Όριο ολοκληρώματος (1)

#1

Έστω $\displaystyle{a_n=\int_{0}^{n}\frac{e^{-x}}{1+\frac{x}{n}}\,dx}$ .

Ας βρεθούν, αν υπάρχουν, τα όρια

$\displaystyle{\lim_{n\to+\infty}a_n}$

$\displaystyle{\lim_{n\to+\infty}n\left(1-a_n\right)}$ και

$\displaystyle{\lim_{n\to+\infty}n^2\left(1-a_n\right)-n}$ .

#2

$\displaystyle{{a_n} = \int\limits_0^n {\frac{{{e^{ - x}}}}{{1 + x/n}}dx} = \mathop = \limits^{x = ny} = n\int\limits_0^1 {\frac{{{e^{ - ny}}}}{{1 + y}}dy} = n\int\limits_0^1 {{e^{ - ny}}\left( {\int\limits_0^\infty {{e^{ - \left( {1 + y} \right)x}}dx} } \right)dy} = n\int\limits_0^\infty {{e^{ - x}}\left( {\int\limits_0^1 {{e^{ - \left( {n + x} \right)y}}dy} } \right)dx} = }$

$\displaystyle{ = n\int\limits_0^\infty {{e^{ - x}}\left( { - \frac{1}{{n + x}}\left[ {{e^{ - \left( {n + x} \right)y}}} \right]_0^1} \right)dx} = n\int\limits_0^\infty {\frac{{{e^{ - x}}}}{{n + x}}\left( {1 - {e^{ - \left( {n + x} \right)}}} \right)dx} = n{e^n}\int\limits_0^\infty {\frac{{{e^{ - \left( {n + x} \right)}}}}{{n + x}}\left( {1 - {e^{ - \left( {n + x} \right)}}} \right)dx} = \mathop = \limits^{n + x = y} = }$

$\displaystyle{ = n{e^n}\int\limits_n^\infty {\frac{{{e^{ - y}}}}{y}\left( {1 - {e^{ - y}}} \right)dy} = n{e^n}\left( {\int\limits_n^\infty {\frac{{{e^{ - y}}}}{y}dy} - \int\limits_n^\infty {\frac{{{e^{ - 2z}}}}{z}dz} } \right) = \mathop = \limits^{2z = x} = n{e^n}\left( {\int\limits_n^\infty {\frac{{{e^{ - y}}}}{y}dy} - \int\limits_{2n}^\infty {\frac{{{e^{ - x}}}}{x}dx} } \right) \Rightarrow \boxed{{a_n} = n{e^n}\int\limits_n^{2n} {\frac{{{e^{ - y}}}}{y}dy} }}$

Όλα τα παραπάνω έγιναν για να βγεί το έξω απ' το ολοκλήρωμα .. τώρα τα πράγματα γίνονται απλά .. (Λύκειο σχεδόν) ..

Αν $\displaystyle{f\left( x \right) = x{e^x}\int\limits_x^{2x} {\frac{{{e^{ - y}}}}{y}dy} }$ τότε επειδή (στοιχειώδες) έχουμε $\displaystyle{\mathop {\lim }\limits_{x \to \infty } \int\limits_x^{2x} {\frac{{{e^{ - y}}}}{y}dy} = 0}$ .

$\displaystyle{\mathop {\lim }\limits_{x \to \infty } x{e^x}\int\limits_x^{2x} {\frac{{{e^{ - y}}}}{y}dy} = \mathop {\lim }\limits_{x \to \infty } \frac{{\int\limits_x^{2x} {\dfrac{{{e^{ - y}}}}{y}dy} }}{{\dfrac{{{e^{ - x}}}}{x}}} = \mathop = \limits^{\left( {\frac{0}{0}} \right)\left( {DLH} \right)} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\dfrac{{{e^{ - 2x}}}}{x} - \dfrac{{{e^{ - x}}}}{x}}}{{\dfrac{{ - x{e^{ - x}} - {e^{ - x}}}}{{{x^2}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\dfrac{{1 - {e^{ - x}}}}{x}}}{{\dfrac{{x + 1}}{{{x^2}}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{x + 1}}{x}\left( {1 - {e^{ - x}}} \right) = 1}$

Οπότε $\displaystyle{\mathop {\lim }\limits_{n \to \infty } {a_n} = 1}$ και $\displaystyle{\mathop {\lim }\limits_{n \to \infty } n\left( {1 - {a_n}} \right) = \mathop {\lim }\limits_{x \to \infty } x\left( {1 - x{e^x}\int\limits_x^{2x} {\frac{{{e^{ - y}}}}{y}dy} } \right) = .. = \mathop {\lim }\limits_{x \to \infty } \frac{{\dfrac{{{e^{ - x}}}}{x} - \int\limits_x^{2x} {\dfrac{{{e^{ - y}}}}{y}dy} }}{{\dfrac{{{e^{ - x}}}}{{{x^2}}}}} = \mathop = \limits^{\left( {\frac{0}{0}} \right)} = ... = 1}$

και $\displaystyle{\mathop {\lim }\limits_{n \to \infty } \left( {{n^2}\left( {1 - {a_n}} \right) - n} \right) = \mathop {\lim }\limits_{x \to \infty } \left[ {{x^2}\left( {1 - x{e^x}\int\limits_x^{2x} {\frac{{{e^{ - y}}}}{y}dy} } \right) - x} \right] = \mathop {\lim }\limits_{x \to \infty } \left[ {{x^3}{e^x}\left( {\frac{{{e^{ - x}}}}{x} - \int\limits_x^{2x} {\frac{{{e^{ - y}}}}{y}dy} } \right) - x} \right] = }$

$\displaystyle{ = \mathop {\lim }\limits_{x \to \infty } {x^3}{e^x}\left( {\frac{{{e^{ - x}}}}{x} - \int\limits_x^{2x} {\frac{{{e^{ - y}}}}{y}dy} - \frac{{{e^{ - x}}}}{{{x^2}}}} \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{\dfrac{{{e^{ - x}}}}{x} - \int\limits_x^{2x} {\dfrac{{{e^{ - y}}}}{y}dy} - \dfrac{{{e^{ - x}}}}{{{x^2}}}}}{{\dfrac{{{e^{ - x}}}}{{{x^3}}}}} = \mathop = \limits^{\left( {\frac{0}{0}} \right)} = .. = - 2}$ .

Όριο ολοκληρώματος (1)

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