Menu Close

e-x-x-dx-




Question Number 129275 by abdurehime last updated on 14/Jan/21
∫e^x^x  dx??????
$$\int\mathrm{e}^{\mathrm{x}^{\mathrm{x}} } \mathrm{dx}?????? \\ $$
Commented by abdurehime last updated on 14/Jan/21
no one????
$$\mathrm{no}\:\mathrm{one}???? \\ $$
Commented by Dwaipayan Shikari last updated on 14/Jan/21
Σ_(n=1) ^∞ (1/(n!))∫(x^x )^n dx = Σ_(n=1) ^∞ (1/(n!))∫(x^x )^n   =Σ_(n=1) ^∞ (1/(n!))∫e^(nxlogx) dx  =Σ_(n=1) ^∞ Σ_(k=1) ^∞ (1/(n!)).(n^k /(k!))∫(xlogx)^k dx=Σ_(n≥1) Σ_(k≥1) (n^k /(n!k!))∫t^k e^((k+1)t) dt     logx=t  =Σ_(n≥1) Σ_(k≥1) (n^k /(n!k!))Σ_(m≥1) ∫((t^k (k+1)^m t^m )/(m!))dt  =Σ_(n≥1) Σ_(k≥1) Σ_(m≥1) ((n^k (k+1)^m (logx)^(k+m+1) )/(n!k!m!(k+m+1)))+C
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int\left({x}^{{x}} \right)^{{n}} {dx}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int\left({x}^{{x}} \right)^{{n}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int{e}^{{nxlogx}} {dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}.\frac{{n}^{{k}} }{{k}!}\int\left({xlogx}\right)^{{k}} {dx}=\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{n}^{{k}} }{{n}!{k}!}\int{t}^{{k}} {e}^{\left({k}+\mathrm{1}\right){t}} {dt}\:\:\:\:\:{logx}={t} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{n}^{{k}} }{{n}!{k}!}\underset{{m}\geqslant\mathrm{1}} {\sum}\int\frac{{t}^{{k}} \left({k}+\mathrm{1}\right)^{{m}} {t}^{{m}} }{{m}!}{dt} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\underset{{k}\geqslant\mathrm{1}} {\sum}\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{{n}^{{k}} \left({k}+\mathrm{1}\right)^{{m}} \left({logx}\right)^{{k}+{m}+\mathrm{1}} }{{n}!{k}!{m}!\left({k}+{m}+\mathrm{1}\right)}+{C} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *