How would i evaluate an integral such as this

[Storm Butler]

Homework Statement

∫ (a^(n+1)-1)/(a-1) da

Homework Equations

i took this equation from the series of sigma with the bound of N k=0 and the argument as a^K (I'm sorry if this is confusing i don't know how to type it out and I'm not sure what all the correct terminology is).

The Attempt at a Solution

i really just don't know where to start with this one, i don't know what would be appropriate to use, by parts or substitution, and if it is by substitution what do i substitute?

 

[azure kitsune]

Are you saying that the original problem was

\int \left( \sum_{k=0}^{n} {a^k} \right) da

?

In that case, you can use the fact that ∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx and apply this over and over again to your summation.

The summation implies that n is a nonnegative integer, but if that is not the case, I don't think this function has an elementary integral.

 

[Storm Butler]

what is a non-elementary integral?
and basically i was just trying to see what would happen if the summation was switched with an integral, because that sum is (when n approaches infinite) is 2. But that's only when the numbers increase at an integer amount, i wanted to see what it would be if it increased at every amount so i thought substituting the integral in would work. (is my logic flawed)

EDIT:forgot to put in the fact that it only approaches two when a=1/2

 

[HallsofIvy]

a^{n+1}- 1= a^{n+1}- 1^{n+1}= (a-1)(a^n+ a^{n-1}+\cdot\cdot\cdot+ 1). Does that help?

 

[Дьявол]

S=a_1+qa_1+q^2a_1+q^3a_1+...+q^na_1

Sq=qa_1+q^2a_1+q^3a_1+q^4a_1+...+q^{n+1}a_1

Sq-S=q^{n+1}a_1-a_1

S(q-1)=a_1(q^{n+1}-1)

S=\frac{a_1(q^{n+1}-1)}{q-1}

Is it enough?

Regards.

 

[azure kitsune]

what is a non-elementary integral?

A non-elementary integral is one that cannot be expressed in elementary functions.

An elementary function is one that can be written in terms of "simple" functions. A more thorough definition can be found at http://mathworld.wolfram.com/ElementaryFunction.html

and basically i was just trying to see what would happen if the summation was switched with an integral, because that sum is (when n approaches infinite) is 2. But that's only when the numbers increase at an integer amount, i wanted to see what it would be if it increased at every amount so i thought substituting the integral in would work. (is my logic flawed)

EDIT:forgot to put in the fact that it only approaches two when a=1/2

So you started with

\lim_{n \to \infty} \sum_{k=0}^{n} {\left( \frac{1}{2} \right)^k} = 2

right? I'm sorry, but I don't understand what you did afterwards.

 

[Storm Butler]

after that i replaced the sigma notation with an integral