Question Number 7637 by Tawakalitu. last updated on 07/Sep/16
$$\int{x}^{{x}} \:{dx} \\ $$
Answered by FilupSmith last updated on 07/Sep/16
$${x}^{{x}} ={e}^{{x}\mathrm{ln}\left({x}\right)} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} \mathrm{ln}\left({x}\right)^{{n}} }{{n}!} \\ $$$$\left(\:\:\because{e}^{{x}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!}\:\:\right) \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int{x}^{{n}} \mathrm{ln}\left({x}\right)^{{n}} {dx} \\ $$$$\mathrm{let}\:{x}={e}^{−\frac{{u}}{{n}+\mathrm{1}}} \\ $$$${u}=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left({x}\right) \\ $$$${dx}=−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\frac{{u}}{{n}+\mathrm{1}}} {du} \\ $$$${dx}=−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\frac{{u}}{{n}+\mathrm{1}}} {du} \\ $$$$ \\ $$$$\int_{{a}} ^{\:{b}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int_{\:−\left({n}+\mathrm{1}\right)\mathrm{ln}\left({a}\right)} ^{\:−\left({n}+\mathrm{1}\right)\mathrm{ln}\left({b}\right)} \left({e}^{−\frac{{u}}{{n}+\mathrm{1}}} \right)^{{n}} \mathrm{ln}\left({e}^{−\frac{{u}}{{n}+\mathrm{1}}} \right)^{{n}} \left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}{e}^{−\frac{{u}}{{n}+\mathrm{1}}} \right){du} \\ $$$$\beta=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left({b}\right),\:\:\:\:\alpha=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left({a}\right) \\ $$$$\int_{{a}} ^{\:{b}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\int_{\:\alpha} ^{\:\beta} {e}^{−\left(\frac{{un}}{{n}+\mathrm{1}}+\frac{{u}}{{n}+\mathrm{1}}\right)} \left(−\frac{{u}}{{n}+\mathrm{1}}\right)^{{n}} \left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right){du} \\ $$$$\int_{{a}} ^{\:{b}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}−\frac{\mathrm{1}}{{n}!}\int_{\:\alpha} ^{\:\beta} {e}^{−\frac{{u}}{{n}+\mathrm{1}}\left({n}+\mathrm{1}\right)} \left(−\mathrm{1}\right)^{{n}} \frac{{u}^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}} }\left(−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right){du} \\ $$$$\int_{{a}} ^{\:{b}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}−\frac{\mathrm{1}}{{n}!}\int_{\:\alpha} ^{\:\beta} {e}^{−{u}} \left(−\mathrm{1}\right)^{{n}} \frac{{u}^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }{du} \\ $$$$\int_{{a}} ^{\:{b}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{−\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} {n}!}\int_{\:\alpha} ^{\:\beta} {e}^{−{u}} {u}^{{n}} {du}\right) \\ $$$$ \\ $$
Commented by FilupSmith last updated on 07/Sep/16
$$\mathrm{for}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx},\:\:\beta=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha=−\left({n}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{0}\right)=\infty \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{−\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} {n}!}\int_{\:\infty} ^{\:\mathrm{0}} {e}^{−{u}} {u}^{{n}} {du}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} {n}!}\int_{\:\mathrm{0}} ^{\:\infty} {e}^{−{u}} {u}^{{n}} {du}\right) \\ $$$${n}!=\int_{\:\mathrm{0}} ^{\:\infty} {e}^{−{u}} {u}^{{n}} {du} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} {n}!}{n}!\right) \\ $$$$\therefore\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \left({n}+\mathrm{1}\right)^{−\left({n}+\mathrm{1}\right)} \\ $$$$\left(\approx\mathrm{0}.\mathrm{783431}\right) \\ $$
Commented by Tawakalitu. last updated on 07/Sep/16
$${Thanks}\:{so}\:{much},\:{i}\:{really}\:{appreciate} \\ $$
Commented by Tawakalitu. last updated on 07/Sep/16
$${Thanks}\:{sir} \\ $$