Question-152722

Question Number 152722 by SANOGO last updated on 31/Aug/21

Answered by Olaf_Thorendsen last updated on 31/Aug/21

• I = ∫_0 ^1 (e^(−(1/x)) /x^2 ) dx Let u = (1/x) : I = ∫_1 ^∞ e^(−u) du = [−e^(−u) ]_1 ^∞ = (1/e) (1) • I = ∫_0 ^1 (e^(−(1/x)) /x^2 ) dx I = lim_(n→∞) (1/n)Σ_(k=0) ^n (e^(−(1/(k/n))) /(k^2 /n^2 )) I = lim_(n→∞) nΣ_(k=0) ^n (e^(−(n/k)) /k^2 ) (2) (1) and (2) : lim_(n→∞) nΣ_(k=0) ^n (e^(−(n/k)) /k^2 ) = (1/e)

$$\bullet\:\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{−\frac{\mathrm{1}}{{x}}} }{{x}^{\mathrm{2}} }\:{dx} \\ $$$$\mathrm{Let}\:{u}\:=\:\frac{\mathrm{1}}{{x}}\:: \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{1}} ^{\infty} {e}^{−{u}} \:{du}\:=\:\left[−{e}^{−{u}} \right]_{\mathrm{1}} ^{\infty} \:=\:\frac{\mathrm{1}}{{e}}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\bullet\:\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{−\frac{\mathrm{1}}{{x}}} }{{x}^{\mathrm{2}} }\:{dx} \\ $$$$\mathrm{I}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{e}^{−\frac{\mathrm{1}}{\frac{{k}}{{n}}}} }{\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} }} \\ $$$$\mathrm{I}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{e}^{−\frac{{n}}{{k}}} }{{k}^{\mathrm{2}} }\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\:: \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{e}^{−\frac{{n}}{{k}}} }{{k}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{{e}} \\ $$

Commented by SANOGO last updated on 31/Aug/21