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I am learning repeated integration with the Tabular method. I see textbooks or web sites prove the method, but most of the times, they either do not explain each step or just skip with (...). I try to prove it myself by going through each step. I am not sure if my manipulations of differential "$dx$" are legitimate, and in general how to justify them.

I started with with (uv)'=u'v+vu'

$\frac {d(uv)} {dx} = \frac {du} {dx}v+ u\frac {dv} {dx}$ $\to$ (Q1: Can I simply multiply both sides by $dx$?)

$d(uv)= vdu+udv$

$\int uv =\int udv+\int vdu$

$\int udv=\int uv -\int vdu$ $ $ (I)

Set $\frac {du} {dx}=u^1$ , $\frac {dv^{-1}}{dx}=v$ $\to$ $du= u^1dx$, $v=\frac {dv^{(-1)}} {dx}$

$\int vdu= \int u^1dx$$\frac {dv^{(-1)}} {dx}$ $\to$ $\int u^1dv^{-1}$ (Q2: Can I directly cancel $dx$ from both numerator and denominator?)

$\int udv= uv -\int u^1dv^{-1} $

Repeating similar steps $\to$

$\int udv=$$\sum_{i=0}^{n-1} (-1)^iu^i v^{-i} +\int (-1)^nu^n dv^{-n}$ $ $

$\int udv=$$\sum_{i=0}^{n} (-1)^iu^i v^{-i} +\int (-1)^{n+1}u^{n+1} dv^{-(n+1)}$ $ $ (II)

which is equivalent to common expression
$\int udv=$$\sum_{i=0}^{n} (-1)^iu^i v^{-i} +\int (-1)^{n+1}u^{n+1} v^{-(n)}dx$ $ $ (II')

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  • $\begingroup$ Thanks! I missed this and corrected accordingly. $\endgroup$ – BStar Feb 27 at 23:45
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Treating dx as a fraction is technically wrong but I don't think you'll make any 'big mistakes' due to it. As in, even though it is an incorrect operation, it is one which produces correct results.

Anyways for repetead integration by parts, there is a faster derivation by a switch of notation, let $f_{i} $ denote the ith derivative of $f$, so for example $f_1$ is first derivative of $f$.

Now, we can write: $$(fg)_{-1}= f_{-1} g - (f_{-1} g_1)_{-1}= f_{-1}g -f_{-2} g_{1} + (f_{-2} g_{2})_{-1} $$

Repeating the above method n times,

$$ (fg)_{-1} = \sum_{i=1}^n f_{-i} g_{i-1}(-1)^{i+1}+ (-1)^n (f_{-n}g_{n})_i$$

refer here page-50 of the Joseph Edward book

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  • $\begingroup$ thanks! I read your paper and still do not understand your first line here. How do you extend combination formula to negative and apply here? $\endgroup$ – BStar Mar 3 at 0:18
  • $\begingroup$ whut, wdym by combination formula to negative @BrightStar? $\endgroup$ – Buraian Mar 3 at 5:33
  • $\begingroup$ I thought you use this formula : miro.medium.com/max/443/1*uVFGIVq5F_r5nWtD4-lXyg.png. $\endgroup$ – BStar Mar 3 at 5:49
  • $\begingroup$ I actually got this technique from an odl calculus book I read refer page-50 of the book $\endgroup$ – Buraian Mar 3 at 5:52
  • $\begingroup$ Got it! BTW, really like your paper, very cool:-) $\endgroup$ – BStar Mar 3 at 17:37

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