Question Number 70696 by Erik_Regiomontan last updated on 07/Oct/19
$$\int{x}^{{x}} {dx}=? \\ $$
Commented by sadimuhmud 136 last updated on 07/Oct/19
$$\mathrm{error}\:\mathrm{question}.. \\ $$$$ \\ $$
Commented by Prithwish sen last updated on 07/Oct/19
Commented by Prithwish sen last updated on 07/Oct/19
Commented by Prithwish sen last updated on 07/Oct/19
$$\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{found}}\:\boldsymbol{\mathrm{it}}.\:\boldsymbol{\mathrm{If}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{helps}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{somehow}}. \\ $$
Answered by mind is power last updated on 07/Oct/19
![∫e^(xln(x)) dx x>0 =∫Σ_(k≥0) (((xln(x))^k )/(k!))dx =∫_0 ^t x^x dx=Σ_(k≥0) ∫(((xln(x))^k )/(k!))dx becauseΣ_(k≥0) (((xln(x))^k )/(k!)) converge uniformuly ⇒e^(xln(x)) ∫_0 ^t x^k ln^j (x)dx=A_(i,k) ∫x^k ln(x)^j dx=((x^(k+1) /(k+1))ln^j (x)]_0 ^t −(j/(k+1))∫x^k ln(x)^(j−1) A_(j,k) =(t^(k+1) /(k+1))ln^j (t)−((j.A_(j−1,k) )/(k+1)) A_(0,k) =∫_0 ^t x^k dx=(t^(k+1) /(k+1)) A_(j,k) =((t^(k+1) ln^j (k))/(k+1))−(j/(k+1)).(((t^(k+1) ln^(j−1) (t))/(k+1))−(((j−1))/(k+1))A_(k,j−2) ) A_(j,k) =Σ_(s=0) ^j (((j)!)/(s!)) .(((−1)^(s−j) t^(k+1) ln^s (t))/(((k+1)^(j−s+1) )) ⇒A_(k,k) =Σ_(s=0) ^k ((j!.(−1)^(s−k) t^(k+1) ln^k (t))/(s!(k+1)^(k−s+1) )) ∫_0 ^t x^x dx=Σ_(k=0) ^(+∞) (1/(k!)).Σ_(s=0) ^k ((j! (−1)^(s−k) t^(k+1) ln^k (x))/(s!(k+1)^(k−s+1) )) We cant find expressin of explicite function cause we need special function like F_2 (a,b,c,x) hyper geometric function](https://www.tinkutara.com/question/Q70712.png)
$$\int{e}^{{xln}\left({x}\right)} {dx}\:\:\:{x}>\mathrm{0} \\ $$$$=\int\sum_{{k}\geqslant\mathrm{0}} \frac{\left({xln}\left({x}\right)\right)^{{k}} }{{k}!}{dx} \\ $$$$=\int_{\mathrm{0}} ^{{t}} {x}^{{x}} {dx}=\underset{{k}\geqslant\mathrm{0}} {\sum}\int\frac{\left({xln}\left({x}\right)\right)^{{k}} }{{k}!}{dx}\:\:\:{because}\sum_{{k}\geqslant\mathrm{0}} \frac{\left({xln}\left({x}\right)\right)^{{k}} }{{k}!}\:{converge}\:{uniformuly}\:\Rightarrow{e}^{{xln}\left({x}\right)} \\ $$$$\int_{\mathrm{0}} ^{{t}} {x}^{{k}} {ln}^{{j}} \left({x}\right){dx}={A}_{{i},{k}} \\ $$$$\int{x}^{{k}} {ln}\left({x}\right)^{{j}} {dx}=\left(\frac{{x}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}{ln}^{{j}} \left({x}\right)\right]_{\mathrm{0}} ^{{t}} −\frac{{j}}{{k}+\mathrm{1}}\int{x}^{{k}} {ln}\left({x}\right)^{{j}−\mathrm{1}} \\ $$$${A}_{{j},{k}} =\frac{{t}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}{ln}^{{j}} \left({t}\right)−\frac{{j}.{A}_{{j}−\mathrm{1},{k}} }{{k}+\mathrm{1}} \\ $$$${A}_{\mathrm{0},{k}} =\int_{\mathrm{0}} ^{{t}} {x}^{{k}} {dx}=\frac{{t}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}} \\ $$$${A}_{{j},{k}} =\frac{{t}^{{k}+\mathrm{1}} {ln}^{{j}} \left({k}\right)}{{k}+\mathrm{1}}−\frac{{j}}{{k}+\mathrm{1}}.\left(\frac{{t}^{{k}+\mathrm{1}} {ln}^{{j}−\mathrm{1}} \left({t}\right)}{{k}+\mathrm{1}}−\frac{\left({j}−\mathrm{1}\right)}{{k}+\mathrm{1}}{A}_{{k},{j}−\mathrm{2}} \right) \\ $$$${A}_{{j},{k}} =\underset{{s}=\mathrm{0}} {\overset{{j}} {\sum}}\:\:\:\:\:\:\:\frac{\left({j}\right)!}{{s}!}\:.\frac{\left(−\mathrm{1}\right)^{{s}−{j}} {t}^{{k}+\mathrm{1}} {ln}^{{s}} \left({t}\right)}{\left(\left({k}+\mathrm{1}\right)^{{j}−{s}+\mathrm{1}} \right.} \\ $$$$\Rightarrow{A}_{{k},{k}} =\sum_{{s}=\mathrm{0}} ^{{k}} \frac{{j}!.\left(−\mathrm{1}\right)^{{s}−{k}} {t}^{{k}+\mathrm{1}} {ln}^{{k}} \left({t}\right)}{{s}!\left({k}+\mathrm{1}\right)^{{k}−{s}+\mathrm{1}} } \\ $$$$\int_{\mathrm{0}} ^{{t}} {x}^{{x}} {dx}=\sum_{{k}=\mathrm{0}} ^{+\infty} \frac{\mathrm{1}}{{k}!}.\underset{{s}=\mathrm{0}} {\overset{{k}} {\sum}}\frac{{j}!\:\left(−\mathrm{1}\right)^{{s}−{k}} {t}^{{k}+\mathrm{1}} {ln}^{{k}} \left({x}\right)}{{s}!\left({k}+\mathrm{1}\right)^{{k}−{s}+\mathrm{1}} } \\ $$$${We}\:{cant}\:{find}\:{expressin}\:{of}\:{explicite}\:{function}\:{cause}\:{we}\:{need}\:{special}\:{function} \\ $$$${like}\:{F}_{\mathrm{2}} \left({a},{b},{c},{x}\right)\:{hyper}\:{geometric}\:{function} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$