Evaluate integral of t/(1 - t^8)dt as a power series.

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MarkSA

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Hello,

Evaluate the integral as a power series. What is the radius of convergence?

I have a problem:
Indefinite Integral of: t/(1 - t^8)dt

Ok, I tried this. I started with 1/(1 - t) = summation n=0 to infinity of: t^n
So, t/(1 - t^8) = t * 1/[1 - (t^8)] and in the same form as the one above. So:
t * 1/[1 - (t^8)] = summation n=0 to infinity of: t^(8n + 1)
At this point if I take the integral of both sides of the equation, I end up with:
Indefinite Integral of: t/(1 - t^8)dt = Summation from n=0 to infinity of: t^(8n + 2)/(8n + 2) + C
This is where my question arises.

How would I solve for C, and if it is necessary to solve for it?
 
Re: Evaluate the integral as a power series.

Basically in such problems as these: there ain't no C.
The history of the “C” is interesting. Antiderivatives of a derivative differ by a constant. So in very basic courses we use it out of habit.

Now recall that you started with a geometric series so |t|<1.
 
Ok thanks, so in these types of problems, I should just ignore C? My TA said that it will usually be zero, but sometimes not.

And since I started with a geo series, 1/1-x represented as a power series, it is unnecessary for me to run the ratio test on it, for instance? I can simply say that it's convergent when |t|<1 and the interval of convergence is (-1,1), Radius=1?
 
Instead of everyone proffering their opinion, will someone who is more erudite than I solve this problem?

I can expand f(x) = t/(1-t^8) as a power series , but the integral?
 
Dr. Flim-Flam said:
Instead of everyone proffering their opinion, will someone who is more erudite than I solve this problem? I can expand f(x) = t/(1-t^8) as a power series , but the integral?
Click to expand...
Well, you are really living up to your screen name.
No one is proffering his or her opinion: fact is fact.
Do you understand that an integral is a number? Many people misuse the word ‘integral’ when what is really meant is 'antiderivative'.
If you had carefully read the original posting you would have realized that the series for the antiderivative is given there: “Summation from n=0 to infinity of: t^(8n + 2)/(8n + 2)”
Or that is \(\displaystyle \sum\limits_{k = 0}^\infty {\frac{{t^{8n + 2} }}{{8t + 2}}}\) which is only valid in the sect \(\displaystyle (-1,1)\).
The question was about the archaic use of the “C” in an antiderivative.
Because antiderivatives differ by at most a constant, there are applications where boundary conditions require that we find that constant.
But in the context of the particular question, the series would be used to evaluate an integral.
 
Dr. Flim-Flam said:
You are all wet.
Click to expand...
To the other contributors to this thread, I apologize for the slightly-hostile tone of the one person.

To the original poster, please reply with any other questions you might have on this thread (the mathematical part, anyway). Or, if your question has been answered to your satisfaction, please let us know, so we can "lock" this thread and prevent any further unpleasantness.

Thank you! :D

Eliz.
 
t/(1-t^8) = -t^-7 -t^-15 -t^-23 -t^-31 +... -t^(1-8n) = Sum -t^(1-8n), n goes from 1 to infinity.


Integral of {Sum -t^(1-8n),n goes from 1 to infinity = Sum [t^(2-8n)/(8n-2)] +C, n goes from 1 to infinity,

Letting t = 0 implies C = 0, Hence integral[t/(1-t^8)] = Sum[t^(2-8n)/(8n-2)], n goes from 1 to infinity,

Interval of convergence = (-infinity,-1) union (1,infinity).
 
Dr. Flim-Flam said:
t/(1-t^8) = -t^-7 -t^-15 -t^-23 -t^-31 +... -t^(1-8n) = Sum -t^(1-8n), n goes from 1 to infinity.


Integral of {Sum -t^(1-8n),n goes from 1 to infinity = Sum [t^(2-8n)/(8n-2)] +C, n goes from 1 to infinity,

Letting t = 0 implies C = 0, Hence integral[t/(1-t^8)] = Sum[t^(2-8n)/(8n-2)], n goes from 1 to infinity,

Interval of convergence = (-infinity,-1) union (1,infinity).
Click to expand...

uhh ... no.

Once again, read the original post ... the power series for the original expression is
\(\displaystyle \frac{t}{1 - t^8} = \sum_{n=0}^{\infty} t^{8n+1}\)
 
skeeter, you summation (after taking the integral) will coverge from (-1,1)

Mine will have a inteval of convergence from (-inf,-1) union [0] union (1,inf).

Hence we can find the indefinite integral by using one or the other summations.
 
How would I solve for C, and if it is necessary to solve for it?

Let t = 0, then 0 = 0 +C, C = 0.
 
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