First, we should notice that $$\sum_{m=1}^{\infty}\frac{1}{x-\pi n} +\sum_{n=0}^{\infty}\frac{1}{x+\pi n}=\cot (x)$$ (I can no longer find the proof of this, sorry) and that if we take the $a$'th derivative of both sides we get $$\Gamma(a+1)(-1)^a[\sum_{m=1}^{\infty}\frac{1}{(x-\pi n)^{a+1}}+\sum_{n=0}^{\infty}\frac{1}{(x+\pi n)^{a+1}}]=\frac{d^a}{dx^a}(\cot x)$$ This is a definition of a fractional derivative of $cot(x)$, but the question I have is does this work for all real numbers? I know it works for rational numbers that have an odd denominator. Thank you.
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Certainly the fractional derivative doesn't exist over each division by zero. – Simply Beautiful Art May 01 '17 at 23:54
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2You have to add in $-1/(\pi n)$ to each term with $n \neq 0$ in the sum or it doesn't converge. Or you can take $$\frac{1}{x} + \sum_{n=1}^{\infty} \left( \frac{1}{x+\pi n} + \frac{1}{x-\pi n} \right),$$ which does converge. – Chappers May 01 '17 at 23:56
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Well $cot(x)$ has asymptotes so it makes sense thst there are divisions by 0, right? – Jacob Claassen May 01 '17 at 23:56
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@Chappers, how is the second suggestion different from what I did? I don't see it. – Jacob Claassen May 01 '17 at 23:59
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It's the same as the difference between $$\int_{-\infty}^{\infty} \frac{dx}{x-i},$$ which does not converge, and $$ \int_0^{\infty} \left( \frac{1}{x-i}+\frac{1}{-x-i} \right) dx = \int_0^{\infty} \frac{2i , dx}{x^2+1}, $$ which does: when the series is not absolutely convergent, the order in which you add the terms matters. – Chappers May 02 '17 at 00:04
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More abstractly, $ \lim_{M \to \infty} \int_{-M}^M f(x) , dx $ can exist when $ \lim_{\substack{M \to \infty \ N \to \infty}} \int_{-M}^N f(x) , dx$ does not. It is the latter that is the definition of an improper integral. – Chappers May 02 '17 at 00:07
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Huh, I've never come across this before. Thank you. – Jacob Claassen May 02 '17 at 00:08
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@Chappers, I just noticed, you're actually wrong on your example, $\int_0^{\infty}\frac{2i}{x^2+1}$ diverges – Jacob Claassen May 02 '17 at 04:22
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@JacobClaassen: Chappers is not wrong, $\int_{0}^{+\infty}\frac{2i}{x^2+1},dx = \pi i.$ – Jack D'Aurizio May 02 '17 at 08:43
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I just plugged it into wolfram alpha. My bad, I'm wrong. – Jacob Claassen May 02 '17 at 14:46
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This problem might be better approached by way of the exponential definition of the cotangent. The fractional calculus of the exponential is well established and is described in detail by Oldham & Spanier, The Fractional Calculus, Dover, Chapter 6. – Cye Waldman May 02 '17 at 16:40