How to prove MeijerG identity I have an expression which involves Meijer G-functions, to be precise
$$
G_{1,4}^{2,1}\left(x \left\vert
 \begin{array}{c}
  1 \\
  1,1,0,\frac{1}{2} \\
 \end{array}
 \right)\right.
$$
After messing around a bit with the Mathematica software I found that the following appears to hold:
$$
G_{1,4}^{2,1}\left(x \left\vert
 \begin{array}{c}
  1 \\
  1,1,0,\frac{1}{2} \\
 \end{array}
 \right)\right. -\frac{1}{\sqrt{\pi}} =  - G_{0,3}^{2,0}\left(x \left \vert
  \begin{array}{c}
   0,1,\frac{1}{2} \\
  \end{array}
  \right)\right.
$$
With "appears to hold" I mean the plots overlay perfectly. But I would like to prove it. I know that you can lower the order of the G function if the same parameters appear in the $a$ and $b$ vectors. But this only applies if the $1$ in $b$ are in its second half. So in my example it does not cancel. Any suggestions on how to proceed are highly appreciated.
PS: Input for Mathematica
Plot[MeijerG[{{1}, {}}, {{0, 1}, {1/2, 1}}, x] - 1/Sqrt[Pi] + MeijerG[{{1}, {}}, {{1, 1}, {0, 1/2}}, x], {x, 0, 4}, PlotRange -> All]
Which just gives machine precision noise.
 A: From the definition,
\begin{align}
 G_{1,4}^{2,1}\left(x \left\vert
 \begin{array}{c}
  1 \\
  1,1,0,\frac{1}{2} \\
 \end{array}
 \right)\right.&=\frac{1}{2i\pi}\int_{\mathcal L}\frac{[\Gamma(1+s)]^2\Gamma(-s)}{\Gamma(1-s)\Gamma(1/2-s)}x^{-s}\,ds
\end{align}
Here, the parameters of the function corresponds to the case (i) of the of the definition of the Meijer function given above. The contour $\mathcal L$ can be a straight line $(\gamma-i\infty,\gamma+i\infty)$ with $-1<\gamma<0$.
Using the functional relation of the Gamma function, simplifications of the integrand leads to the alternative expression
\begin{equation}
 G_{1,4}^{2,1}\left(x \left\vert
 \begin{array}{c}
  1 \\
  1,1,0,\frac{1}{2} \\
 \end{array}
 \right)\right.
 =-\frac{1}{2i\pi}\int_{\mathcal L}\frac{\Gamma(1+s)\Gamma(s)}{\Gamma(1/2-s)}x^{-s}\,ds
\end{equation}
Under the simplified form the integrand could correspond to another Meijer function, but the contour must be checked carefully. Indeed,
\begin{equation}
  G_{0,3}^{2,0}\left(x \left \vert
  \begin{array}{c}
   0,1,\frac{1}{2} \\
  \end{array}
  \right)\right.=\frac{1}{2i\pi}\int_{\mathcal L'}\frac{\Gamma(1+s)\Gamma(s)}{\Gamma(1/2-s)}x^{-s}\,ds
\end{equation}
where, $\mathcal L'$ can be the straight line  $(\gamma'-i\infty,\gamma'+i\infty)$, but here $0<\gamma'<1$. From the residue theorem, we can shift the contour to make it correspond to $\mathcal L$ by including the residue at $s=0$:
\begin{equation}
  G_{0,3}^{2,0}\left(x \left \vert
  \begin{array}{c}
   0,1,\frac{1}{2} \\
  \end{array}
  \right)\right.=\frac{1}{2i\pi}\int_{\mathcal L}\frac{\Gamma(1+s)\Gamma(s)}{\Gamma(1/2-s)}x^{-s}\,ds+\frac{1}{\sqrt{\pi}}
\end{equation}
Then,
\begin{equation}
 G_{1,4}^{2,1}\left(x \left\vert
 \begin{array}{c}
  1 \\
  1,1,0,\frac{1}{2} \\
 \end{array}
 \right)\right.= -G_{0,3}^{2,0}\left(x \left \vert
  \begin{array}{c}
   0,1,\frac{1}{2} \\
  \end{array}
  \right)\right.+\frac{1}{\sqrt{\pi}}
\end{equation}
as expected.
