# Simple proof of integration formula from geometric series

• Mar 6th 2010, 11:03 AM
zg12
Simple proof of integration formula from geometric series
The last 2 lines of the proof for the formula for integration of x^a, for positive a, not the second half which deals with negative alpha. I don't understand how he gets that sum from (q-1)/(q^(1+a)-1)). And then I don't know how he gets the final result 1/(1+a). The attached file is from page 133 of Courant and John's Intro to analysis and calculus.
• Mar 6th 2010, 02:35 PM
HallsofIvy
Presumably you know that the sum of a geometric series $\sum_{n= 1}^\alpha ar^n= \frac{a(1- r^{\alpha+ 1}{1- r}$. With a= 1, r= q, that is $\sum_{n=1}^{\alpha} q^n= \frac{q^{\alpha+1}- 1}{q- 1}$.

Taking the reciprocal of both sides gives
$\frac{1}{\sum_{n=1}^{\alpha} q^n}= \frac{q-1}{q^{\alpha+1}- 1}$
the next to the last formula. Notice that there are $\alpha+ 1$ terms in that sum. As the book says, as n goes to infinity, all those powers of q go to 1 so we are adding 1 $\alpha+ 1$ times- the sum is $\alpha+ 1$ so the left side is $\frac{1}{\alpha+1}$.

Putting that into the fourth formula on the page,
$F_n= (b^{\alpha+ 1}- a^{\alpha+1})q^\alpha\frac{q- 1}{q^{\alpha+1}- 1}$
gives the final result.