Question Number 99403 by I want to learn more last updated on 20/Jun/20
$$\int\:\mathrm{x}^{\mathrm{x}} \:\:\mathrm{dx} \\ $$
Commented by PRITHWISH SEN 2 last updated on 20/Jun/20
$$\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{possible}\:\mathrm{soln}.\:\mathrm{for}\:\mathrm{this}\:\mathrm{integral}. \\ $$
Answered by smridha last updated on 21/Jun/20
![∫x^x dx=∫e^(xln(x)) dx=Σ_(n=0) ^∞ (1/(n!))∫x^n (ln[x])^n dx let x=e^k so=Σ_(n=0) ^∞ (1/(n!))∫e^((n+1)k) k^n dk =Σ_(n=0) ^∞ (1/(n!))[Σ_(m=0) ^∞ (1/(m!)) (n+1)^m ∫k^(m+n) dk] =Σ_(n,m=0) ^∞ (1/(n!m!)).(((n+1)^m )/((m+n+1))).[ln(x)]^(m+n+1) +c](https://www.tinkutara.com/question/Q99561.png)
$$\int\boldsymbol{{x}}^{\boldsymbol{{x}}} \boldsymbol{{dx}}=\int\boldsymbol{{e}}^{\boldsymbol{{xln}}\left(\boldsymbol{{x}}\right)} \boldsymbol{{dx}}=\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{{n}}!}\int\boldsymbol{{x}}^{\boldsymbol{{n}}} \left(\boldsymbol{{ln}}\left[\boldsymbol{{x}}\right]\right)^{\boldsymbol{{n}}} \boldsymbol{{dx}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{x}}=\boldsymbol{{e}}^{\boldsymbol{{k}}} \:\boldsymbol{{so}}=\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{{n}}!}\int\boldsymbol{{e}}^{\left(\boldsymbol{{n}}+\mathrm{1}\right)\boldsymbol{{k}}} \boldsymbol{{k}}^{\boldsymbol{{n}}} \boldsymbol{{dk}} \\ $$$$=\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{{n}}!}\left[\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{{m}}!}\:\left(\boldsymbol{{n}}+\mathrm{1}\right)^{\boldsymbol{{m}}} \int\boldsymbol{{k}}^{\boldsymbol{{m}}+\boldsymbol{{n}}} \boldsymbol{{dk}}\right] \\ $$$$=\underset{\boldsymbol{{n}},\boldsymbol{{m}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{{n}}!\boldsymbol{{m}}!}.\frac{\left(\boldsymbol{{n}}+\mathrm{1}\right)^{\boldsymbol{{m}}} }{\left(\boldsymbol{{m}}+\boldsymbol{{n}}+\mathrm{1}\right)}.\left[\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)\right]^{\boldsymbol{{m}}+\boldsymbol{{n}}+\mathrm{1}} \:\:+\boldsymbol{{c}} \\ $$
Commented by I want to learn more last updated on 23/Jun/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$