I've run across several integral representations of $H_z$ such as the following (see Harmonic Number Integral Representations).
(1) $\quad H_z=\int\limits_0^1 \frac{1-t^z}{1-t} \, dt\,,\qquad\qquad\qquad\qquad\Re(z)>-1$
(2) $\quad H_z=\int\limits_0^\infty\left(\frac{e^{-t}}{t}-\frac{(t+1)^{-z-1}}{t}\right)\,dt+\gamma\,,\quad\Re(z)>-1$
(3) $\quad H_z=\int\limits_0^\infty\frac{e^{-t}-e^{-(z+1)\,t}}{1-e^{-t}}\,dt\,,\quad\qquad\qquad\quad\Re(z)>-1$
(4) $\quad H_z=\int\limits_0^1\left(\frac{t^z}{t-1}-\frac{1}{\log (t)}\right)\,dt+\gamma\,,\quad\quad\Re(z)>-1$
I believe the following integral representation of $H_z$ is also valid but can't seem to translate it to any of the integral representations illustrated in (1) to (4) above.
(5) $\quad H_z=-(z+1)\int\limits_0^\infty\frac{\log(t)}{(t+1)^{z+2}}\,dt\,,\qquad\quad\Re(z)>-1$
The digamma function $\psi(z)$ defined in (6) below is related to $H_z$ as illustrated in (7) below.
(6) $\quad\psi(z)=\frac{\partial\,\log(\Gamma(z))}{\partial\,z}=\frac{\Gamma'(z)}{\Gamma(z)}$
(7) $\quad H_z=\psi(z+1)+\gamma$
The integral representations of $\psi(z)$ illustrated in (8) to (12) below (see PolyGamma Integral Representations) are related to the integral representations of $H_z$ illustrated in (1) to (4) above.
(8) $\quad\psi(z)=\int\limits_0^1\frac{1-t^{z-1}}{1-t}\,dt-\gamma\,,\qquad\quad\Re(z)>0$
(9) $\quad\psi(z)=\int\limits_0^\infty\left(\frac{e^{-t}}{t}-\frac{(t+1)^{-z}}{t}\right)\,dt\,,\quad\Re(z)>0$
(10) $\quad\psi(z)=\int\limits_0^\infty\frac{e^{-t}-e^{-z\,t}}{1-e^{-t}}\,dt-\gamma\,,\qquad\Re(z)>0$
(11) $\quad\psi(z)=\int\limits_0^1\left(\frac{t^{z-1}}{t-1}-\frac{1}{\log (t)}\right)\,dt\,,\quad\Re(z)>0$
The following integral representation of $\psi(z)$ follows from (5) and (7) above but again I can't seem to translate it to any of the integral representations illustrated in (8) to (11) above.
(12) $\quad\psi(z)=-z\int_0^\infty\frac{\log(t)}{(t+1)^{z+1}}\,dt-\gamma\,,\quad \Re(z)>0$
Question: What is the proof of the integral representations of $H_z$ and $\psi(z)$ illustrated in (5) and (12) above?
Best Answer
For $z \in \mathbb{C}$ with $\Re (z) > -1$ let $f(z) = (z+1) \int_0^\infty \frac{-\log(t)}{(1+t)^{z+2}} \, \mathrm{d} t$. Then $f(0) = 0$, so $$ f(z) = f(z) - f(0) = \int \limits_0^\infty (-\log(t)) \left[\frac{z+1}{(1+t)^{z+2}} - \frac{1}{(1+t)^2}\right] \, \mathrm{d} t \, .$$ Now we can integrate by parts and let $s = \frac{1}{1+t}$ to find $$ f(z) = \int \limits_0^\infty \frac{1}{t(1+t)} \left[1 - \frac{1}{(1+t)^z}\right] \, \mathrm{d} z = \int \limits_0^1 \frac{s}{\frac{1}{s} - 1} (1 - s^z) \, \frac{\mathrm{d} s}{s^2} = \int \limits_0^1 \frac{1 - s^z}{1-s} \, \mathrm{d} s = H_z$$ from your first identity.
Alternatively, we can prove the desired result directly using differentiation under the integral sign: \begin{align} f(z) &= - (z+1) \frac{\mathrm{d}}{\mathrm{d} u} \int \limits_0^\infty \frac{t^{u-1}}{(1+t)^{z+2}} \, \mathrm{d} t \, \Bigg\vert_{u=1} \stackrel{(1+t)^{-1} = s}{=} - (z+1) \frac{\mathrm{d}}{\mathrm{d} u} \int \limits_0^1 s^{z+1-u} (1-s)^{u-1} \, \mathrm{d} s \, \Bigg\vert_{u=1}\\ &= - (z+1) \frac{\mathrm{d}}{\mathrm{d} u} \frac{\Gamma(z+2-u)\Gamma(u)}{\Gamma(z+2)} \, \Bigg\vert_{u=1} = \frac{(z+1)\Gamma(z+1) \Gamma(1)}{\Gamma(z+2)} [\psi (z+1) - \psi(1)] \\ &= \psi(z+1) + \gamma = H_z \, . \end{align}