How i can integrate this by using powe series
int from 0 to 2x ((e^t)-1)/(t)) dt
thanx
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.
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