# An integral involving the argument of the Gamma function and the Riemann Hypothesis

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

• @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 $\bmod \pi$ – juan Dec 18 '18 at 22:00
• @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. – juan Dec 18 '18 at 22:13
• @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 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. – juan Dec 18 '18 at 22:19
• @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 – juan Dec 18 '18 at 22:23

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 $$I=-\frac14 \bigl\{\pi\gamma+2\pi-2\pi+2\pi\log2\bigr\}=-\frac{\pi}{4}(\gamma+\log 4).$$
• 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
• 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