<s>[INST] How to estimate the following integral: $\int_0^1 \frac{1-\cos x}{x}\,dx$ 
How to estimate the following integral?
  $$
\int_0^1 \frac{1-\cos x}{x}\,dx
$$

 [/INST] First note that the integral exists since $$0 \leq \dfrac{1-\cos(x)}x = \dfrac{2 \sin^2(x/2)}x \leq \dfrac{x}2$$ Hence, the integral is between $0$ and $1/4$. To compute the integral, proceed as follows. We have
$$\cos(x) = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} \mp = \sum_{k=0}^{\infty}(-1)^k \dfrac{x^{2k}}{(2k)!}$$
Hence,
$$1-\cos(x) = \dfrac{x^2}{2!} - \dfrac{x^4}{4!} + \dfrac{x^6}{6!} \pm = \sum_{k=1}^{\infty} (-1)^{k-1} \dfrac{x^{2k}}{(2k)!} $$
This gives us
$$\dfrac{1-\cos(x)}x = \sum_{k=1}^{\infty} (-1)^{k-1} \dfrac{x^{2k-1}}{(2k)!} $$
Now lets get back to the integral.
\begin{align}
\int_0^1 \dfrac{1-\cos(x)}x dx & = \int_0^1 \sum_{k=1}^{\infty} (-1)^{k-1}\dfrac{x^{2k-1}}{(2k)!} dx = \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{(2k)!}\int_0^1 x^{2k-1} dx\\
& = \sum_{k=1}^{\infty}\dfrac{(-1)^{k-1}}{(2k)!} \cdot \dfrac1{2k}
\end{align}
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