1. ## Power series integrastion

How i can integrate this by using powe series

int from 0 to 2x ((e^t)-1)/(t)) dt

thanx

2. ## Re: Power series integrastion

is this $e^t-1$ or $e^{t-1}$ ?

3. ## Re: Power series integrastion

The first one (e^t)-1

4. ## Re: Power series integrastion

Originally Posted by Haytham1111
The first one (e^t)-1
so you want

$\displaystyle{\int_0^{2x}}\dfrac{e^t-1}{t}~dt$

First thing you do is express the integrand in terms of a power series.

$e^t-1=\displaystyle{\sum_{k=1}^\infty}\dfrac {t^k}{k!}$

so

$\dfrac {e^t-1}{t} = \displaystyle{\sum_{k=1}^\infty}\dfrac {t^{k-1}}{k!}$

now just integrate that term for term. I leave this to you.

5. ## Re: Power series integrastion

The express u did is just for (e^t) and not for (e^t)-1 . I got the wrong answer

6. ## Re: Power series integrastion

Originally Posted by Haytham1111
The express u did is just for (e^t) and not for (e^t)-1 . I got the wrong answer
did you note that the series starts at $k=1$ and not $k=0$ ?

7. ## Re: Power series integrastion

What is that mean?

8. ## Re: Power series integrastion

When u have e^t -1. Is not mean that u need to trekk 1 from each ledd in the series?

So it will be ((x^n)/n!) - 1 = ((x^n-n!))/n!. I understand it like that

And when u make reindeksing, Then thats mean that u put n-1 in the serie for each n, Then you will start from n=1 and not n=0

9. ## Re: Power series integrastion

Ja ja. I understood it noe. You remove just one from Hole series, and not from each ledd. Then u r right, and i got the right answer. Tanx very much

10. ## Re: Power series integrastion

Can u tell me how u can write the series index? I use the ipad, do u know an another way to write it?