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Evaluate $$I=\int_{0}^{\infty} \frac{t\arg \Gamma(\frac{1}{4}+\frac{it}{2})}{(\frac{1}{4}+t^2)^2}\mathrm{d}t$$ where $\Gamma(s)=\int_{0}^{\infty}e^{-x}x^{s-1}\mathrm{d}x.$

Note that $I$ converges since $\Gamma(s)\sim s\log s$. 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

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    $\begingroup$ @OneTwoOne No. This not disproof the Riemann hypothesis. when one uses $\arg\zeta(1/2+ix)$ one refers to the "discontinuous" arg defined in a certain specific way. Your formula for $\arg\zeta(1/2+it)$ is not correct. It is only an equality $\bmod \pi$ $\endgroup$ – juan Dec 18 '18 at 22:00
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    $\begingroup$ @OneTwoOne It is not so easy to explain. It is defined in the book by Titchmarsh Section 9.3. $\arg\zeta(1/2+iT)$ 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 $\arg\Gamma(1/4+it/2)$ is usually meaning as the continuous argument. So the relation you writes between then is not true. $\endgroup$ – juan Dec 18 '18 at 22:13
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    $\begingroup$ @OneTwoOne This discontinuous function that uses user 64494 is not the same as you consider. RH is much more difficult than you think. $\endgroup$ – juan Dec 18 '18 at 22:16
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    $\begingroup$ @OneTwoOne What you have done have nothing to do with the Riemann Hypothesis. $\arg\zeta(1/2+it)$ contains many information about the zeros. $\Gamma(1/4+it/2)$ almost nothing. $\endgroup$ – juan Dec 18 '18 at 22:19
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    $\begingroup$ @OneTwoOne $\frac{x}{2}\log\pi-\arg\Gamma(1/4+it/2)$ is a very simple function, continuous and indefinitely differentiable. $\arg\zeta(1/2+ix)$ 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 $\endgroup$ – juan Dec 18 '18 at 22:23
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We prove that $$I=-\frac{\pi}{4}(\gamma+\log 4).$$ $$I=\int_0^\infty\frac{t\arg\Gamma(\frac14+\frac{it}{2})}{(\frac14+t^2)^2}\,dt.$$ $I$ is the imaginary part of the complex integral $$\int_0^\infty \frac{t\log\Gamma(\frac14+\frac{it}{2})}{(\frac14+t^2)^2}\,dt$$ using the usual branch of the logarithm of $\Gamma(s)$. Integrating by parts $$=\frac12\int_0^\infty \log\Gamma(\tfrac14+\tfrac{it}{2})\,d\Bigl\{-\frac{1}{\frac14+t^2}\Bigr\}= \Bigl.-\frac{1}{2(\frac14+t^2)}\log \Gamma(\tfrac14+\tfrac{it}{2})\Bigr|_{t=0}^\infty+\frac12\int_0^\infty \frac{1}{\frac14+t^2}\frac{i}{2}\frac{\Gamma'(\frac14+\frac{it}{2})}{\Gamma(\frac14+\frac{it}{2})}\,dt$$ $$=2\log\Gamma(1/4)+\frac{i}{4}\int_0^\infty \frac{\Gamma'(\frac14+\frac{it}{2})}{\Gamma(\frac14+\frac{it}{2})}\frac{dt}{\frac14+t^2}.$$ We have $$-\frac{\Gamma'(s)}{\Gamma(s)}=\gamma+\frac{1}{s}+\sum_{n=1}^\infty\Bigl(\frac{1}{s+n}-\frac{1}{n}\Bigr).$$ It is easy to justify the interchange here so that $$=2\log\Gamma(1/4)-\frac{i}{4}\Bigl\{\gamma\int_0^\infty \frac{dt}{\frac14+t^2}+ \int_0^\infty \frac{1}{\frac14+\frac{it}{2}}\frac{dt}{\frac14+t^2}+\sum_{n=1}^\infty \int_0^\infty \Bigl(\frac{1}{\frac14+\frac{it}{2}+n}-\frac{1}{n}\Bigr)\frac{dt}{\frac14+t^2} \Bigr\}$$ With Mathematica we find $$\int_0^\infty \frac{dt}{\frac14+t^2}=\pi,\quad \int_0^\infty \frac{1}{\frac14+\frac{it}{2}}\frac{dt}{\frac14+t^2}=2\pi-4i,$$ $$\int_0^\infty \Bigl(\frac{1}{\frac14+\frac{it}{2}+n}-\frac{1}{n}\Bigr)\frac{dt}{\frac14+t^2}=-\frac{\pi+i\log(1+4n)}{n(2n+1)}.$$ Taking the imaginary part we obtain $$I=-\frac14\Bigl\{\pi\gamma+2\pi-\sum_{n=1}^\infty \frac{\pi}{n(2n+1)}\Bigr\}$$ For the sum in $n$ we get with Mathematica $$\sum_{n=1}^\infty\frac{1}{n(2n+1)}=2-2\log 2.$$ It follows that $$I=-\frac14 \bigl\{\pi\gamma+2\pi-2\pi+2\pi\log2\bigr\}=-\frac{\pi}{4}(\gamma+\log 4).$$

The evaluations with Mathematica are not difficult to prove.

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  • $\begingroup$ 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$. $\endgroup$ – user64494 Dec 18 '18 at 18:30
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    $\begingroup$ @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. $\endgroup$ – juan Dec 18 '18 at 18:39
  • $\begingroup$ 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 $\endgroup$ – user64494 Dec 18 '18 at 18:49
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    $\begingroup$ @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. $\endgroup$ – juan Dec 18 '18 at 18:53
  • $\begingroup$ 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. $\endgroup$ – user64494 Dec 18 '18 at 18:58

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