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?

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Recall the series for $\sin\,t$, divide out $t$ for each term, and integrate each term. – J. M. Dec 31 '11 at 12:27
Pls try to format your math, it's not difficult to learn (right-click->"show source" over the equation to see) – leonbloy Dec 31 '11 at 13:11
@J.M. I think you should put your comment as an answer. – a.r. Dec 31 '11 at 16:27
Yes, JM this one belongs to you. I shall defer. – ncmathsadist Dec 31 '11 at 19:08
Sometimes "sinc" means this one $\mathrm{sinc}(x) = \sin(\pi x)/(\pi x)$. – GEdgar Jan 9 '13 at 14:28

To settle this:

We have the Maclaurin expansion

$$\sin\,t=t\left(1-\frac{t^2}{3!}+\frac{t^4}{5!}-\frac{t^6}{7!}+\cdots\right)$$

Upon obtaining the expansion of $\dfrac{\sin\,t}{t}$ from this, integrate each term of this series expansion, using the formula

$$\int_0^x t^k \mathrm dt=\frac{x^{k+1}}{k+1}$$

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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.

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