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Question Number 91054 by Mikael_786 last updated on 27/Apr/20

Commented by MJS last updated on 27/Apr/20

$$\mathrm{I}\mathrm{can}\mathrm{solve}\mathrm{the}\mathrm{integral}\int {\omega }^{1/x}dx\mathrm{but}\mathrm{I}\mathrm{cannot}$$$$\mathrm{solve}\mathrm{the}\mathrm{limit}.\mathrm{it}\mathrm{seems}L=1$$

Commented by MJS last updated on 28/Apr/20

$$\int {\omega }^{1/x}dx=$$$$\left[\mathrm{by}\mathrm{parts}\right]$$$$={\omega }^{1/x}x+\ln \omega \int \frac{{\omega }^{1/x}}{x}dx=$$$$$$$$\int \frac{{\omega }^{1/x}}{x}dx=$$$$[t=\frac{1}{x}\to dx=-x^{2}dt]$$$$=-\int \frac{{\omega }^t}{t}dt=\mathrm{Ei}\left(t\ln \omega \right)=\mathrm{Ei}\frac{\ln \omega }{x}$$$$$$$$={\omega }^{1/x}x-\ln \omega \mathrm{Ei}\frac{\ln \omega }{x}+C$$$$$$$$\frac{1}{\omega }\underset{1}{\stackrel{\omega }{\int }}{\omega }^{1/x}dx=\frac{1}{\omega }{[{\omega }^{1/x}x-\ln \omega \mathrm{Ei}\frac{\ln \omega }{x}]}_1^{\omega }=$$$$=\frac{1}{\omega }({\omega }^{1/\omega +1}-\omega -\ln \omega \mathrm{Ei}\frac{\ln \omega }{\omega }+\ln \omega \mathrm{Ei}\ln \omega )$$$$={\omega }^{\frac{1}{\omega }}-1+\frac{\ln \omega }{\omega }(\mathrm{Ei}\ln \omega -\mathrm{Ei}\frac{\ln \omega }{\omega })$$$$\underset{\omega \to \mathrm{\infty }}{\lim }{\omega }^{\frac{1}{\omega }}-1=0$$$$\underset{\omega \to \mathrm{\infty }}{\lim }\frac{\ln \omega }{\omega }(\mathrm{Ei}\ln \omega -\mathrm{Ei}\frac{\ln \omega }{\omega })=?$$$$\mathrm{my}\mathrm{knowledge}\mathrm{ends}\mathrm{here}$$

Answered by ~blr237~ last updated on 28/Apr/20

$$assumingthatw\to \mathrm{\infty },w⩾1\Rightarrow w>0$$$$letstatet=w^{\frac{1}{x}}thenx=\frac{1}{{log}_w\left(t\right)}=\frac{lnw}{lnt}$$$$f\left(w\right)={\int }_1^w{w}^{\frac{1}{x}}dx={\int }_w^{w^{\frac{1}{w}}}td\left(\frac{lnw}{lnt}\right)$$$$bypartf\left(w\right)=lnw({\left[\frac{t}{lnt}\right]}_w^{w^{\frac{1}{w}}}-{\int }_w^{w^{\frac{1}{w}}}\frac{dt}{lnt})$$$$f\left(w\right)=lnw([\frac{w^{\frac{1}{w}}}{\frac{lnw}{w}}-\frac{w}{lnw}]-li\left(w^{\frac{1}{w}}\right)+li\left(w\right))$$$$f\left(w\right)=w^{\frac{1}{w}+1}-w-li\left(w^{\frac{1}{w}}\right)lnw+li\left(w\right)lnw$$$$\frac{1}{w}f\left(w\right)=w^{\frac{1}{w}}-1-li\left(w^{\frac{1}{w}}\right)\frac{lnw}{w}+li\left(w\right)\frac{lnw}{w}$$$$wehave\underset{w\to \mathrm{\infty }}{\lim }{w}^{\frac{1}{w}}=1and\underset{w\to \mathrm{\infty }}{\lim }li\left(w\right)\frac{lnw}{w}=1$$$$SirmrWassumingyouransweriscorrect$$$$Howcanweprovethat\underset{w\to \mathrm{\infty }}{\lim }li\left(w^{\frac{1}{w}}\right)\frac{lnw}{w}=0$$$$knowingthat\underset{x\to 1}{\lim }li\left(x\right)=-\mathrm{\infty }withli\left(x\right)={\int }_0^x\frac{dt}{lnt}$$$$$$$$canweprovethat\underset{x\to 1}{\lim }li\left(x\right)ln\left(x\right)=0??$$$$$$

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