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I am trying to determine whether the integral $$\int_0^1 \frac{\sin x}{x} dx$$ can be calculated analytically.

I am aware of the definition of the sine integral function $\text{Si}(x)$, but I haven't been able to find a reference stating whether such an integral can be calculated without resorting to numerical methods.

The usual methods employed to compute the improper integral over the real line, such as via contour integration or by defining $$G(t) = \int_0^1 \frac{\sin x}{x}e^{-tx} dx$$ do not seem to work here.

Can someone give me a hint here?

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  • $\begingroup$ What's wrong with $\operatorname{Si}$? Yes, it needs to be calculated using "numerical methods". But so does everything else. $\sin$ is calculated using a power series. $\pi$ is calculated with an infinite series. I don't see how they are any different to something like $\operatorname{Si}$. $\endgroup$ – K.defaoite Sep 18 '20 at 8:41
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Since the range of integration is restricted to $[0, 1]$ you can make use of Taylor Series for the sine function:

$$\sin(x) = \sum_{k = 0}^{+\infty} (-1)^{k}\frac{x^{2k+1}}{(2k+1)!}$$

Hence

$$\sum_{k = 0}^{+\infty} \dfrac{(-1)^{k}}{(2k+1)!} \int_0^1 \dfrac{x^{2k+1}}{x}\ \text{d}x$$

The integration is trivial and gives you

$$\sum_{k = 0}^{+\infty} \dfrac{(-1)^{k}}{(2k+1)!} \frac{1}{2 k+1}$$

Then you can go ad libitum with numerical.

Notice that that series is nothing but the expression of the series of \text{Si}(1) (SineIntegral).

The numerical value is, for the sake of completeness, $0.946083(...)$

The value of the series, for some values of $k$ is:

$$k = 2 \to 0.94611$$ $$k = 5 \to 0.946083$$

It takes few terms to obtain a high numerical precision.

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$$\int\limits_0^1 \frac{\sin x}{x} dx=\int\limits_0^1\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-...\right)dx=1-\frac{1}{3\cdot3!}+\frac{1}{5\cdot5!}-...$$

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Using @Turing's answer, it could ba amaring to know the numbers of terms to be added for a given accuracy.

Writing $$\sum_{k = 0}^{+\infty} \dfrac{(-1)^{k}}{(2k+1)(2k+1)!}= \sum_{k = 0}^{p} \dfrac{(-1)^{k}}{(2k+1)(2k+1)!}+\sum_{k = p+1}^{+\infty} \dfrac{(-1)^{k}}{(2k+1)(2k+1)!} $$ since it is al alternating series, we search for $p$ such that $$\dfrac{1}{(2p+3)(2p+3)!}\leq \epsilon \implies (2p+3)(2p+3)!\geq \frac 1\epsilon$$ We shall simply the problem (since we look for an approximation) to $$(2p+4)!\geq \frac 1\epsilon$$

If you have a look here, you will find a superb approxiamtion of the inverse factorial. Applied to the present case, this would give

$$p \simeq -\frac{\log \left(\sqrt{2 \pi } \epsilon \right)}{2 W\left(-\frac{\log \left(\sqrt{2 \pi } \epsilon \right)}{e}\right)}-\frac{9}{4}$$ where $W(.)$ is Lambert function.

Since the argument is quite large, you can use the approximation $$W(t)=L_1-L_2+\frac{L_2}{L_1}+\frac{(L_2-2) L_2}{2 L_1^2}+\frac{(2 L_2^2-9L_2+6) L_2}{6 L_1^3}+ ...$$ where $L_1=\log(t)$ and $L_2=\log(L_1)$ and, for sure, you will use $\lceil p\rceil$.

For $\epsilon=10^{-10}$, this would give $p=4.59018$ that is to say $6$. Computing, this would give $$\dfrac{1}{13 \ 13!}=1.235\times 10^{-11}$$

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