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

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

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Have you tried anything? –  Brian Fitzpatrick Apr 22 '13 at 5:59
yes i did. i got sin^2x/ (x)(1+cosx) and then simplified into sinx/x and sinx/(1+cosx) and now im lost :( –  parker Apr 22 '13 at 6:02

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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can you show us step by steps??cause im confused :( –  parker Apr 22 '13 at 6:10
Once you have this you can use the alternating series test to find a bound for the error in approximating by the $n$th partial sum. –  Brian Fitzpatrick Apr 22 '13 at 6:11
you guys are so smart lol –  parker Apr 22 '13 at 6:13
@parker Have added the intermediate details. –  user17762 Apr 22 '13 at 6:15
@parker In the summation take the first $4$ terms, i.e., $$\dfrac1{2 \cdot 2!} - \dfrac1{4 \cdot 4!} + \dfrac1{6 \cdot 6!} - \dfrac1{8 \cdot 8!}$$ to get an accuracy of $1e-5$. –  user17762 Apr 22 '13 at 6:31

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 $\frac1{4}$.

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I Understood the computation of the integral. however when i try to find the estimate for the integral that is accurate to within 10^-5, i got 1/4-1/96+1/4320-1/322560+1/36288000 approximate to 0.2398117422. is 0.239811722 accurate to within 10^-5? im just confused again..

so my question would be

1st. did i computate the integral right?

2nd. is there other way to do the integral? not in terms of summation.