# 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.

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

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