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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

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010