=for timestamp
Fr Nov 24 13:08:14 CET 2006

=head2 115. Hausaufgabe

=head3 Analysis-Buch Seite 256, Aufgabe 15h

=for latex
{\small

M<\displaystyle\int\limits_0^{\infty} x^2 e^{-x} \,\mathrm{d}x =
{}\int\limits_0^{\infty} x^2 \left(-e^{-x}\right)' \,\mathrm{d}x =
{}\lim\limits_{\alpha \to \infty} \left[-x^2 e^{-x} - \int 2x \left(-e^{-x}\right) \mathrm{d}x\right]_0^{\alpha} = \\
{}\lim\limits_{\alpha \to \infty} \left[-x^2 e^{-x} + 2 \int x \left(-e^{-x}\right)' \mathrm{d}x\right]_0^{\alpha} =
{}\lim\limits_{\alpha \to \infty} \left[-x^2 e^{-x} + 2 \left(-x e^{-x} - \int -e^{-x} \,\mathrm{d}x\right)\right]_0^{\alpha} =
{}\lim\limits_{\alpha \to \infty} \left[e^{-x} \left(-x^2 - 2 x - 2\right)\right]_0^{\alpha} = 2;>

=for latex
}

=head3 Analysis-Buch Seite 256, Aufgabe 16

=over

=item a)

Für M<n \neq 1>:

M<\int\limits_1^a \frac{\ln x}{x^n} \mathrm{d}x =
{}\int\limits_1^a \left(\frac{x^{1-n}}{1-n}\right)' \ln x \,\mathrm{d}x =
{}\left[\frac{x^{1-n}}{1 - n} \ln x - \int \underbrace{\frac{x^{1-n}}{1 - n} \cdot \frac{1}{x}}_{\frac{x^{-n}}{1-n}} \mathrm{d}x\right]_1^a =
{}\left[\frac{x^{1-n}}{1-n} \left(\ln x - \frac{1}{1-n}\right)\right]_1^a =
{}\frac{a^{1-n}}{1-n} \left(\ln a - \frac{1}{1-n}\right) +
\left(\frac{1}{1-n}\right)^2;>

Für M<n = 1>: M<\int \frac{\ln x}{x} \,\mathrm{d}x = \int I(\ln x) \left(\ln
x\right)' \,\mathrm{d}x = \int I(t) \,\mathrm{d}t = \frac{1}{2} \ln^2 x> mit
M<I(t) = t;>

=item c)

M<\int\limits_0^a x^n \ln x \,\mathrm{d}x =
{}\left[\int \frac{\ln x}{x^{-n}} \mathrm{d}x\right]_0^a =
{}\left[\frac{x^{1+n}}{1+n} \left(\ln x - \frac{1}{1+n}\right)\right]_0^a =
{}\frac{a^{1+n}}{1+n} \left(\ln a - \frac{1}{1+n}\right);>

Speziell für M<a = 1>:

M<\int\limits_0^1 x^n \ln x \,\mathrm{d}x = -\left(\frac{1}{1 + n}\right)^2;>

=back