Improper integral of $\frac{1}{x} \sin \frac{1}{x}$

Question

Let $f(x)= \frac{1}{x} \sin \frac{1}{x}$ when $x \in (0, 1]$and $f(0)=0$. Prove that $\int_{0}^1 f(x)dx$ exists.

Can someone give me a hint to solution?

@zhw. Sorry I'm tired :( – Hippalectryon Jan 14 '16 at 20:50 — Hippalectryon, Jan 14 '16 at 20:50

score 3 · Accepted Answer · answered Jan 14 '16 at 20:47

3

Hint: Make the change of variable $x = 1/y.$

answered Jan 14 '16 at 20:47

zhw.

Transforms to $\int_1^{\infty} \sin x / x dx < \infty$ +1 – RRL Jan 14 '16 at 20:53
Oh, thanks xD I see – mrnobody Jan 14 '16 at 20:56

1 Answers1