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Question Number 195436 by MrGHK last updated on 02/Aug/23

$$\underset{\mathbf{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-\mathbf{a})}^{-\mathbf{n}}}{{(\mathbf{n}+2)}^2}(\mathit{\psi }\left(\frac{\mathbf{n}+4}{2}\right)-\mathit{\psi }\left(\frac{n+3}{2}\right))=???$$

Answered by witcher3 last updated on 02/Aug/23

$$\mathrm{\Psi }(1+\mathrm{x})=-\gamma +{\int }_0^1\frac{1-{\mathrm{x}}^{\mathrm{s}}}{1-\mathrm{x}}\mathrm{dx}$$$$\mathrm{S}=\sum _{\mathrm{n}⩾0}\frac{{(-\mathrm{a})}^{-\mathbf{n}}}{{(\mathbf{n}+2)}^2}(\mathbf{\Psi }\left(\frac{\mathrm{n}+4}{2}\right)-\mathrm{\Psi }\left(\frac{\mathrm{n}+3}{2}\right))$$$$=\mathrm{\Sigma }\frac{{(-\mathrm{a})}^{-\mathrm{n}}}{(\mathrm{n}+2)}({\int }_0^1\frac{{\mathrm{x}}^{\frac{\mathrm{n}+1}{2}}-{\mathrm{x}}^{\frac{\mathrm{n}+2}{2}}}{1-\mathrm{x}}\mathrm{dx}$$$$={\mathrm{a}}^2\sum _{\mathrm{n}⩾0}\frac{{(-\frac{1}{\mathrm{a}})}^{\mathrm{n}+2}}{{(\mathrm{n}+2)}^2}{\int }_0^1\frac{{\mathrm{x}}^{\frac{\mathrm{n}+1}{2}}}{1+\sqrt{\mathrm{x}}}\mathrm{dx}$$$$={\mathrm{a}}^2\sum _{\mathrm{n}⩾0}(\frac{{(-\frac{1}{\mathrm{a}})}^{\mathrm{n}+2}}{{(\mathrm{n}+2)}^2}{\int }_0^1\frac{{\mathrm{x}}^{\frac{\mathrm{n}+2}{2}}}{\sqrt{\mathrm{x}}(1+\sqrt{\mathrm{x}})}\mathrm{dx},\sqrt{\mathrm{x}}=\mathrm{y},\mathrm{\forall }\mid \mathrm{a}\mid ⩾1$$$$={2\mathrm{a}}^2{\int }_0^1\sum _{\mathrm{n}⩾0}\frac{{(-\frac{\mathrm{y}}{\mathrm{a}})}^{\mathrm{n}+2}}{{(\mathrm{n}+2)}^2}.\frac{\mathrm{dy}}{1+\mathrm{y}}$$$$\sum _{\mathrm{n}⩾1}\frac{{\mathrm{x}}^{\mathrm{n}}}{{\mathrm{n}}^2}={\mathrm{Li}}_2\left(\mathrm{x}\right)$$$$={2\mathrm{a}}^2{\int }_0^1\frac{{\mathrm{Li}}_2(-\frac{\mathrm{y}}{\mathrm{a}})+\frac{\mathrm{y}}{\mathrm{a}}}{1+\mathrm{y}}\mathrm{dy}$$$$={2\mathrm{a}}^2{\int }_0^1\frac{{\mathrm{Li}}_2(-\frac{\mathrm{y}}{\mathrm{a}})}{1+\mathrm{y}}\mathrm{dy}+2\mathrm{a}{\int }_0^1\frac{\mathrm{y}}{1+\mathrm{y}}\mathrm{dy}$$$$={2\mathrm{a}}^3{\int }_0^{\frac{1}{\mathrm{a}}}\frac{{\mathrm{Li}}_2(-\mathrm{y})}{1+\mathrm{ay}}...\mathrm{verry}\mathrm{long}\mathrm{calculate}$$$$$$

Commented by MathedUp last updated on 02/Aug/23

$$\mathrm{Wow}\mathrm{You}\mathrm{R}\mathrm{Hero}\mathrm{LOL}$$$$\mathrm{u}\mathrm{always}\mathrm{come}\mathrm{out}\mathrm{everywhere}:)$$

Commented by MrGHK last updated on 03/Aug/23

$$\underset{\mathbf{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-\mathbf{a})}^{-\mathbf{n}}}{{(\mathbf{n}+2)}^2}(\mathit{\psi }\left(\frac{\mathbf{n}+4}{2}\right)-\mathit{\psi }\left(\frac{n+3}{2}\right))$$$$\mathbf{Y}ouarereallyabrilliantman$$

Commented by witcher3 last updated on 04/Aug/23

$$\mathrm{thak}\mathrm{You}\mathrm{god}\mathrm{bless}\mathrm{You}\mathrm{hard}[\mathrm{worck}\mathrm{and}\mathrm{passion}$$

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