Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 189980 by sonukgindia last updated on 25/Mar/23

Answered by aleks041103 last updated on 26/Mar/23

$$I={\int }_0^{\mathrm{\infty }}{x}^{-\frac{lnx}{n}}dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-t\frac{t}{n}}{e}^{t}dt$$$$x=e^t$$$$I={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-(t^2/n-t)}dt$$$$\frac{t^2}{n}-t={\left(\frac{t}{\sqrt{n}}\right)}^2-2\frac{t}{\sqrt{n}}\frac{\sqrt{n}}{2}+\frac{n}{4}-\frac{n}{4}=$$$$={(\frac{t}{\sqrt{n}}-\frac{\sqrt{n}}{2})}^2-\frac{n}{4}$$$$I=e^{n/4}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-{(\frac{t}{\sqrt{n}}-\frac{\sqrt{n}}{2})}^2}dt=$$$$=\sqrt{n}{e}^{n/4}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-t^2}dt=$$$$=\sqrt{n\pi }{e}^{n/4}$$$$\Rightarrow S=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\sqrt{n\pi }}{\sqrt{n\pi }{e}^{n/4}}=\frac{1}{e^{1/4}}\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left(e^{-1/4}\right)}^n=$$$$=\frac{1}{e^{1/4}}\frac{1}{1-e^{-1/4}}=\frac{1}{e^{1/4}-1}=S$$

Commented by sonukgindia last updated on 26/Mar/23

$$thNks$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com