-frac14 bigl{pigamma+2pi-2pi+2pilog2bigr}=-frac{pi}{4}(gamma+log 4).$$
The evaluations with Mathematica are not difficult to prove.
answered Dec 18 '18 at 17:37
juan
5,36512830
answered Dec 18 '18 at 17:37
juan
5,36512830
answered Dec 18 '18 at 17:37
juan
5,36512830
answered Dec 18 '18 at 17:37
juan
5,36512830
5,36512830
Unfortunately, the output of the Mathematica's code N[-Pi/4*(EulerGamma + Log[4])] equals $-1.54214$. This is not in accordance with the numeric value NIntegrate[t * Arg[Gamma[1/4 + I*t/2]]/(1/4 + t^2)^2, {t, 0, Infinity}, AccuracyGoal -> 3, WorkingPrecision -> 15], i. e. $ -1.62060929754175$.
– user64494
Dec 18 '18 at 18:30
2
@user64494 You can not use here Arg[Gamma] because this is not the continuous argument. use instead Im[LogGamma[ and you will get my value.
– juan
Dec 18 '18 at 18:39
Can you elaborate your comment? The plots Plot[Arg[Gamma[1/4 + I * t/2]], {t, 0, 10}] and Plot[Im[LogGamma[1/4 + I * t/2]], {t, 0, 10}] are identical
– user64494
Dec 18 '18 at 18:49
1
@user64494 Plot it from 0 to 20, the problem is that the argument of gamma can be defined as a continuous function that sometimes cross pi, 3pi and so on. Of course you can consider the integral with the discontinuous argument. But usually when one speaks about the argument of Gamma one uses the continuous one.
– juan
Dec 18 '18 at 18:53
The question is unclearly formulated: usually $arg z$ stands for the main value of the argument of a complex number $z$ and the Arg[z] command realizes it.
– user64494
Dec 18 '18 at 18:58
add a comment |
Unfortunately, the output of the Mathematica's code N[-Pi/4*(EulerGamma + Log[4])] equals $-1.54214$. This is not in accordance with the numeric value NIntegrate[t * Arg[Gamma[1/4 + I*t/2]]/(1/4 + t^2)^2, {t, 0, Infinity}, AccuracyGoal -> 3, WorkingPrecision -> 15], i. e. $ -1.62060929754175$.
– user64494
Dec 18 '18 at 18:30
2
@user64494 You can not use here Arg[Gamma] because this is not the continuous argument. use instead Im[LogGamma[ and you will get my value.
– juan
Dec 18 '18 at 18:39
Can you elaborate your comment? The plots Plot[Arg[Gamma[1/4 + I * t/2]], {t, 0, 10}] and Plot[Im[LogGamma[1/4 + I * t/2]], {t, 0, 10}] are identical
– user64494
Dec 18 '18 at 18:49
1
@user64494 Plot it from 0 to 20, the problem is that the argument of gamma can be defined as a continuous function that sometimes cross pi, 3pi and so on. Of course you can consider the integral with the discontinuous argument. But usually when one speaks about the argument of Gamma one uses the continuous one.
– juan
Dec 18 '18 at 18:53
The question is unclearly formulated: usually $arg z$ stands for the main value of the argument of a complex number $z$ and the Arg[z] command realizes it.
– user64494
Dec 18 '18 at 18:58
Unfortunately, the output of the Mathematica's code N[-Pi/4*(EulerGamma + Log[4])] equals $-1.54214$. This is not in accordance with the numeric value NIntegrate[t * Arg[Gamma[1/4 + I*t/2]]/(1/4 + t^2)^2, {t, 0, Infinity}, AccuracyGoal -> 3, WorkingPrecision -> 15], i. e. $ -1.62060929754175$.
– user64494
Dec 18 '18 at 18:30
Unfortunately, the output of the Mathematica's code N[-Pi/4*(EulerGamma + Log[4])] equals $-1.54214$. This is not in accordance with the numeric value NIntegrate[t * Arg[Gamma[1/4 + I*t/2]]/(1/4 + t^2)^2, {t, 0, Infinity}, AccuracyGoal -> 3, WorkingPrecision -> 15], i. e. $ -1.62060929754175$.
– user64494
Dec 18 '18 at 18:30
2
2
@user64494 You can not use here Arg[Gamma] because this is not the continuous argument. use instead Im[LogGamma[ and you will get my value.
– juan
Dec 18 '18 at 18:39
@user64494 You can not use here Arg[Gamma] because this is not the continuous argument. use instead Im[LogGamma[ and you will get my value.
– juan
Dec 18 '18 at 18:39
Can you elaborate your comment? The plots Plot[Arg[Gamma[1/4 + I * t/2]], {t, 0, 10}] and Plot[Im[LogGamma[1/4 + I * t/2]], {t, 0, 10}] are identical
– user64494
Dec 18 '18 at 18:49
Can you elaborate your comment? The plots Plot[Arg[Gamma[1/4 + I * t/2]], {t, 0, 10}] and Plot[Im[LogGamma[1/4 + I * t/2]], {t, 0, 10}] are identical
– user64494
Dec 18 '18 at 18:49
1
1
@user64494 Plot it from 0 to 20, the problem is that the argument of gamma can be defined as a continuous function that sometimes cross pi, 3pi and so on. Of course you can consider the integral with the discontinuous argument. But usually when one speaks about the argument of Gamma one uses the continuous one.
– juan
Dec 18 '18 at 18:53
@user64494 Plot it from 0 to 20, the problem is that the argument of gamma can be defined as a continuous function that sometimes cross pi, 3pi and so on. Of course you can consider the integral with the discontinuous argument. But usually when one speaks about the argument of Gamma one uses the continuous one.
– juan
Dec 18 '18 at 18:53
The question is unclearly formulated: usually $arg z$ stands for the main value of the argument of a complex number $z$ and the Arg[z] command realizes it.
– user64494
Dec 18 '18 at 18:58
The question is unclearly formulated: usually $arg z$ stands for the main value of the argument of a complex number $z$ and the Arg[z] command realizes it.
– user64494
Dec 18 '18 at 18:58
add a comment |
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2
@OneTwoOne No. This not disproof the Riemann hypothesis. when one uses $argzeta(1/2+ix)$ one refers to the "discontinuous" arg defined in a certain specific way. Your formula for $argzeta(1/2+it)$ is not correct. It is only an equality $bmod pi$
– juan
Dec 18 '18 at 22:00
3
@OneTwoOne It is not so easy to explain. It is defined in the book by Titchmarsh Section 9.3. $argzeta(1/2+iT)$ is obtained by continuous variation along the straight lines joining 2, $2+iT$, $1/2+iT$, starting with the value $0$. It is a discontinuous function. While $argGamma(1/4+it/2)$ is usually meaning as the continuous argument. So the relation you writes between then is not true.
– juan
Dec 18 '18 at 22:13
2
@OneTwoOne This discontinuous function that uses user 64494 is not the same as you consider. RH is much more difficult than you think.
– juan
Dec 18 '18 at 22:16
4
@OneTwoOne What you have done have nothing to do with the Riemann Hypothesis. $argzeta(1/2+it)$ contains many information about the zeros. $Gamma(1/4+it/2)$ almost nothing.
– juan
Dec 18 '18 at 22:19
2
@OneTwoOne $frac{x}{2}logpi-argGamma(1/4+it/2)$ is a very simple function, continuous and indefinitely differentiable. $argzeta(1/2+ix)$ is continuous except at zeros of zeta (to simplify I assume the Riemann Hypothesis here) the function has jumps at the zeros of zeta. The equality you write is only true $mod pi$. But if you consider the function of user 64494
– juan
Dec 18 '18 at 22:23