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Let $\displaystyle P_n (x) = 1 + \frac{x}{1!} + \frac{x^2 }{2!} + \cdots + \frac{x^n }{n!} \ $ and $$ I(x) = \int \frac{2n!\sin x + x^n }{e^x + \sin x + \cos x + P_n (x)}\, dx $$ (where $\ n \to \infty \ $).

This problem is REALLY frustrating to me at the moment, it's 6 AM here and I've been trying to sort it out since 4:30 AM. First of all, I don't get the use of the $P_n(x)$ notation, isn't that just $e^x$ ? Anyhow...None of my approaches yielded any useful results, so I'm reaching out to you. Can someone suggest anything, at all ?

It would be much appreciated, thanks a lot!

EDIT : I managed to solve part of it, I'm now stuck with $ I(x) = n!(x - \log [2e^x + \sin (x) + \cos (x)] + \int {\frac{{x^n }}{{e^x + \sin x + \cos x + P_n (x)}}} dx $.

Can't really figure out if this is much better, but that's all I could get up until this point.

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    $\begingroup$ $$\lim_{n\to\infty}P_n(x)=e^x$$ $\endgroup$ – lab bhattacharjee Sep 14 '14 at 3:01
  • $\begingroup$ Uh yeah I already mentioned that in the body of my question $\endgroup$ – Victor Sep 14 '14 at 3:05
  • $\begingroup$ Just to clarify, is that $(2n)!$ or $ 2(n!)$? $\endgroup$ – ClassicStyle Sep 14 '14 at 3:14
  • $\begingroup$ Hey @TylerHG that's $2(n!)$ $\endgroup$ – Victor Sep 14 '14 at 3:14
  • $\begingroup$ That last integral can be simplified to $$\int \frac{x^n}{2 e^x+\sqrt2 \sin(x+\frac{\pi}{4})}$$ but I don't know if it really is a simplification. $\endgroup$ – UserX Sep 14 '14 at 3:47
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This is a tricky integral!

Let's prove that

$$ \begin{align} \int \frac{2n!\sin x + x^n }{e^x + \sin x + \cos x + P_n (x)} {\rm d} x & = n! \:x-n!\:\log |e^x+\cos x+\sin x+P_n(x)| +C, \end{align} $$

where $C$ is any constant (depending on $n$).

Observe that $$ P_n (x) = 1 + \frac{x}{1!} + \frac{x^2 }{2!} + \cdots + \frac{x^n }{n!} $$ is such that $$ P_n '(x) = 1 + \frac{x}{1!} + \frac{x^2 }{2!} + \cdots + \frac{x^{n-1} }{(n-1)!}= P_n (x) -\frac{x^n}{n!}. $$ Setting $$ f(x):=e^x+\cos x+\sin x+P_n(x) $$ we then have $$ f'(x)=e^x-\sin x+\cos x+P_n(x)-\frac{x^n}{n!} $$ and $$ f(x)-f'(x)=2\sin x+\frac{x^n}{n!} $$ Hence your integral may be rewritten as $$ \begin{align} \int \frac{2n!\sin x + x^n }{e^x + \sin x + \cos x + P_n (x)} {\rm d} x &= n! \int \frac{f(x)-f'(x) }{f(x)}{\rm d} x \\\\ &= n! \int {\rm d} x-n! \int \frac{f'(x) }{f(x)}{\rm d} x \\\\ & = n! \:x-n!\:\log |f(x)| +C \\\\ & = n! \:x-n!\:\log |e^x+\cos x+\sin x+P_n(x)| +C, \end{align} $$ where $C$ is a constant (depending on $n$).

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    $\begingroup$ Wow!That was really, I mean really cool.. +1 $\endgroup$ – Victor Sep 14 '14 at 12:35
  • $\begingroup$ Awesome answer +1 $\endgroup$ – Mike Miller Dec 30 '14 at 22:47
  • $\begingroup$ impressive :) (+1) $\endgroup$ – tired Feb 11 '15 at 12:09
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Is the $n\rightarrow\infty$ apply to the $n$ in the integral numerator as well as to $P_{n}(x)$? If so, then my guess is that either this is an incorrectly posed problem or the answer is totally divergent to infinity due to the $n!$ in the numerator. I have been working to find ways to get rid of this but the first integral shows that the first part diverges as $n\rightarrow\infty$.

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