Series Problem Help: Evaluating \int \frac{\sin x}{x} with Convergence Analysis

[courtrigrad]

I want to evaluate $$\int \frac{\sin x}{x}$$.

So $$\sin x = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{(2n+1)!}$$. Therefore $$\frac{\sin x}{x} = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n}}{(2n+1)!}$$. So would that mean:

$$\int \frac{\sin x}{x} = C + \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{2n+1}}{2n+1(2n+1)!}$$ would be absolutely convergent (i.e. $$R = \infty$$)?

Thanks

 

[StatusX]

How do you know that series is absolutely convergent?

 

[courtrigrad]

I would use the ratio test $$|\frac{ a_{n+1}}{a_{n}}|$$. If the limit as $$n\rightarrow \infty$$ is less than 1, then the series is absolutely convergent. Ok so I guess it is then.