DERIVATIVES OF GAMMA FUNCTION @ 1/2

             We have previously showed the values of the derivatives of the Gamma function at 1. Today our goal is to calculate the values of the derivatives of the Gamma function at  \frac12 , in other words, we will evaluate the following integrals


\begin{aligned} &\int_0^{\infty} \frac{e^{-z}\ln z}{\sqrt{z}} dz=-\sqrt{\pi}\left(\gamma +\ln 4 \right) \\ & \\ &\int_0^{\infty} \frac{e^{-z}\ln^2 z}{\sqrt{z}} dz=\frac{\sqrt{\pi}}{2}\left(\pi^2+2\left(\gamma+ \ln 4 \right)^2 \right)\\ & \\ &\int_0^{\infty} \frac{e^{-z}\ln^3 z}{\sqrt{z}} dz=-\frac{\sqrt{\pi}}{2}\left(28\zeta(3)+2\gamma^3+2\ln^3(4)+3\gamma(\pi^2+2\ln 4)+6\gamma^2\ln 4+\pi^2\ln 64 \right)\\ & \\ &\int_0^{\infty} \frac{e^{-z}\ln^4 z}{\sqrt{z}} dz=\frac{\sqrt{\pi}}{4}\left(224 \zeta(3)(\gamma+\ln 4)+4 \gamma^{4}+7 \pi^{4}+4 \ln ^{4} 4 +16 \gamma \ln ^{3}(4)+12 \pi^{2} \ln ^{2} 4+12 \gamma^{2}\left(\pi^{2}+2 \ln ^{2} 4\right)+16 \gamma^{3} \ln 4+8 \gamma \pi^{2} \ln 64\right) \end{aligned}

We will follow the same steps from our previous post. Lets start from the integral representation of the Gamma function and take it´s derivatives


\Gamma(z)=\int_0^\infty e^{-x}x^{z-1} \,dx(1)

Differentiating (1) w.r. to z

\begin{aligned} \int_0^\infty e^{-x}x^{z-1}\ln x \,dx&=\Gamma^{\prime}(z)\\ &=\Gamma(z)\,\psi(z) \end{aligned}

Setting z=\frac{1}{2} we obtain

\begin{aligned} \int_0^\infty \frac{e^{-x}\ln x}{\sqrt{x}} \,dx&=\Gamma\left(\frac{1}{2}\right)\,\psi\left(\frac{1}{2}\right)\\ &=\sqrt{\pi}\left(-\gamma-2 \ln 2 \right)\\ &=-\sqrt{\pi}\left(\gamma +\ln 4 \right) \qquad \blacksquare\\ \end{aligned}

Where the values of the Digamma function were proved in this post.

Differentiating twice  (1)  w.r. to z

\begin{aligned} \int_0^\infty e^{-x}x^{z-1}\ln^2 x \,dx&=\Gamma(z)\left(\psi^{\prime}(z)+\psi^2(z) \right)\\ \end{aligned}

Setting z=\frac{1}{2} we obtain

\begin{aligned} \Gamma^{\prime \prime}\left(\frac{1}{2}\right)=\int_0^\infty \frac{e^{-z}\ln^2 z}{\sqrt{x}} \,dx&=\Gamma(z)\left(\psi^{\prime}(z)+\psi^2(z) \right)\\ &=\sqrt{\pi}\left(3\zeta(2)+\left(\gamma+2 \ln 2 \right)^2 \right)\\ &=\frac{\sqrt{\pi}}{2}\left(\pi^2+2\left(\gamma+ \ln 4 \right)^2 \right) \qquad \blacksquare\\ \end{aligned}

Differentiating  (1) three times w.r. to z


\int_0^{\infty} x^{z-1}e^{-x}\ln^3 x dx=\Gamma^{\prime\prime\prime}(z)=\Gamma(z)\psi^3(z)+3\Gamma(z)\psi(z)\psi^{\prime}(z)+\Gamma(z)\psi^{\prime\prime}(z)

Setting z=\frac{1}{2} we obtain


\begin{aligned} \int_0^{\infty} \frac{e^{-z}\ln^3 z}{\sqrt{z}} dz&=\Gamma\left(\frac{1}{2}\right)\psi^3\left(\frac{1}{2}\right)+3\Gamma\left(\frac{1}{2}\right)\psi\left(\frac{1}{2}\right)\psi^{\prime}\left(\frac{1}{2}\right)+\Gamma\left(\frac{1}{2}\right)\psi^{\prime\prime}\left(\frac{1}{2}\right)\\ &=-\frac{\sqrt{\pi}}{2}\left(2\left(\gamma+2 \ln 2 \right)^3+18\left(\gamma+2 \ln 2 \right)\zeta(2)+28\zeta(3) \right)\\ &=-\frac{\sqrt{\pi}}{2}\left(28\zeta(3)+2\gamma^3+2\ln^3(4)+3\gamma(\pi^2+2\ln 4)+6\gamma^2\ln 4+\pi^2\ln 64 \right) \qquad \blacksquare\\ \end{aligned}

Differentiating (1) four times w.r. to z


\int_0^{\infty} x^{z-1}e^{-x}\ln^4 x dx=\Gamma^{(4)}\left( z\right)=\Gamma(z)\psi^4(z)+6\Gamma(z)\psi^{2}(z)\psi^{\prime}(z)+3\Gamma(z)\psi^{\prime}(z)^2+4\Gamma(z)\psi(z)\psi^{\prime\prime}(z)+\Gamma(z)\psi^{\prime\prime \prime}(z)

Setting z=\frac{1}{2} we obtain


\begin{aligned} \int_0^{\infty} \frac{e^{-z}\ln^4 z}{\sqrt{z}} dz&=\Gamma\left(\frac{1}{2}\right)\psi^4\left(\frac{1}{2}\right)+6\Gamma\left(\frac{1}{2}\right)\psi^{2}\left(\frac{1}{2}\right)\psi^{\prime}\left(\frac{1}{2}\right)+3\Gamma\left(\frac{1}{2}\right)\psi^{\prime}\left(\frac{1}{2}\right)^2+4\Gamma\left(\frac{1}{2}\right)\psi\left(\frac{1}{2}\right)\psi^{\prime\prime}\left(\frac{1}{2}\right)+\Gamma\left(\frac{1}{2}\right)\psi^{\prime\prime \prime}\left(\frac{1}{2}\right)\\ &=\sqrt{\pi}\left(\left(\gamma +\ln 4 \right)^4+18\zeta(2)\left(\gamma +\ln 4 \right)^2 +\frac{3\pi^2}{4}+56\zeta(3)\left(\gamma +\ln 4 \right)+90\zeta(4)\right)\\ &=\frac{\sqrt{\pi}}{4}\left(224 \zeta(3)(\gamma+\ln 4)+4 \gamma^{4}+7 \pi^{4}+4 \ln ^{4} 4 +16 \gamma \ln ^{3}(4)+12 \pi^{2} \ln ^{2} 4+12 \gamma^{2}\left(\pi^{2}+2 \ln ^{2} 4\right)+16 \gamma^{3} \ln 4+8 \gamma \pi^{2} \ln 64\right)\qquad \blacksquare\\ \end{aligned}


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