7
$\begingroup$

Let $$f_n(x)=\frac{1}{\sqrt{\pi n}}e^{-x^2/n}.$$ Note that $f_n(x)\to 0$ uniformly as $n\to\infty$. [Proof: $0\leq f_n(x)\leq\frac{1}{\sqrt{\pi n}}$; given any $\epsilon > 0$, let $M=\left\lceil\frac{1}{\pi\epsilon^2}\right\rceil$. This guarantees that $\forall n>M:\forall x:|f_n(x)-0|<\epsilon$.]

Uniform convergence justifies taking the limit $n\to\infty$ under the integral sign:

$$\lim_{n\to\infty}\int_{-\infty}^{+\infty} f_n(x)\,dx = \int_{-\infty}^{+\infty} \lim_{n\to\infty} f_n(x)\,dx$$

The left-hand side is $1$, because

$$\forall n>0:\int_{-\infty}^{+\infty} f_n(x)\,dx = 1,$$

whereas the right-hand side is $0$, because

$$\forall x\in\mathbb{R}:\lim_{n\to\infty} f_n(x) = 0.$$

Therefore,

$$1=0.$$

$\endgroup$
12
$\begingroup$

Uniform convergence justifies taking the limit under the integral sign for functions with bounded domain, not for functions whose domain is $\mathbb R$.

$\endgroup$
4
$\begingroup$

Uniform convergence does not justify the interchange for an integral over an infinite interval. As an example, take $f_n(x) =1/n$ for $0\leqslant x\leqslant n$ and $f_n(x) =0$ for $x>n$.

If the improper integral is also uniformly convergent then it is permissible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.