The Taylor series of $\int_0^x \operatorname{sinc}(t) dt$  I tried to find what is the Taylor series of the function $$\int_0^x \frac{\sin(t)}{t}dt .$$
Any suggestions?

I just want to extend J.M.'s solution to a full solution of the exercise:
$$\int_0^x \frac{\sin(t)}{t} \,dt$$
Is with Maclaurin expansion ($\sin\,t=t\left(1-t^2/3!+t^4/5!-t^6/7!+\cdots\right)$) equals to
$$\int_0^x \frac{t}{t}\left(1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}+\cdots\right) \,dt = \int_0^x \left(1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}+\cdots\right) \,dt $$
We do now integrate this series to: 
$$\int_0^x \left(1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}+\cdots\right) \,dt = 
x-\frac{x^3}{3 * 3!}+\frac{x^5}{5*5!}-\frac{x^7}{7 * 7!}+\cdots$$ 
Above we did use the flowing fact:
$$\int_0^x t^k \, dt=\left.\frac{t^{k+1}}{k+1}\right|_0^x = \frac{x^{k+1}}{k+1}-\frac{0^{k+1}}{k+1} = \frac{x^{k+1}}{k+1}$$
We can now write the result as a series:
$$x-\frac{x^3}{3 * 3!}+\frac{x^5}{5*5!}-\frac{x^7}{7 * 7!}+\cdots = \sum_{n=0}^\infty (-1)^n\frac{x^{(2n+1)}}{(2n+1) * (2n+1)!} = Si(x)$$
And see that we get the Sine Integral.
