# 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

$$\left|\frac 1 x \int_0^x t \cos \frac 1 t \, dt\right| \le \frac 1 x \int_0^x t \, dt = \frac x 2 \to 0$$
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 $|\eta_{x}\cos(1/\eta_{x})|\leq\eta_{x}$.