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Question Number 194236 by MrGHK last updated on 01/Jul/23

Answered by witcher3 last updated on 03/Jul/23

$$\mathrm{S}=\sum _{\mathrm{n}⩾0}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}!(\mathrm{zn}+1){\mathrm{k}}^{\mathrm{n}}}$$$$=\sum _{\mathrm{n}⩾0}\int {\left(\frac{-1}{\mathrm{k}}\right)}^{\mathrm{n}}.\frac{1}{\mathrm{n}!}{\int }_0^1{\mathrm{t}}^{\mathrm{zn}}\mathrm{dt}$$$$={\int }_0^1\sum _{\mathrm{n}⩾0}{(-\frac{{\mathrm{t}}^{\mathrm{z}}}{\mathrm{k}})}^{\mathrm{n}}.\frac{1}{\mathrm{n}!}\mathrm{dt}$$$$={\int }_0^1{\mathrm{e}}^{-\frac{{\mathrm{t}}^{\mathrm{z}}}{\mathrm{k}}}\mathrm{dt}$$$$\mathrm{t}={\left(\mathrm{kw}\right)}^{\frac{1}{\mathrm{z}}}\Rightarrow \mathrm{dt}=\frac{1}{\mathrm{z}}{\mathrm{k}}^{\frac{1}{\mathrm{z}}}.{\mathrm{w}}^{\frac{1}{\mathrm{z}}-1}\mathrm{dw}$$$$=\frac{{\mathrm{k}}^{\frac{1}{\mathrm{z}}}}{\mathrm{z}}{\int }_0^{\frac{1}{\mathrm{k}}}{\mathrm{e}}^{-\mathrm{w}}{\mathrm{w}}^{\frac{1}{\mathrm{z}}-1}\mathrm{dw}$$$$\mathrm{\Gamma }(\mathrm{a},\mathrm{z})={\int }_{\mathrm{z}}^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{a}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}..\mathrm{complet}\mathrm{Gamma}\mathrm{function}$$$$\mathrm{S}.{\mathrm{zk}}^{-\frac{1}{\mathrm{z}}}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{w}}{\mathrm{w}}^{\frac{1}{\mathrm{z}}-1}\mathrm{dw}-{\int }_{\frac{1}{\mathrm{k}}}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{w}}{\mathrm{w}}^{\frac{1}{\mathrm{z}}-1}\mathrm{dw}$$$$=\mathrm{\Gamma }\left(\frac{1}{\mathrm{z}}\right)-\mathrm{\Gamma }(\frac{1}{\mathrm{z}},\frac{1}{\mathrm{k}})$$$$\mathrm{S}=\frac{{\mathrm{k}}^{\frac{1}{\mathrm{z}}}}{\mathrm{z}}(\mathrm{\Gamma }\left(\frac{1}{\mathrm{z}}\right)-\mathrm{\Gamma }(\frac{1}{\mathrm{z}},1))$$

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