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I am given the following problem :-

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I know that I have to first find the Taylor series of the polynomial and then integrate each term, however I am having trouble finding the Taylor series before integrating because the derivative of $\frac{(e^t-e)}{(t-1)}\to \frac{e^t(t-1)-1(e^t-e)}{(t-1)^2}$ at $t=1$ does not exist $\frac{0}{0}$. How do I go about finding the Taylor series of $\frac{(e^t-e)}{(t-1)}?$

Thanks for any and all help!

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  • $\begingroup$ The integrand is only undefined at $x=0$, in a removable way, so its antiderivative is defined everywhere and turns out to be differentiable. $\endgroup$
    – user65203
    Commented Aug 5, 2020 at 14:57

1 Answer 1

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Hint:

Set $t=1+u\enspace(u\to 0)\,$ first. The integrand becomes $$\frac{\mathrm e^t-\mathrm e}{t-1}=\mathrm e\,\frac{\mathrm e^u-1}u.$$ Can you take it from here?

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  • $\begingroup$ Sorry, I'm still a little bit confused. If I take the derivative of this function for the first term in the taylor series, wouldn't u^2 be the denominator (leading to a problem when you plug in 0)? $\endgroup$
    – Meemer222
    Commented Aug 5, 2020 at 15:11
  • $\begingroup$ Why do you want to take the derivative? The numerator has a well-known expansion, and everything simplifies. If you compute the derivative and consider the expansion of the numerator , the terms with order less than $2$ disappear, so it's not indeterminate. $\endgroup$
    – Bernard
    Commented Aug 5, 2020 at 15:20
  • $\begingroup$ Oh! I guess I am a bit confused about rules of dividing and adding taylor series... what gives us the ability to just find the taylor series of e^u and then subtract 1 from it and divide it by u ?(I was under the impression that we would have to find the taylor series for each term and then do operations on them (subtract/add/divide)) $\endgroup$
    – Meemer222
    Commented Aug 5, 2020 at 16:10
  • $\begingroup$ Never mind, I get it now! Thanks for all the walking through! $\endgroup$
    – Meemer222
    Commented Aug 5, 2020 at 16:16

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