Unbounded Lebesgue integrable and continuous function

Question

Problem

Show that there exists a function $f$ integrable on $[0,\infty)$, continuous and such that there is a sequence $(x_n)_{n \in \mathbb N}$ for which $$\lim_{n \to \infty}x_n=\infty \space \space \text{and} \space \lim_{n \to \infty} f(x_n)=\infty$$

I thought of picking the ordinary divergent sequence $x_n=n$ so as to construct a continuous lebesgue integrable function $f$ for which $f(x_n)=n$. I don't know if this is possible but my idea was to make the function take the value $0$ except on a very small neighborhood around each natural number $n$, I've attached a horrible drawing of the function.

I'm awfully sorry about my drawing skills, I hope the idea of the function I have in mind is clear. I would like to know if there is a way of making the slopes tend to infinity so that the integral is finite. If this is not possible, I would appreciate suggestions to construct a function that satisfies the conditions of the problem. Thanks in advance.

Answer 1

Hint.

Your idea is good.

The integral of $f$ is the sum of the areas of the triangles. Find the slope in order to have the area of triangle $n$ equal to $\frac{1}{n^2}$. As $\sum \frac{1}{n^2}$ converges you're done.

For more details, you can hve a look here.

Answer 2

Here's an example with an explicit formula: $f(x) = x[\cos^2(x)]^{x^5}.$

Sketch of proof: Easy part: $f(n\pi) = n\pi \to \infty.$ Harder part: From Laplace's method

$$\int_0^\pi [\cos^2x]^p\ dx \le Cp^{-1/2}$$ as $p\to \infty.$ Hence

$$\int_{n\pi}^{(n+1)\pi} x[\cos^2x]^{x^5}\ dx \le (n+1)\pi\cdot C[(n\pi)^{5}]^{-1/2} \le C'/n^{3/2}$$

as $n\to \infty.$ Because $\sum 1/n^{3/2} < \infty,$ we see

$$\int_0^\infty x[\cos^2x]^{x^5}\ dx <\infty.$$