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Question Number 121680 by aristarque last updated on 10/Nov/20

$$pleaseevaluate\int x^{x}dx$$

Answered by Dwaipayan Shikari last updated on 11/Nov/20

$$\int e^{xlogx}dx$$$$=\int \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(xlogx\right)}^n}{n!}dx$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\int \frac{{\left(xlogx\right)}^n}{n!}dx$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}\int x^n{log}^{n}xdxx=e^{-t}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}{(-1)}^{n+1}\int e^{-(n+1)t}{\left(t\right)}^{n}dt$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}{(-1)}^{n+1}\int e^{-u}\frac{u^n}{{(n+1)}^n}du(n+1)t=u$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}.\frac{{(-1)}^{n+1}}{{(n+1)}^n}\int e^{-u}{u}^{n}du$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}.\frac{{(-1)}^{n+1}}{{(n+1)}^n}\int (\underset{r=0}{\sum ^{\mathrm{\infty }}}\frac{{(-u)}^r}{r!}.u^n)du$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1+r}}{n!{(n+1)}^n}\underset{r=0}{\sum ^{\mathrm{\infty }}}\frac{u^{r+n+1}}{r!(r+n+1)}$$$$$$$$Ifitwas{\int }_0^1{x}^{x}dxthen$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{n!}.\frac{{(-1)}^{n+2}}{{(n+1)}^{n+1}}{\int }_0^{\mathrm{\infty }}{e}^{-u}{u}^{n}du$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(n+1)}^{(n+1)}}\frac{\mathrm{\Gamma }(n+1)}{n!}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(n+1)}^{(n+1)}}$$

Commented by aristarque last updated on 11/Nov/20

$$pleaseevaluate\int x^{x}dx$$$$thanksverymuchsir$$

Commented by Dwaipayan Shikari last updated on 11/Nov/20

$$Ihaveeditedtheanswer.kindlygothroughit$$

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