An integral rewritten (from “Almost impossible integrals“, p.59 in Valean)

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Pnin
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I want to understand where the minus 1 in the first line in the RHS term comes from.
I assume the little apostrophe means taking a derivative. But the antiderivative of x^(n-1) is (1/n)x^n. Why the -1?

thank you

Handwritten-style derivation from a printed text: integration by parts of I_n = integral from 0 to 1 of x^(n-1) log(1-x) dx, rewritten as (1/n) times the integral of (x^n - 1)' log(1-x), then a boundary term that vanishes, leading to -(1/n) sum of 1/k from k=1 to n equals -H_n/n.
 
It has ##(1-x^n)## in the numerator while the previous expression has ##(x^n-1)'##.
 
This I understand. But I do not understand the first line.

How does the -1 come up in the RHS giving the LHS in the first line?
 
It's a trick. The -1 vanishes upon differentiation. You could have written any number there instead.
 
SchroedingersLion said:
It's a trick. The -1 vanishes upon differentiation. You could have written any number there instead.

Ok. I never saw that trick before. But if I had written another number I would not have gotten the desired result, the harmonic series. So is that really a legal trick which allows you getting different values from a definite integral?
 
Pnin said:
Ok. I never saw that trick before. But if I had written another number I would not have gotten the desired result, the harmonic series. So is that really a legal trick which allows you getting different values from a definite integral?
Well, me neither, but this is what happens. From the first integral to the second integral, they simply rewrite ##x^{n-1}## as ##\frac 1 n (x^n - C)'## with ##C## as a constant (i.e. 1 in your case). If you take the derivative of ##\frac 1 n (x^n - C)## you obtain ##x^{n-1}## for any constant ##C##, so the RHS and LHS of your first line are equal. They are choosing a particular ##C##, i.e. 1, because it helps in the following steps. As an exercise, try to do it with another constant, then you might see why there was just one logical choice that simplifies the calculation.
 
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