How to estimate the following integral: $\int_0^1 \frac{1-\cos x}{x}\,dx$ 
How to estimate the following integral?
  $$
\int_0^1 \frac{1-\cos x}{x}\,dx
$$

 A: 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}
A: Since $\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$,
$1-\cos(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n}}{(2n)!}$,
so 
$\frac{1-\cos(x)}{x} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n-1}}{(2n)!}$.
Therefore
$\begin{align}
\int_0^1 \frac{1-\cos(x)}{x} dx
&= \int_0^1 dx \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{2n-1}}{(2n)!}\\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{(2n)!} \int_0^1 x^{2n-1} dx\\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{(2n)!} \frac1{2n} \\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n+1} }{2n(2n)!}\\
\end{align}
$
This is an alternating series with terms decreasing in absolute value,
so its sum is between any two consecutive terms.
The first two terms are
$\frac1{4}$ and $\frac{-1}{4\cdot 4!} = \frac{-1}{96}$,
so the result is slightly less than $\frac1{4}$.
