(New page: '''a)'''<math>\lim_{t\rightarrow 0} \int_0^1 \frac{e^{-t \ln(x)} - 1}{t}dx = \lim_{t\rightarrow 0} \ \frac{1}{1-t} = 1</math> by direct integration, since the integrand has an continuous a...) |
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Since <math>f_n(x)\chi_{(1,n^2)} \geq 0</math>, apply Fatou to obtain: | Since <math>f_n(x)\chi_{(1,n^2)} \geq 0</math>, apply Fatou to obtain: | ||
| − | <math>\liminf_{n\rightarrow \infty} \int_1^{n^2} f_n(x) dx \geq \int_1^{\infty} \liminf_{n\rightarrow \infty}f_n(x)\chi_{(1,n^2)} dx = \infty | + | <math>\liminf_{n\rightarrow \infty} \int_1^{n^2} f_n(x) dx \geq \int_1^{\infty} \liminf_{n\rightarrow \infty}f_n(x)\chi_{(1,n^2)} dx = \int_1^{\infty} \frac{1}{\ln(x)} dx \geq \int_1^{\infty} \frac{1}{x-1} dx = \infty </math> |
-pw | -pw | ||
Latest revision as of 14:21, 22 July 2008
a)$ \lim_{t\rightarrow 0} \int_0^1 \frac{e^{-t \ln(x)} - 1}{t}dx = \lim_{t\rightarrow 0} \ \frac{1}{1-t} = 1 $ by direct integration, since the integrand has an continuous antiderivative for $ t<1 $.
b)Call $ f_n(x) = \frac{n\cos(xn^{-2})}{1+n\ln(x)} $
$ \lim_{n\rightarrow \infty} \int_1^{n^2} f_n(x) dx= \lim_{n\rightarrow \infty} \int_1^{\infty} f_n(x)\chi_{(1,n^2)} dx $
Since $ f_n(x)\chi_{(1,n^2)} \geq 0 $, apply Fatou to obtain:
$ \liminf_{n\rightarrow \infty} \int_1^{n^2} f_n(x) dx \geq \int_1^{\infty} \liminf_{n\rightarrow \infty}f_n(x)\chi_{(1,n^2)} dx = \int_1^{\infty} \frac{1}{\ln(x)} dx \geq \int_1^{\infty} \frac{1}{x-1} dx = \infty $
-pw
