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La [tex]a,A[/tex] Være et positivt tall. Evaluer
[tex]\lim_{j\rightarrow \infty }\int_{0}^{a}\frac{1}{j!}\left [ ln(\frac{a}{x}) \right ]^jdx.[/tex]

Klarte å skrive feil, det skal stå:

[tex]\lim_{j\rightarrow \infty }\int_{0}^{a}\frac{1}{j!}\left [ ln(\frac{A}{x}) \right ]^jdx.[/tex]

Nullvei wrote:Klarte å skrive feil, det skal stå:

[tex]\lim_{j\rightarrow \infty }\int_{0}^{a}\frac{1}{j!}\left [ ln(\frac{A}{x}) \right ]^jdx.[/tex]

[tex]\lim_{j\rightarrow \infty }\int_{0}^{a}\frac{1}{j!}\left [ ln(\frac{A}{x}) \right ]^jdx.[/tex]
[tex]La \, oss\, evaluere\, I=\int_{0}^{a}\frac{1}{j!}\left [ ln(\frac{A}{x}) \right ]^jdx=\lim_{y\rightarrow 0}\int_{y}^{a}\left [ ln(\frac{A}{x}) \right ]^jdx.\: Ved\: intregrasjon\: har \: vi\: at:\\u=\frac{1}{j!}\left [ ln(\frac{A}{x}) \right ]^j\: \: \: \: \: \: \: dv=dx\\du=\frac{-1}{(j-1)!}\left [ ln(\frac{A}{x}) \right ]^{j-1}\frac{dx}{x}\: \:\: \: \: \: \: \: \: \:v=x[/tex]


[tex]I_j(y)=\int_{y}^{a}\frac{1}{j!}\left [ln\frac{A}{x} \right ]^jdx=\frac{x}{j!}\left [ ln(\frac{A}{x}) \right ]^j|_{y}^{a}+\int_{y}^{a}\frac{1}{(j-1)!}\left [ ln(\frac{A}{x}) \right ]^{j-1}dx\\\\=\frac{a\left [ ln(\frac{A}{a}) \right ]^j}{j!}-\frac{y\left [ ln\frac{A}{a} \right ]^j}{j!}+I_{j-1}(y).\\Ved\: induksjon\: kan\: vi\: utlede:[/tex]


[tex]I_j(y)=a\sum_{k=0}^{j}\frac{\left [ ln(\frac{A}{a}) \right ]^k}{k!}-\sum_{k=0}^{j}(\frac{y\left [ ln(\frac{A}{y}) \right ]^k}{k!})\\men\lim_{y\rightarrow 0}y\left [ ln(\frac{A}{y}) \right ]^k=\lim_{y\rightarrow 0}(lnA-lny)^k=0\: siden\: \lim_{y\rightarrow 0}y(lny)^{p}=0\: derfor,\\\\\\ \int_{0}^{a}\frac{\left [ ln(\frac{A}{x}^j) \right ]}{j!}dx=a\sum_{k=0}^{j}\frac{\left [ ln(\frac{A}{a}) \right ]^k}{k!}[/tex]

[tex]Tilsutt,\\\\\lim_{j\rightarrow \infty }\int_{0}^{a}\frac{\left [ ln(\frac{A}{x}) \right ]}{j!}dx\\=a\sum_{k=0}^{\infty }\frac{\left [ ln(\frac{A}{a}) \right ]^k}{k!}\\=ae^{ln(\frac{A}{a})}=a\cdot \frac{A}{a}=A[/tex]