how to integrate $\int \frac{\sin x}{x}$ in $[0,1]$

Question

I would appreciate if somebody could help me with this problem $$\int_{0}^{1}\frac{\sin{x}}{x}dx$$ here using Taylor series I got $\sum_{0}^{\infty} $$\frac{(-1)^n}{(2n+1)!(2n+1)} $ then what to do? is it the final answer ? please explain

So this is a problem from the book where you are supposed to find the exact value? — imranfat, Feb 03 '14 at 17:08
possible duplicate of What is the integral of function $f(x) = (\sin x)/x$ — qwr, Feb 03 '14 at 17:16

score 1 · Accepted Answer · answered Feb 03 '14 at 16:58

1

You can first expand $\frac {\sin x} x$ as a Taylor series and then integrate that, or you can check out this link:

http://mathworld.wolfram.com/SineIntegral.html

answered Feb 03 '14 at 16:58

bgfriend0

246

1

That's $\large {\rm Si}\left(1\right)$ where $\large{\rm Si}\left(z\right)$ is the Sine Integral function. – Felix Marin Feb 03 '14 at 17:05

how to integrate $\int \frac{\sin x}{x}$ in $[0,1]$

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