Question Number 189980 by sonukgindia last updated on 25/Mar/23
Answered by aleks041103 last updated on 26/Mar/23
$${I}=\int_{\mathrm{0}} ^{\infty} {x}^{−\frac{{lnx}}{{n}}} {dx}=\int_{−\infty} ^{\:\infty} {e}^{−{t}\frac{{t}}{{n}}} {e}^{{t}} {dt} \\ $$$${x}={e}^{{t}} \\ $$$${I}=\int_{−\infty} ^{\:\infty} {e}^{−\left({t}^{\mathrm{2}} /{n}\:−{t}\right)} {dt} \\ $$$$\frac{{t}^{\mathrm{2}} }{{n}}−{t}=\left(\frac{{t}}{\:\sqrt{{n}}}\right)^{\mathrm{2}} −\mathrm{2}\frac{{t}}{\:\sqrt{{n}}}\:\frac{\sqrt{{n}}}{\mathrm{2}}\:+\frac{{n}}{\mathrm{4}}−\frac{{n}}{\mathrm{4}}= \\ $$$$=\left(\frac{{t}}{\:\sqrt{{n}}}−\frac{\sqrt{{n}}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{{n}}{\mathrm{4}} \\ $$$${I}={e}^{{n}/\mathrm{4}} \int_{−\infty} ^{\:\infty} {e}^{−\left(\frac{{t}}{\:\sqrt{{n}}}−\frac{\sqrt{{n}}}{\mathrm{2}}\right)^{\mathrm{2}} } {dt}= \\ $$$$=\sqrt{{n}}{e}^{{n}/\mathrm{4}} \int_{−\infty} ^{\:\infty} {e}^{−{t}^{\mathrm{2}} } {dt}= \\ $$$$=\sqrt{{n}\pi}{e}^{{n}/\mathrm{4}} \\ $$$$\Rightarrow{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\sqrt{{n}\pi}}{\:\sqrt{{n}\pi}{e}^{{n}/\mathrm{4}} }=\frac{\mathrm{1}}{{e}^{\mathrm{1}/\mathrm{4}} }\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left({e}^{−\mathrm{1}/\mathrm{4}} \right)^{{n}} = \\ $$$$=\frac{\mathrm{1}}{{e}^{\mathrm{1}/\mathrm{4}} }\:\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}/\mathrm{4}} }=\:\frac{\mathrm{1}}{{e}^{\mathrm{1}/\mathrm{4}} −\mathrm{1}}\:=\:{S} \\ $$
Commented by sonukgindia last updated on 26/Mar/23
$${thNks} \\ $$