Reduce $\frac{d^{n-1}}{dw^{n-1}}\frac{4^{-n/{\sqrt w}}}{\sqrt w}\Big|_1=\frac1{\sqrt\pi}G^{3,0}_{1,3}\left(^{3/2-n}_{0,1/2,1/2};(\ln(2)n)^2\right)$

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answer to Is there any valid complex or just real solution to $\sin(x)^{\cos(x)} = 2$?,

one must calculate $$\frac{d^{n-1}}{dw^{n-1}}\left.\frac{4^{-\frac{n}{\sqrt w}}}{\sqrt w}\right|_1=\sum_{m=0}^\infty\frac{\Gamma\left(\frac12-\frac m2\right)(-\ln(4)n)^m}{\Gamma\left(\frac32-n-\frac m2\right)m!}=\text H^{1,1}_{1,2}\left(^{\left(\frac12,-\frac12\right)}_{(0,1),\left(n-\frac12,-\frac12\right)};\ln(4)n\right)=\frac1{\sqrt\pi}\text G^{3,0}_{1,3}\left(^{\frac32-n}_{0,\frac12,\frac12};(\ln(2)n)^2\right)\tag1$$

where one can convert Fox H into Meijer G functions using the Wolfram repository function FoxHToMeijerG. @Mariusz Iwaniuk simplified it further using Maple:

$$\frac{d^{n-1}}{dw^{n-1}}\left.\frac{4^{-\frac{n}{\sqrt w}}}{\sqrt w}\right|_1 = \frac1{\sqrt\pi}\text G^{3,0}_{1,3}\left(^{\frac32-n}_{0,\frac12,\frac12};(\ln(2)n)^2\right) \\ =\frac{\sqrt\pi}{\Gamma\left(\frac32-n\right)}\,_1\text F_2\left(n-\frac12;\frac12,\frac12;(\ln(2)n)^2\right)\\ +\ln(4)(-1)^n n!\,_1\text F_2\left(n;1,\frac32;(\ln(2)n)^2\right)\tag2$$

which was tested and it matches the derivatives.

However, there is seemingly no other way to find this result. MeijerGToHypergeometricPFQ did not work. Also, there is a formula for converting $\text G^{3,0}_{1,3}\left(^{\ \ \ \ \ a_1}_{b_1,b_2,b_3};z\right)$ into a sum of $_1\text F_2$ functions, but part of it involves $\csc(\pi (b_2-b_3))$ which is undefined if $b_3=b_2$, like in $(1)$, so $\lim\limits_{b_2\to\frac12}$ must be taken. This problem occurs in other cases where $b_{m+1}=b_{m+j}$, so understanding how to reduce Meijer G in these cases helps.

Is there any way to find $(2)$ without using Maple, like maybe with a Wolfram function or a reduction formula?

Answer 1

Using Mathematica: $$\underset{w\to 1}{\text{lim}}\frac{\partial ^{n-1}}{\partial w^{n-1}}\frac{\exp \left(-\frac{n \log (4)}{\sqrt{w}}\right)}{\sqrt{w}}=\\\frac{G_{1,3}^{3,0}\left(n \log (2),\frac{1}{2}| \begin{array}{c} \frac{3}{2}-n \\ 0,\frac{1}{2},\frac{1}{2} \\ \end{array} \right)}{\sqrt{\pi }}=\\\mathcal{M}_s^{-1}\left[\frac{n^{-s} \Gamma \left(\frac{1}{2}+\frac{s}{2}\right) \Gamma (s) (2 \log (2))^{-s}}{\Gamma \left(\frac{3}{2}-n+\frac{s}{2}\right)}\right](1)=\\\mathcal{M}_s^{-1}\left[\frac{n^{-s} \left(\int_0^{\infty } e^{-x} x^{-\frac{1}{2}+\frac{s}{2}} \, dx\right) \Gamma (s) (2 \log (2))^{-s}}{\Gamma \left(\frac{3}{2}-n+\frac{s}{2}\right)}\right](1)=\\\int_0^{\infty } \mathcal{M}_s^{-1}\left[\frac{n^{-s} e^{-x} x^{-\frac{1}{2}+\frac{s}{2}} \Gamma (s) (2 \log (2))^{-s}}{\Gamma \left(\frac{3}{2}-n+\frac{s}{2}\right)}\right](1) \, dx=\int_0^{\infty } e^{-x} \left(\frac{\, _1F_1\left(-\frac{1}{2}+n;\frac{1}{2};-\frac{n^2 \log ^2(2)}{x}\right)}{\sqrt{x} \Gamma \left(\frac{3}{2}-n\right)}-\frac{n \, _1F_1\left(n;\frac{3}{2};-\frac{n^2 \log ^2(2)}{x}\right) \log (4)}{x \Gamma (1-n)}\right) \, dx=-\frac{\cos (n \pi ) \Gamma \left(-\frac{1}{2}+n\right) \, _1F_2\left(-\frac{1}{2}+n;\frac{1}{2},\frac{1}{2};n^2 \log ^2(2)\right)}{\sqrt{\pi }}+\cos (n \pi ) \Gamma (1+n) \, _1F_2\left(n;1,\frac{3}{2};n^2 \log ^2(2)\right) \log (4)-\frac{n \sqrt{\pi } \log (4) G_{1,3}^{2,1}\left(n^2 \log ^2(2)| \begin{array}{c} 1-n \\ 0,0,-\frac{1}{2} \\ \end{array} \right)}{2 \Gamma (1-n) \Gamma (n)}=\\-\frac{\cos (n \pi ) \Gamma \left(-\frac{1}{2}+n\right) \, _1F_2\left(-\frac{1}{2}+n;\frac{1}{2},\frac{1}{2};n^2 \log ^2(2)\right)}{\sqrt{\pi }}+\cos (n \pi ) \Gamma (1+n) \, _1F_2\left(n;1,\frac{3}{2};n^2 \log ^2(2)\right) \log (4)$$

The last term: $\frac{\sqrt{\pi } n \log (4) G_{1,3}^{2,1}\left(n^2 \log ^2(2)| \begin{array}{c} 1-n \\ 0,0,-\frac{1}{2} \\ \end{array} \right)}{2 \Gamma (1-n) \Gamma (n)}=0$ for: $n\in \mathbb{Z}\land n>0$