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.
Presumably you know that the sum of a geometric series. With a= 1, r= q, that is
.
Taking the reciprocal of both sides gives
the next to the last formula. Notice that there areterms 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
.
Putting that into the fourth formula on the page,
q^\alpha\frac{q- 1}{q^{\alpha+1}- 1})
gives the final result.
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