. Since is continuous over our interval, is bounded. Thus as .
Hence .
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This question was rather deceiving! I was using very complicated formulas for and before the hint.
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Random observation: . The whole time I was working on this problem I thought the answer was because of the decimal approximation Wolfram|Alpha gave me. Weird coincidence!
This proof doesn't work in general since is continuous does not imply that exists or even if it is differentiable it need not be bounded! Take with the obvious continuation of . It does work though if you consider . Try coming up with an alternative.Or, proving that . Meanwhile, here is my solution (probably won't be the same as yours, but works out fine):
Spoiler:
Those sums are known to be easily calculated with the method of residues in complex analysis, and I assume you didn't mean to use this method. I've never seen them calculated in other ways nor do I get an idea of another way from their values. I will, however, keep thinking about doing them in a method not involving complex analysis. It must be interesting to calculate them in such a way.
Magic differentiation kills it
We want
For this, note the integral is where
Calculated by Weierstrass substitution ( ).
To calculate the integral of put and you'll easily get:
So our integral equals with
Yes, the obvious way is to use CA. One question, if you know CA how do you not know multiple integrals?
Woah, your answer looks completely different than mine. Are you sure it's correct? Not doubting your, but it's just really different.
Magic differentiation kills it
We want
For this, note the integral is where
Calculated by Weierstrass substitution ( ).
To calculate the integral of put and you'll easily get:
So our integral equals with