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Question-194236

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Question Number 194236 by MrGHK last updated on 01/Jul/23
Answered by witcher3 last updated on 03/Jul/23
S=Σ_(n≥0) (((−1)^n )/(n!(zn+1)k^n )) =Σ_(n≥0) ∫(((−1)/k))^n .(1/(n!))∫_0 ^1 t^(zn) dt =∫_0 ^1 Σ_(n≥0) (−(t^z /k))^n .(1/(n!))dt =∫_0 ^1 e^(−(t^z /k)) dt t=(kw)^(1/z) ⇒dt=(1/z)k^(1/z) .w^((1/z)−1) dw =(k^(1/z) /z)∫_0 ^(1/k) e^(−w) w^((1/z)−1) dw Γ(a,z)=∫_z ^∞ t^(a−1) e^(−t) dt..complet Gamma function S.zk^(−(1/z)) =∫_0 ^∞ e^(−w) w^((1/z)−1) dw−∫_(1/k) ^∞ e^(−w) w^((1/z)−1) dw =Γ((1/z))−Γ((1/z),(1/k)) S=(k^(1/z) /z)(Γ((1/z))−Γ((1/z),1))
$$\mathrm{S}=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}!\left(\mathrm{zn}+\mathrm{1}\right)\mathrm{k}^{\mathrm{n}} } \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\int\left(\frac{−\mathrm{1}}{\mathrm{k}}\right)^{\mathrm{n}} .\frac{\mathrm{1}}{\mathrm{n}!}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{t}^{\mathrm{zn}} \mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(−\frac{\mathrm{t}^{\mathrm{z}} }{\mathrm{k}}\right)^{\mathrm{n}} .\frac{\mathrm{1}}{\mathrm{n}!}\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\frac{\mathrm{t}^{\mathrm{z}} }{\mathrm{k}}} \mathrm{dt} \\ $$$$\mathrm{t}=\left(\mathrm{kw}\right)^{\frac{\mathrm{1}}{\mathrm{z}}} \Rightarrow\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{z}}\mathrm{k}^{\frac{\mathrm{1}}{\mathrm{z}}} .\mathrm{w}^{\frac{\mathrm{1}}{\mathrm{z}}−\mathrm{1}} \mathrm{dw} \\ $$$$=\frac{\mathrm{k}^{\frac{\mathrm{1}}{\mathrm{z}}} }{\mathrm{z}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{k}}} \mathrm{e}^{−\mathrm{w}} \mathrm{w}^{\frac{\mathrm{1}}{\mathrm{z}}−\mathrm{1}} \mathrm{dw} \\ $$$$\Gamma\left(\mathrm{a},\mathrm{z}\right)=\int_{\mathrm{z}} ^{\infty} \mathrm{t}^{\mathrm{a}−\mathrm{1}} \mathrm{e}^{−\mathrm{t}} \mathrm{dt}..\mathrm{complet}\:\mathrm{Gamma}\:\mathrm{function} \\ $$$$\mathrm{S}.\mathrm{zk}^{−\frac{\mathrm{1}}{\mathrm{z}}} =\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{w}} \mathrm{w}^{\frac{\mathrm{1}}{\mathrm{z}}−\mathrm{1}} \mathrm{dw}−\int_{\frac{\mathrm{1}}{\mathrm{k}}} ^{\infty} \mathrm{e}^{−\mathrm{w}} \mathrm{w}^{\frac{\mathrm{1}}{\mathrm{z}}−\mathrm{1}} \mathrm{dw} \\ $$$$=\Gamma\left(\frac{\mathrm{1}}{\mathrm{z}}\right)−\Gamma\left(\frac{\mathrm{1}}{\mathrm{z}},\frac{\mathrm{1}}{\mathrm{k}}\right) \\ $$$$\mathrm{S}=\frac{\mathrm{k}^{\frac{\mathrm{1}}{\mathrm{z}}} }{\mathrm{z}}\left(\Gamma\left(\frac{\mathrm{1}}{\mathrm{z}}\right)−\Gamma\left(\frac{\mathrm{1}}{\mathrm{z}},\mathrm{1}\right)\right) \\ $$

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