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This is NOT the same topic as the post "Infinite Integration by Parts", though it is similar.

I've noticed some truly fascinating results of expressing infinite integration by parts as an infinite series. For example,

$$\int \sin(x)\,dx = C +\sum_{n=1}^\infty{(-1)^{n-1}\frac{x^{2n-1}\sin(x)}{(2n-1)!}+(-1)^n\frac{x^{2n}\cos(x)}{(2n)!}}$$

Simply decrementing the index by one gets you

$$C+\sum_{n=0}^\infty{(-1)^n\frac{x^{2n+1}\sin(x)}{(2n+1)!}+(-1)^{n+1}\frac{x^{2n}\cos(x)}{(2n)!}}=-\cos(x)+C+\sum_{n=1}^\infty{(-1)^{n-1}\frac{x^{2n-1}\sin(x)}{(2n-1)!}+(-1)^n\frac{x^{2n}\cos(x)}{(2n)!}}$$ The sum is convergent for any finite value of $x$, and the antiderivative of $\sin(x)$ is indeed $-\cos(x)+C$. I tried the same process on a few other functions that have elementary antiderivatives, with similar results.

Here's where things get really interesting. For $\frac{1}{\log(x)}$, whose antiderivative is not elementary, I got $$\frac{x}{\log(x)}+C+\sum_{n=1}^\infty{\frac{(n-1)!x}{\log^{n}(x)}}$$ and for $x$, which is not infinitely differentiable, I got $$C+x^2 \sum_{n=0}^\infty{(-1)^n}$$ by this same process of infinite integration by parts and decrementing the initial index of the resulting series.

Various summation methods for divergent series, such as Cesaro summation, assign $\sum_{n=0}^\infty{(-1)^n}$ a value of $\frac{1}{2}$, and $\frac{1}{2}x^2$ is the antiderivative of $x$. Notice also that the zeroth term of the series for $\frac{1}{\log(x)}$ is $\frac{x}{\log(x)}$, and that $\frac{x}{\log(x)}$ ~ $li(x)+C = \int \frac{dx}{\log(x)}$.

It appears that this process yields convergent series for functions with elementary antiderivatives and divergent series for those without them (I tried it on other functions with non-elementary antiderivatives as well, such as $x^x$.) Maybe a divergent summation method for $\sum_{n=1}^\infty{\frac{(n-1)!x}{\log^{n}(x)}}$ could yield a function that equals $li(x) - \frac{x}{\log(x)}$, which I'm sure would be of intense interest to number theorists, if it hasn't already been found. Has any of this been investigated by anyone else? Is it actually of any use, or am I seeing patterns where they don't exist?

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The hypothesis that repeated integration by parts yields a divergent series for non-elementary functions is false, since it yields a convergent one for $e^{x^2}$. It would perhaps be more fruitful to try repeated integration by parts on $\sin^{m}{x}$ since it yields some interesting results (Wallis formula etc).

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  • $\begingroup$ I will correct this and provide my exact method soon. I apologise. $\endgroup$ Dec 7, 2017 at 22:46
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    $\begingroup$ @IsaacSaffold There's an out of copyright treatise on integral calculus available on the internet, written in the days when calculating hard integrals by hand was an artform. You will find many interesting results on the integrals of trigonometric/hyperbolic functions there, and some remarkable techniques. Search for A Treatise On The Integral Calculus - Joseph Edwards. $\endgroup$
    – Jack
    Dec 7, 2017 at 22:55
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    $\begingroup$ @IsaacSaffold Okay, I see you knew the one about the log was divergent. However, the hypothesis that infinite/repeated integration by parts yields a divergent series for non-elementary functions is false, since it yields a convergent one for $e^{x^2}$. $\endgroup$
    – Jack
    Dec 7, 2017 at 23:02
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    $\begingroup$ Ah. Thank you. I haven't tested this on all that many functions. I figured some of my hypotheses were wrong. That's one of the reasons I posted this. $\endgroup$ Dec 7, 2017 at 23:05
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    $\begingroup$ @IsaacSaffold No worries. I think any infinite series you find by repeated integration by parts will be the usual Maclaurin/Taylor series in disguise, since it's one of the ways of deriving Taylor's formula. So it might be more economical to just use series expansion straight away and integrate term by term if you want to tinker with it. $\endgroup$
    – Jack
    Dec 7, 2017 at 23:41

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