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Question-124116




Question Number 124116 by danielasebhofoh last updated on 30/Nov/20
Answered by Dwaipayan Shikari last updated on 01/Dec/20
∫x^(−x) dx  =Σ_(n≥0) ^∞ (−1)^n ∫(((xlogx)^n )/(n!))dx  =Σ_(n≥0) ^∞ (−1)^n (1/(n!))∫x^n log^n x d3  =Σ_(n≥0) ^∞ (−1)^n (1/(n!))∫e^((n+1)t) t^n dt       logx=t  =Σ_(n≥0) ^∞ (−1)^n (1/(n!(n+1)^(n+1) ))∫e^u u^n du     (n+1)t=u  =Σ_(n≥0) ^∞ (−1)^n (1/(n!(n+1)^(n+1) ))Σ_(k≥0) ^∞ ∫(u^(n+k) /(k!))du  =Σ_(n≥0) ^∞ (−1)^n (1/(n!(n+1)^(n+1) ))Σ_(k≥0) ^∞ (u^(n+k+1) /(k!(n+k+1)))  =Σ_(n≥0) ^∞ Σ_(k≥0) ^∞ (((−1)^n u^(n+k+1) )/(n!k!(n+1)(n+k+1)))
$$\int{x}^{−{x}} {dx} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int\frac{\left({xlogx}\right)^{{n}} }{{n}!}{dx} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{1}}{{n}!}\int{x}^{{n}} {log}^{{n}} {x}\:{d}\mathrm{3} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \frac{\mathrm{1}}{\boldsymbol{{n}}!}\int{e}^{\left({n}+\mathrm{1}\right){t}} {t}^{{n}} {dt}\:\:\:\:\:\:\:{logx}={t} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{1}}{{n}!\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\int{e}^{{u}} {u}^{{n}} {du}\:\:\:\:\:\left({n}+\mathrm{1}\right){t}={u} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{1}}{{n}!\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\underset{{k}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\int\frac{{u}^{{n}+{k}} }{{k}!}{du} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{1}}{{n}!\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\underset{{k}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\frac{{u}^{{n}+{k}+\mathrm{1}} }{{k}!\left({n}+{k}+\mathrm{1}\right)} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\underset{{k}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {u}^{{n}+{k}+\mathrm{1}} }{{n}!{k}!\left({n}+\mathrm{1}\right)\left({n}+{k}+\mathrm{1}\right)} \\ $$

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