Express the indefinite integral as a power series.

[Sabricd]

Hello,

I'm kind of stuck in this problem. I have to express the integral as a power series.

the integral of (e^x -1)/x

I thought about evaluating it as f(x)=(e^x -1)/x and treating it as a Taylor series is that correct? Could I have any other hints?

I would really appreciate it!
-Sabrina

 

[rock.freak667]

Rewrite it as e<sup>x[/su]/x - 1/x

and you know the series form for ex.</sup>

 

[Sabricd]

I'm sorry I'm not sure I understood that :(

-Sabrina

 

[Sabricd]

Would it be pointless to treat it as f(x)=(e^x -1)/x and then take its fourth derivative and use Taylor series then?

 

[rock.freak667]

Would it be pointless to treat it as f(x)=(e^x -1)/x and then take its fourth derivative and use Taylor series then?

It would be the same thing, but it would take more work to find the series for (e^x-1)/x than taking subbing the series for e^x into

∫(e^x/x - 1/x) dx

 

[adpc]

This one doesn't have an antiderivative so you have to express first (exp(x)-1)/x as a Taylor series and then integrate term by term, and you have to make sure that the resultant series converges.

You already know the series expansion of exp(x), so it's easy to find the series of exp(x)/x just divide each term by x, and then substract the series expansion of 1/x.

 

[Sabricd]

Hi,
OK. So would it be correct if I do:
f(x)=(e^x-1)/x
f(x)=e^x/x - 1/x
\Sigma(e^x)/x -\Sigma(1/x)

(1/x)\Sigma(e^x) -\Sigma(1/x)
(1/x)\Sigma(x^n/n!) - \Sigma(1/x)

...is this correct? that's what I have so far and I'm not really sure if it's right.

Thank you,
-Sabrina