# How to find ${\lim\limits_{x\to 0^+} \frac{1}{x} \int_0^{x} \sin(\frac{1}{t})}\,\mathrm dt$

$$\def\d{\mathrm{d}}$$There was a hint in the book, use intregation by parts in this way: $$\lim_{x\to 0^+} \frac{1}{x} \int_0^{x} \sin\frac{1}{t} \,\d t = \lim_{x\to 0^+} \frac{1}{x} \int_0^{x} t^2 \left(\frac{1}{t^2} \sin\frac{1}{t}\right)\,\d t.$$

When we integrate by parts we find this integral:

$$\int_0^{x} t\cos\frac{1}{t}\,\d t .$$

In every symbolic calculator it says is a special function and gives the value of the integral as $$\mathrm{Ci}(x)$$, but I'm using calculus 1 knowledge, any hint or help? Thanks in advance.

• What is the full expression (including the limit) that you get after the integration by parts? Can you see how this might be nicer? – B. Mehta Feb 12 '18 at 0:20

## 2 Answers

You don't need to compute the integral, you need only to show that the average is tending towards zero. Notice that

$$\left|\frac 1 x \int_0^x t \cos \frac 1 t \, dt\right| \le \frac 1 x \int_0^x t \, dt = \frac x 2 \to 0$$

as desired. Hence, the limit is zero.

P.S. Don't forget about the boundary terms when you integrate by parts. They're easy to deal with, but you still have to!

Also, just to be sure, here is a rough idea of why we could know the limit is zero in advance. The integral $\frac 1 x \int_0^x f(t) \, dt$ is the average of $f$ on $[0, x]$. If $f$ is oscillatory like $\sin 1 / t$ is, and spends an equal amount of time above and below the axis, the average is zero.

\begin{align*} \dfrac{1}{x}\int_{0}^{x}\sin(1/t)dt&=-x\cos(1/x)+\dfrac{2}{x}\int_{0}^{x}t\cos(1/t)dt\\ &=-x\cos(1/x)+2\cdot\eta_{x}\cos(1/\eta_{x}), \end{align*} by Integral Mean Value Theorem, now note that $0<\eta_{x}<x$ and hence $\eta_{x}\rightarrow 0$ as $x\rightarrow 0$ and appeal to Squeeze Theorem to that $|\eta_{x}\cos(1/\eta_{x})|\leq\eta_{x}$.