Ngô Quốc Anh

October 9, 2007

Một bài “khủng” về tích phân không xác định

Filed under: Các Bài Tập Nhỏ, Giải Tích 2 — Ngô Quốc Anh @ 23:36

Evaluate

\displaystyle\int \dfrac{1}{\sqrt { 1 - \sqrt {2 - \sqrt {3 - \sqrt { 4 - \sqrt x } } } }} dx.

Solution. Denote

\displaystyle u = \sqrt {1 - \sqrt {2 - \sqrt {3 - \sqrt {4 - \sqrt {x}}}}}

one can solve  x in terms of u and then try to evaluate

\displaystyle\int\frac {1}{u}dx.

Note that the integrand is just a polynomial.

Thuộc L^1 nhưng không thuộc L^2

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 23:21

Problem. Construct a function in L^1 (-\infty,+\infty) which is not in L^2(a,b) for any a<b.

Solution. Let g(x)=|x|^{-2/3}e^{-x^2}. The particular thing to note about this function is that g\in L^1(\mathbb{R}) but if 0\in[a,b], then \int_a^bg(x)^2\,dx=\infty. (The exact details of g can be varied; all we really need are those two properties). Let r_k be an enumeration of the rational numbers (any other countable dense subset of \mathbb{R} would do just as well). Let

f(x)=\sum_{k=1}^{\infty}2^{-k}g(x-r_k).

Then

 \int_{\mathbb{R}}f=\sum_{k=1}^{\infty}  <br> \int_{\mathbb{R}}2^{-k}g(x-r_k)\,dx =\sum_{k=1}^{\infty}2^{-k}\int_{\mathbb{R}}g,

which converges.

(The interchange of integral and sum is justified by the Monotone Convergence Theorem since g is nonnegative).

But on any interval (a,b), chose some r_m\in(a,b). On that interval,

f(x)>2^{-m}g(x-r_m)

and 

\int_a^b(f(x))^2\,dx\ge2^{-2m}\int_a^b(g(x-r_m))^2\,dx=\infty.

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