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I am trying to find some relevant functional identities involving Meijer G-functions in order to prove

$$ \int_0^\infty\frac{\log(x+1)}{x}\mathrm{e}^{-zx}\,\mathrm{d}x = G^{3,1}_{2,3}\left(z \middle| \begin{array}{c} 0,1 \\ 0,0,0 \\ \end{array} \right), \quad (z>0). $$

This equality came out of Mathematica, in whose syntax the right-hand side reads MeijerG[{{0}, {1}}, {{0, 0, 0}, {}}, z].

In addition, it would be very nice to further express this Meijer G-function in terms of some simpler functions, such as hypergeometric functions (which could then be expressed as an infinite series). What I really need are proof hints rather than Mathematica magic, as I am not too familiar with complex analysis integration techniques.

NB: This question is related to an earlier question of mine: Closed form of $\int_0^\infty \frac{\log(x)-\log(a)}{x-a}e^{-x} \mathrm{d}x$. to which @Jason has given a valid answer

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First express the logarithm as a G-function, namely, \begin{align} \ln(1+x) = \large{G}_{2,2}^{1,2}\left( x \left| \begin{array}{cc} 1,1 \\ 1,0 \end{array} \right. \right). \end{align} Now the integral in question can be seen by the slightly more general integral \begin{align} I = \int_{0}^{\infty} t^{-\alpha -1} \ \ln(1+t) \ e^{-st} \ dt \end{align} which is given by \begin{align} I= \int_{0}^{\infty} e^{-st} \ t^{-\alpha -1} \ \large{G}_{2,2}^{1,2}\left( t \left| \begin{array}{cc} 1,1 \\ 1,0 \end{array} \right. \right) \ dt. \end{align} This last integral is seen to be the Laplace transform of the G-function and yields \begin{align} I &= s^{\alpha} \large{G}_{3,2}^{1,3}\left( \frac{1}{s} \left| \begin{array}{cc} \alpha + 1,1,1 \\ 1,0 \end{array} \right. \right) \\ &= s^{\alpha} \large{G}_{2,3}^{3,1}\left( s \left| \begin{array}{cc} 0,1 \\ -\alpha, 0,0 \end{array} \right. \right) \\ &= s^{\alpha} \large{G}_{1,2}^{3,0}\left( s \left| \begin{array}{cc} 1 \\ -\alpha, 0 \end{array} \right. \right) \\ \end{align} Hence, \begin{align} \int_{0}^{\infty} t^{-\alpha -1} \ \ln(1+t) \ e^{-st} \ dt = s^{\alpha} \large{G}_{1,2}^{3,0} \left( s \left| \begin{array}{cc} 1 \\ -\alpha, 0 \end{array} \right. \right). \end{align}

It is of note that \begin{align} \large{G}_{1,2}^{3,0}\left( s \left| \begin{array}{cc} 1 \\ -\alpha, 0 \end{array} \right. \right) = \large{G}_{2,2}^{3,1}\left( s \left| \begin{array}{cc} 0,1 \\ -\alpha, 0,0 \end{array} \right. \right) \end{align}

Many of the transformation formulas used above can be found in the set given in: http://en.wikipedia.org/wiki/Meijer_G-function .

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