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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

ADDENDUM (20/12/18): This is a reply to @juan and could've been too long as a comment.

The Riemann functional equation states that \begin{equation} \zeta(s)=\chi(s)\zeta(1-s), \end{equation} where $\chi(s)=\pi^{s-\frac{1}{2}}\frac{\Gamma(\frac{1}{2}-\frac{1}{2}s)}{\Gamma(\frac{1}{2}s)}$ and $\Gamma(s)=\int_{0}^{\infty} e^{-x}x^{s-1}\mathrm{d}x$. Let be $s$ be not a zero of $\zeta$, so that taking logarithms gives \begin{equation} \log \zeta(s)-\log \zeta(1-s) = (s-1/2)\log \pi + \log \Gamma\Big(\frac{1}{2}-\frac{1}{2}s\Big)-\log \Gamma \Big(\frac{1}{2}s \Big). \end{equation} Putting $s=1/2 + ix$, where $x\in \mathbb{R}$ and considering the imaginary parts yields \begin{equation} 2\arg \zeta\Big(\frac{1}{2}+ix \Big) = x\log \pi - 2 \arg \Gamma \Big(\frac{1}{4}+\frac{ix}{2} \Big). \end{equation} Define $0<x_1< x_2<x_3< \cdots$ to be the infinitely many positive zeros of $\zeta(z)$ on the line $\Re(z)=1/2$. For every $\delta>0$, consider \begin{equation} F_\delta=\Bigg(\int_{0}^{x_{1}-\delta} + \sum_{n=1}^{\infty} \int_{x_{n}+\delta}^{x_{n+1}-\delta}\Bigg) \frac{2x \arg \zeta(\frac{1}{2}+ix)}{(\frac{1}{4}+t^2)^2}\mathrm{d}x. \end{equation} Since $\zeta(1/2 + ix)\neq 0$ for $x\in[0, x_{1}-\delta]$ and $x\in[x_{n}+\delta, x_{n+1}-\delta]$ for every positive integer $n$, we can insert the above expression for $2\arg \zeta(1/2 + ix)$ and obtain \begin{equation} F_\delta=\Bigg(\int_{0}^{x_{1}-\delta} + \sum_{n=1}^{\infty} \int_{x_{n}+\delta}^{x_{n+1}-\delta}\Bigg) \frac{x^{2}\log \pi-2x \arg \Gamma(\frac{1}{4}+\frac{ix}{2}) }{(\frac{1}{4}+t^2)^2}\mathrm{d}x \end{equation}where $\arg$ denotes the principal value of the argument, hence

\begin{align} \lim_{\delta \rightarrow 0^+} F_\delta=\int_{0}^{\infty} \frac{x^{2}\log \pi}{(\frac{1}{4}+x^2)^2} \mathrm{d}x-\int_{0}^{\infty} \frac{2x \arg \Gamma(\frac{1}{4}+\frac{ix}{2})}{(\frac{1}{4}+x^2)^2} \mathrm{d}x>\frac{1}{2}\pi \log \pi, \end{align} which is incompatible with the RH, according to the Volchkov criterion. $\square$

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

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

ADDENDUM (20/12/18): This is a reply to @juan and could've been too long as a comment.

The Riemann functional equation states that \begin{equation} \zeta(s)=\chi(s)\zeta(1-s), \end{equation} where $\chi(s)=\pi^{s-\frac{1}{2}}\frac{\Gamma(\frac{1}{2}-\frac{1}{2}s)}{\Gamma(\frac{1}{2}s)}$ and $\Gamma(s)=\int_{0}^{\infty} e^{-x}x^{s-1}\mathrm{d}x$. Let be $s$ be not a zero of $\zeta$, so that taking logarithms gives \begin{equation} \log \zeta(s)-\log \zeta(1-s) = (s-1/2)\log \pi + \log \Gamma\Big(\frac{1}{2}-\frac{1}{2}s\Big)-\log \Gamma \Big(\frac{1}{2}s \Big). \end{equation} Putting $s=1/2 + ix$, where $x\in \mathbb{R}$ and considering the imaginary parts yields \begin{equation} 2\arg \zeta\Big(\frac{1}{2}+ix \Big) = x\log \pi - 2 \arg \Gamma \Big(\frac{1}{4}+\frac{ix}{2} \Big). \end{equation} Define $0<x_1< x_2<x_3< \cdots$ to be the infinitely many positive zeros of $\zeta(z)$ on the line $\Re(z)=1/2$. For every $\delta>0$, consider \begin{equation} I_\delta=\Bigg(\int_{0}^{x_{1}-\delta} + \sum_{n=1}^{\infty} \int_{x_{n}+\delta}^{x_{n+1}-\delta}\Bigg) \frac{2x \arg \zeta(\frac{1}{2}+ix)}{(\frac{1}{4}+t^2)^2}\mathrm{d}x. \end{equation} Since $\zeta(1/2 + ix)\neq 0$ for $x\in[0, x_{1}-\delta]$ and $x\in[x_{n}+\delta, x_{n+1}-\delta]$ for every positive integer $n$, we can insert the above expression for $2\arg \zeta(1/2 + ix)$ and obtain \begin{equation} I_\delta=\Bigg(\int_{0}^{x_{1}-\delta} + \sum_{n=1}^{\infty} \int_{x_{n}+\delta}^{x_{n+1}-\delta}\Bigg) \frac{x^{2}\log \pi-2x \arg \Gamma(\frac{1}{4}+\frac{ix}{2}) }{(\frac{1}{4}+t^2)^2}\mathrm{d}x \end{equation}where $\arg$ denotes the principal value of the argument, hence

\begin{align} \lim_{\delta \rightarrow 0^+} I_\delta=I=\int_{0}^{\infty} \frac{x^{2}\log \pi}{(\frac{1}{4}+x^2)^2} \mathrm{d}x-\int_{0}^{\infty} \frac{2x \arg \Gamma(\frac{1}{4}+\frac{ix}{2})}{(\frac{1}{4}+x^2)^2} \mathrm{d}x>\frac{1}{2}\pi \log \pi, \end{align} which is incompatible with the RH, according to the Volchkov criterion. $\square$

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

ADDENDUM (20/12/18): This is a reply to @juan and could've been too long as a comment.

The Riemann functional equation states that \begin{equation} \zeta(s)=\chi(s)\zeta(1-s), \end{equation} where $\chi(s)=\pi^{s-\frac{1}{2}}\frac{\Gamma(\frac{1}{2}-\frac{1}{2}s)}{\Gamma(\frac{1}{2}s)}$ and $\Gamma(s)=\int_{0}^{\infty} e^{-x}x^{s-1}\mathrm{d}x$. Let be $s$ be not a zero of $\zeta$, so that taking logarithms gives \begin{equation} \log \zeta(s)-\log \zeta(1-s) = (s-1/2)\log \pi + \log \Gamma\Big(\frac{1}{2}-\frac{1}{2}s\Big)-\log \Gamma \Big(\frac{1}{2}s \Big). \end{equation} Putting $s=1/2 + ix$, where $x\in \mathbb{R}$ and considering the imaginary parts yields \begin{equation} 2\arg \zeta\Big(\frac{1}{2}+ix \Big) = x\log \pi - 2 \arg \Gamma \Big(\frac{1}{4}+\frac{ix}{2} \Big). \end{equation} Define $0<x_1< x_2<x_3< \cdots$ to be the infinitely many positive zeros of $\zeta(z)$ on the line $\Re(z)=1/2$. For every $\delta>0$, consider \begin{equation} F_\delta=\Bigg(\int_{0}^{x_{1}-\delta} + \sum_{n=1}^{\infty} \int_{x_{n}+\delta}^{x_{n+1}-\delta}\Bigg) \frac{2x \arg \zeta(\frac{1}{2}+ix)}{(\frac{1}{4}+t^2)^2}\mathrm{d}x. \end{equation} Since $\zeta(1/2 + ix)\neq 0$ for $x\in[0, x_{1}-\delta]$ and $x\in[x_{n}+\delta, x_{n+1}-\delta]$ for every positive integer $n$, we can insert the above expression for $2\arg \zeta(1/2 + ix)$ and obtain \begin{equation} F_\delta=\Bigg(\int_{0}^{x_{1}-\delta} + \sum_{n=1}^{\infty} \int_{x_{n}+\delta}^{x_{n+1}-\delta}\Bigg) \frac{x^{2}\log \pi-2x \arg \Gamma(\frac{1}{4}+\frac{ix}{2}) }{(\frac{1}{4}+t^2)^2}\mathrm{d}x \end{equation}where $\arg$ denotes the principal value of the argument, hence

\begin{align} \lim_{\delta \rightarrow 0^+} F_\delta=\int_{0}^{\infty} \frac{x^{2}\log \pi}{(\frac{1}{4}+x^2)^2} \mathrm{d}x-\int_{0}^{\infty} \frac{2x \arg \Gamma(\frac{1}{4}+\frac{ix}{2})}{(\frac{1}{4}+x^2)^2} \mathrm{d}x>\frac{1}{2}\pi \log \pi, \end{align} which is incompatible with the RH, according to the Volchkov criterion. $\square$

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

ADDENDUM (20/12/18): This is a reply to @juan and could've been too long as a comment.

The Riemann functional equation states that \begin{equation} \zeta(s)=\chi(s)\zeta(1-s), \end{equation} where $\chi(s)=\pi^{s-\frac{1}{2}}\frac{\Gamma(\frac{1}{2}-\frac{1}{2}s)}{\Gamma(\frac{1}{2}s)}$ and $\Gamma(s)=\int_{0}^{\infty} e^{-x}x^{s-1}\mathrm{d}x$. Let be $s$ be not a zero of $\zeta$, so that taking logarithms gives \begin{equation} \log \zeta(s)-\log \zeta(1-s) = (s-1/2)\log \pi + \log \Gamma\Big(\frac{1}{2}-\frac{1}{2}s\Big)-\log \Gamma \Big(\frac{1}{2}s \Big). \end{equation} Putting $s=1/2 + ix$, where $x\in \mathbb{R}$ and considering the imaginary parts yields \begin{equation} 2\arg \zeta\Big(\frac{1}{2}+ix \Big) = x\log \pi - 2 \arg \Gamma \Big(\frac{1}{4}+\frac{ix}{2} \Big). \end{equation} Define $0<x_1< x_2<x_3< \cdots$ to be the infinitely many positive zeros of $\zeta$ with real part $1/2$. For every $\delta>0$, consider \begin{equation} I_\delta=\Bigg(\int_{0}^{x_{1}-\delta} + \sum_{n=1}^{\infty} \int_{x_{n}+\delta}^{x_{n+1}-\delta}\Bigg) \frac{2x \arg \zeta(\frac{1}{2}+ix)}{(\frac{1}{4}+t^2)^2}\mathrm{d}x. \end{equation} Since $\zeta(1/2 + ix)\neq 0$ for $x\in[0, x_{1}-\delta]$ and $x\in[x_{n}+\delta, x_{n+1}-\delta]$ for every positive integer $n$, we can insert the above expression for $2\arg \zeta(1/2 + ix)$ and obtain \begin{equation} I_\delta=\Bigg(\int_{0}^{x_{1}-\delta} + \sum_{n=1}^{\infty} \int_{x_{n}+\delta}^{x_{n+1}-\delta}\Bigg) \frac{x^{2}\log \pi-2x \arg \Gamma(\frac{1}{4}+\frac{ix}{2}) }{(\frac{1}{4}+t^2)^2}\mathrm{d}x \end{equation}where $\arg$ denotes the principal value of the argument, hence

\begin{align} \lim_{\delta \rightarrow 0^+} I_\delta=I=\int_{0}^{\infty} \frac{x^{2}\log \pi}{(\frac{1}{4}+x^2)^2} \mathrm{d}x-\int_{0}^{\infty} \frac{2x \arg \Gamma(\frac{1}{4}+\frac{ix}{2})}{(\frac{1}{4}+x^2)^2} \mathrm{d}x>\frac{1}{2}\pi \log \pi, \end{align} which is incompatible with the RH, according to the Volchkov criterion. $\square$

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

ADDENDUM (20/12/18): This is a reply to @juan and could've been too long as a comment.

The Riemann functional equation states that \begin{equation} \zeta(s)=\chi(s)\zeta(1-s), \end{equation} where $\chi(s)=\pi^{s-\frac{1}{2}}\frac{\Gamma(\frac{1}{2}-\frac{1}{2}s)}{\Gamma(\frac{1}{2}s)}$ and $\Gamma(s)=\int_{0}^{\infty} e^{-x}x^{s-1}\mathrm{d}x$. Let be $s$ be not a zero of $\zeta$, so that taking logarithms gives \begin{equation} \log \zeta(s)-\log \zeta(1-s) = (s-1/2)\log \pi + \log \Gamma\Big(\frac{1}{2}-\frac{1}{2}s\Big)-\log \Gamma \Big(\frac{1}{2}s \Big). \end{equation} Putting $s=1/2 + ix$, where $x\in \mathbb{R}$ and considering the imaginary parts yields \begin{equation} 2\arg \zeta\Big(\frac{1}{2}+ix \Big) = x\log \pi - 2 \arg \Gamma \Big(\frac{1}{4}+\frac{ix}{2} \Big). \end{equation} Define $0<x_1< x_2<x_3< \cdots$ to be the infinitely many positive zeros of $\zeta(z)$ on the line $\Re(z)=1/2$. For every $\delta>0$, consider \begin{equation} I_\delta=\Bigg(\int_{0}^{x_{1}-\delta} + \sum_{n=1}^{\infty} \int_{x_{n}+\delta}^{x_{n+1}-\delta}\Bigg) \frac{2x \arg \zeta(\frac{1}{2}+ix)}{(\frac{1}{4}+t^2)^2}\mathrm{d}x. \end{equation} Since $\zeta(1/2 + ix)\neq 0$ for $x\in[0, x_{1}-\delta]$ and $x\in[x_{n}+\delta, x_{n+1}-\delta]$ for every positive integer $n$, we can insert the above expression for $2\arg \zeta(1/2 + ix)$ and obtain \begin{equation} I_\delta=\Bigg(\int_{0}^{x_{1}-\delta} + \sum_{n=1}^{\infty} \int_{x_{n}+\delta}^{x_{n+1}-\delta}\Bigg) \frac{x^{2}\log \pi-2x \arg \Gamma(\frac{1}{4}+\frac{ix}{2}) }{(\frac{1}{4}+t^2)^2}\mathrm{d}x \end{equation}where $\arg$ denotes the principal value of the argument, hence

\begin{align} \lim_{\delta \rightarrow 0^+} I_\delta=I=\int_{0}^{\infty} \frac{x^{2}\log \pi}{(\frac{1}{4}+x^2)^2} \mathrm{d}x-\int_{0}^{\infty} \frac{2x \arg \Gamma(\frac{1}{4}+\frac{ix}{2})}{(\frac{1}{4}+x^2)^2} \mathrm{d}x>\frac{1}{2}\pi \log \pi, \end{align} which is incompatible with the RH, according to the Volchkov criterion. $\square$